NCERT Exemplar Class 10 Maths Solutions Chapter 4 Quadratic Equations

NCERT Exemplar Class 10 Maths Solutions Chapter 4 Quadratic Equations

Edited By Safeer PP | Updated on Sep 05, 2022 01:22 PM IST | #CBSE Class 10th

NCERT exemplar Class 10 Maths solutions chapter 4 provides the understanding for forming quadratic equations by the given set of information in the problems. These NCERT exemplar Class 10 Maths chapter 4 solutions are created by our experienced faculties of Mathematics and have undergone a strict quality check so that the students can get accurate and relevant solutions while reviewing the questions of NCERT Solutions for Class 10 Maths.

These NCERT exemplar Class 10 Maths chapter 4 solutions due to their comprehensive nature provide useful insights to the students regarding the concepts of Quadratic Equations given in the NCERT. These NCERT exemplar Class 10 Maths solutions chapter 4 follow the prescribed CBSE syllabus for Class 10.

Question:1

$(i)x^{2}-3x+4=0$
$(ii)2x^{2}+x-1=0$
$(iii)2x^{2}-6x+\frac{9}{2}=0$
$(iv)3x^{2}-4x+1=0$
$(v)(x+4)^{2}-8x=0$
$(vi)(x-\sqrt{2})^{2}-2(x+1)=0$
$(vii)\sqrt{2}x^{2}-\frac{3}{\sqrt{2}}x+\frac{1}{2}=0$
$(viii)x(1-x)-2=0$
$(ix)(x-1)(x+2)+2=0$
$(x)(x+1)(x-2)+x=0$

(i) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$ Where a, b and c are real numbers with $a\neq 0$
The given equation is $x^{2}-3x+4=0$
Here a=1,b=-3,c=4
$D=b^{2}-4ac\\ =(-3)^{2}-4(1)(4)\\ =9-16=-7\\ b^{2}-4ac<0$
Hence the equation has no real roots.
(ii) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$ Where a, b and c are real numbers with $a\neq 0$
The given equation is $2x^{2}+x-1=0$
Here a=2,b=-1,c=-1
$D=b^{2}-4ac\\ =(1)^{2}-4(2)(-1)\\ =1+8=9\\ b^{2}-4ac>0$
Here for the equation $2x^{2}+x-1=0$, $b^{2}-4ac>0$
Hence it is a quadratic equation with two distinct real roots
(iii) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$ Where a, b and c are real numbers with $a\neq 0$
The given equation is $2x^{2}-6x+\frac{9}{2}=0$
Here $a= 2,b=-6,c=\frac{9}{2}$
$D=b^{2}-4ac\\ =(-6)^{2}-4(2)(\frac{9}{4})\\ =36-36=0\\ b^{2}-4ac=0$
here $b^{2}-4ac=0$
Hence the equation have two real and equal roots.
(iv) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$ Where a, b and c are real numbers with $a\neq 0$
The given equation is $3x^{2}-4x+1=0$
Here a=3,b=-4,c=1
$D=b^{2}-4ac\\ =(-4)^{2}-4(3)(1)\\ =16-12=4\\ b^{2}-4ac>0$
Here $b^{2}-4ac>0$
Hence the equation has distinct and real roots.
(v) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$ Where a, b and c are real numbers with $a\neq 0$
The given equation is $(x+4)^{2}-8x=0$
$(x)^{2}+(4)^{2}+2(x)(4)-8x=0$ $\left (using (a+b)^{2}=(a)^{2}+(b)^{2}+2ab \right )$
$(x)^{2}+16+8x-8x=0\\ x^{2}+16=0$
$a=1,b=0,c=16\\ =(0)^{2}-4(1)(16)\\ =-64\\ b^{2}-4ac<0$
Here $b^{2}-4ac<0$
Hence the equation has no real roots.
(vi) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$ Where a, b and c are real numbers with $a\neq 0$
The given equation is $(x-\sqrt{2})^{2}-2(x+1)=0$
$(x)^{2}+(\sqrt{2})^{2}+2(x)(\sqrt{2})-2x-2=0$ $\left (using (a+b)^{2}=(a)^{2}+(b)^{2}+2ab \right )$
$(x)^{2}+2-2\sqrt{2}x-2x-2=0\\ x^{2}-(2\sqrt{2}+2)x=0$
Compare with $ax^{2}+bx+c=0$ where $a\neq 0$
$a=1,b=-(2\sqrt{2}+2),c=0\\b^{2}-4ac =(-(2\sqrt{2}+2))^{2}-4(1)(0)\\ =(2\sqrt{2}+2)^{2}-0$
$=(2\sqrt{2})^{2}+(2)^{2}+2(2\sqrt{2})(2)$ $\left (using (a+b)^{2}=(a)^{2}+(b)^{2}+2ab \right )$
$=8+4+8\sqrt{2}\\ =12+8\sqrt{2}\\$
$b^{2}-4ac>0$
Here $b^{2}-4ac>0$
Hence the equation has two distinct real roots.
(vii) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$ Where a, b and c are real numbers with $a\neq 0$
The given equation is $\sqrt{2}x^{2}-\frac{3}{\sqrt{2}}x+\frac{1}{2}=0$
Here $a=\sqrt{2},b=\frac{3}{\sqrt{2}},c=\frac{1}{2}$
$b^{2}-4ac =\left (- \frac{3}{\sqrt{2}} \right )^{2}-4(\sqrt{2})\left (\frac{1}{2} \right )\\ =\frac{9}{2}-2\sqrt{2}\\=4.5-3.8=1.7\\ b^{2}-4ac>0$
Here $b^{2}-4ac>0$
Hence the equation has distinct real roots.
(viii) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$ Where a, b and c are real numbers with $a\neq 0$
The given equation is $x(1-x)-2=0$
$x-(x)^{2}-2=0\\ -x^{2}+x-2=0$
compare with $ax^{2}+bx+c=0$ where $a\neq 0$
$a=-1,b=1,c=-2\\ b^{2}-4ac=(1)^{2}-4(-1)(-2)\\1-8=-7\\ b^{2}-4ac<0$
Here $b^{2}-4ac<0$
Hence the equation has no real roots.
(ix) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$ Where a, b and c are real numbers with $a\neq 0$
The given equation is $(x-1)(x+2)+2=0$
$x(x+2)-1(x+2)+2=0\\ x^{2}+2x-x-2+2=0\\ x^{2}+x=0\\$
$a=1,b=1,c=0\\b^{2}-4ac =(1)^{2}-4(1)(0)\\ =1$
$b^{2}-4ac>0$
Here $b^{2}-4ac>0$
Hence the equation has two distinct real roots.
(x) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$ Where a, b and c are real numbers with $a\neq 0$
The given equation is $(x+1)(x-2)+x=0$
$x(x-2)+1(x-2)+x=0\\ x^{2}-2x+x-2+x=0\\ x^{2}-2x+2x-2=0\\x^{2}-2=0$
Compare with $ax^{2}+bx+c=0$ where $a\neq 0$
$a=1,b=0,c=-2\\b^{2}-4ac =(0)^{2}-4(1)(-2)\\ =8$
$b^{2}-4ac>0$
Here $b^{2}-4ac>0$
Hence the equation has two distinct real roots.

Question:2

Write whether the following statements are true or false. Justify your answers.
(i) Every quadratic equation has exactly one root.
(ii) Every quadratic equation has at least one real root.
(iii) Every quadratic equation has at least two roots.
(iv) Every quadratic equations has at most two roots.
(v) If the coefficient of x2 and the constant term of a quadratic equation have opposite signs, then the quadratic equation has real roots.
(vi)If the coefficient of x2 and the constant term have the same sign and if the coefficient of x term is zero, then the quadratic equation has no real roots.

(i) False
Solution
Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, $ax^{2} + bx + c = 0$
Where a, b and c are real numbers with $a \neq 0$
Root: If $ax^{2} + bx + c = 0$ …..(1)
Is a quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.
Let us consider a quadratic equation $x^{2}-4=0$
$\\x^{2}-4=0\\ x^{2}=4\\ x=\pm \sqrt{4}\\ x=2 \: \: \: \: \: \: \: x=-2$
Here – 2, 2 are the two roots of the equation.
Hence it is false that every quadratic equation has exactly one root.
(ii) False
Solution
Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, $ax^{2} + bx + c = 0$
Where a, b and c are real numbers with $a \neq 0$
Root: If $ax^{2} + bx + c = 0$ …..(1)
Is a quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.
Let us consider a quadratic equation
$x^{2}-x+2=0$
$x^{2}-x+2=0$
compare with $ax^{2} + bx + c = 0$ where $a \neq 0$
$\\a=1,b=-1,c=2\\ b^{2}-4ac=(-1)^{2}-4(1)(2)\\ =1-8\\ =-7\\ b^{2}-4ac<0$
Hence both the roots of the equation are imaginary.
Hence the statement every quadratic equation has at least one real root is False.
(iii) False
Solution
Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, $ax^{2} + bx + c = 0$
Where a, b and c are real numbers with $a \neq 0$
Root: If $ax^{2} + bx + c = 0$ …..(1)
Is a quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.
Let us consider a quadratic equation $x^{2}-4x+4=0$
$\\x^{2}-4x+4=0\\ x^{2}+(2)^{2}-2(x)(2)\\ (x-2)^{2}=0\: \: \: \: \: \: \: \: \: \: \: \: \: \: (using\: (a-b)^{2}=a^{2}+b^{2}-2ab)\\ (x-2)=0\\ x=2$
The equation $x^{2}-4x+4=0$ has only one root which is x = 2
Hence the given statement is False
(iv) True
Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, $ax^{2} + bx + c = 0$
Where a, b and c are real numbers with $a \neq 0$
Root: If $ax^{2} + bx + c = 0$ …..(1)
Is a quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.
As we know that the standard form of a quadratic equation is $ax^{2} + bx + c = 0$
It is a polynomial of degree 2.
As per the power of x is 2. There is at most 2 values of x exist that satisfy the equation.
Hence the given statement every quadratic equation has at most two roots is true.
(v) True
Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, $ax^{2} + bx + c = 0$
Where a, b and c are real numbers with $a \neq 0$
Root: If $ax^{2} + bx + c = 0$ …..(1)
Is a quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.
Let us suppose a quadratic equation $x^{2} + x - 2 = 0$
$x^{2} + x - 2 = 0$ is a quadratic equation in which the coefficient of x2 and constant term have opposite signs.
$x^{2} + x - 2 = 0$
compare with $ax^{2} + bx + c = 0$ where $a \neq 0$
$\\a=1,b=1,c=-2\\ b^{2}-4ac=(1)^{2}-4(1)(-2)\\ =1+8\\ =9\\ b^{2}-4ac>0$
Here $b^{2}-4ac>0$
Hence these types of equations have real roots.
(vi) True
Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, $ax^{2} + bx + c = 0$
Where a, b and c are real numbers with $a \neq 0$
Root: If $ax^{2} + bx + c = 0$ …..(1)
Is a quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.
Let us suppose a quadratic equation $x^{2} + 4= 0$
In $x^{2} + 4= 0$, the coefficient of x2 and the constant term has the same sign and coefficient of x is 0.
$x^{2} + 4= 0$
Compare with $ax^{2} + bx + c = 0$ where $a \neq 0$
$\\a=1,b=0,c=4\\ b^{2}-4ac=(0)^{2}-4(1)(4)\\ =16\\ b^{2}-4ac<0$
Here $b^{2}-4ac<0$
Hence these types of equations have no real roots.

Question:3

False
Solution
Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, $ax^{2} + bx + c = 0$
Where a, b and c are real numbers with $a \neq 0$
Root: If $ax^{2} + bx + c = 0$ …..(1)
Is a quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.
Let us take a quadratic equation with integral coefficient
$2x^{2}-3x-5=0$
compare with $ax^{2} + bx + c = 0$ where $a \neq 0$
(a=2,b=-3,c=-5)
let us find the roots of the equation
$\\x=\frac{-b+\sqrt{b^{2}-4ac}}{2a}\\ x=\frac{3 \pm \sqrt{9-4(2)(-5)}}{2 \times 2}\\ x=\frac{3 \pm \sqrt{9+40}}{4}\\ x=\frac{3+\sqrt{49}}{4}\\ x=\frac{3+7}{4}=\frac{5}{7} \; \; \; \; \; \; \; \; x=\frac{3-7}{4}=-1$
Here we found that 5/2 is not on integral
Hence the given statement is false

Question:4

Does there exist a quadratic equation whose coefficients are rational but both of its roots are irrational? Justify your answer.

True
Let us suppose a quadratic equation with rational coefficients.
$x^{2}-2x+4=0$
compare with $ax^{2} + bx + c = 0$ where $a \neq 0$
a=-1,b=-2,c=4
let us find the roots of the equation
$\\x=\frac{-b+\sqrt{b^{2}-4ac}}{2a}\\ x=\frac{2 \pm \sqrt{4-4(-1)(4)}}{2 (-1)}\\ x=\frac{2 \pm \sqrt{20}}{-2}\\ x=\frac{2\pm 2\sqrt{5}}{-2}\\ x=\frac{2+2\sqrt{5}}{-2} \; \; \; \; \; \; \; \; x=\frac{2-2\sqrt{5}}{-2}\\$
Both of its roots are irrational
Hence the given statement is true

Question:5

Does there exist a quadratic equation whose coefficients are all distinct irrationals but both the roots are rational? Why?

True
Let us consider an equation with distinct irrational numbers
$\sqrt{2}x^{2}-5\sqrt{2}x+4\sqrt{2}=0$
compare with $ax^{2} + bx + c = 0$ where $a \neq 0$
$a=\sqrt{2},b=-5\sqrt{2},c=4\sqrt{2}$
let us find the roots of the equation
$\\x=\frac{-b+\sqrt{b^{2}-4ac}}{2a}\\ x=\frac{5\sqrt{2 }\pm \sqrt{50-4(\sqrt{2})(4\sqrt{2})}}{2 \sqrt{2}}\\ x=\frac{5 \sqrt{2} \pm \sqrt{50-32}}{2\sqrt{2}}\\ x=\frac{5 \sqrt{2} \pm \sqrt{18}}{2\sqrt{2}}\\ x=\frac{\sqrt{2}(5\pm 3)}{2\sqrt{2}}=\frac{5 \pm 3}{2}\\ x=\frac{5+3}{2}=4 \; \; \; \; \; \; \; \; x=\frac{5-3}{2}=1$
The roots are rational. Hence the given statement is true

Question:6

Is 0.2 a root of the equation x2 – 0.4 = 0? Justify

False
Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, ax2 + bx + c = 0 Where a, b and c are real numbers with a $\neq$ 0
Roots : If ax2 + bx + c = 0 …..(1)
Is a quadratic equation then the values of x which satisfy equation 1 are the roots of the equation.
Here the given equation is x2 – 0.4 = 0 …..(2)
If 0.2 is a root of equation 2 then it should satisfy its equation
Put x = 0.2 in (2)
$\\(0.2)^{2}-0.4=0\\ \left (\frac{2}{10} \right )-0.4=0\\ \frac{4}{100}-0.4=0\\ 0.04-0.4=0\\ -0.36 \neq 0$
Here 0.2 is not satisfying the equation (2)
Hence 0.2 is not a root of the equation $x^{2}-0.4=0$

Question:7

If b = 0, c < 0, is it true that the roots of x2 + bx+ c = 0 are numerically equal and opposite in sign? Justify.

True
Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, ax2 + bx + c = 0 Where a, b and c are real numbers with a $\neq$ 0
Roots : If ax2 + bx + c = 0 …..(1)
Is a quadratic equation then the values of x which satisfy equation 1 are the roots of the equation.
Here the given equation is x2 + bx+ c = 0 …..(2)
It is also given that b = 0, c < 0.
Let c=-y
Put b = 0, c = – y in (2)
$\\x^{2}+0(x)-y=0\\ x^{2}=y\\ x=\pm \sqrt{y}\\ x=+\sqrt{y}\; \; \; \; \; \; \; \; \; \; x=-\sqrt{y}$
Hence both the roots are equal and opposite in sign. Hence the given statement is true.

Question:1

Find the roots of the quadratic equations by using the quadratic formula in each of the following:
$(i)2x^{2}-3x-5=0$
$(ii)5x^{2}+13x+8=0$
$(iii)-3x^{2}+5x+12=0$
$(iv)-x^{2}+7x-10=0$
$(v)x^{2}+2\sqrt{2}x-6=0$
$(vi)x^{2}+3\sqrt{5}x+10=0$
$(vii)\frac{1}{2}x^{2}-\sqrt{11}x+1=0$

(i) $\frac{5}{2},-1$
$2x^{2}-3x-5=0$
Compare with $ax^{2}+bx+c=0$ where $a \neq 0$
Here a=2, b=-3, c=-5
$\\x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\\ x=\frac{-(-3)\pm \sqrt{(-3)^{2}-4(2)(-5)}}{2 \times 2}\\ x=\frac{3 \pm \sqrt{9+40}}{4}\\ x=\frac{3 \pm 7}{4}\\ x=\frac{3+7}{4}=\frac{10}{4}=\frac{5}{2 }\;\; \; \;\;\; x=\frac{3-7}{4} =\frac{-4}{4} =-1$
$x=\frac{5}{2},-1$
$\left (\frac{5}{2},-1 \right )$ are the roots of the equation $2x^{2}-3x-5=0$

(ii) $-1,\frac{-8}{5}$
$5x^{2}+13x+8=0$
Compare with $ax^{2}+bx+c=0$ where $a \neq 0$
Here a=5, b=13, c=8
$\\x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\\ x=\frac{-13\pm \sqrt{(13)^{2}-4(5)(8)}}{2 \times 5}\\ x=\frac{-13 \pm \sqrt{169-160}}{10}\\ x=\frac{-13 \pm 3}{10}\\ x=\frac{-13+3}{10}=\frac{-10}{10}=-1\;\; \; \;\;\; x=\frac{-13-3}{10} =\frac{-16}{10} =\frac{-8}{5}$
$x=-1,\frac{-8}{5}$
$\left (-1,\frac{-8}{5} \right )$ are the roots of the equation $5x^{2}+13x+8=0$
(iii) $\frac{-4}{3},3$
$-3x^{2}+5x+12=0$
Compare with $ax^{2}+bx+c=0$ where $a \neq 0$
Here a=-3, b=5, c=12
$\\x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\\ x=\frac{-5\pm \sqrt{(5)^{2}-4(-3)(12)}}{2 \times (-3)}\\ x=\frac{-5 \pm \sqrt{25+144}}{-6}\\ x=\frac{-5 \pm 13}{-6}\\ x=\frac{-5+13}{-6}=\frac{8}{-6}=\frac{-4}{3 }\;\; \; \;\;\; x=\frac{-5-13}{-6} =\frac{-18}{-6} =3$
$x=\frac{-4}{3},3$
$\left (\frac{-4}{3},3 \right )$ $\text{are the roots of the equation}$ $-3x^{2}+5x+12=0$
(iv) 5,2
$-x^{2}+7x-10=0$
Compare with $ax^{2}+bx+c=0$ where $a \neq 0$
Here a=-1, b=7, c=-10
$\\x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\\ x=\frac{-7\pm \sqrt{49-4(-1)(-10)}}{2 \times (-1)}\\ x=\frac{-7 \pm \sqrt{9}}{-2}\\ x=\frac{-7 \pm 3}{-2}\\ x=\frac{-7+3}{-2}=\frac{-4}{-2}=2\;\; \; \;\;\; x=\frac{-7-3}{-2} =\frac{-10}{-2} =5$
x=5,2
(5,2) are the roots of the equation $-x^{2}+7x-10=0$
(v) $\sqrt{2},-3\sqrt{2}$
$x^{2}+2\sqrt{2}x-6=0$
Compare with $ax^{2}+bx+c=0$ where $a \neq 0$
Here $a=1, b=2\sqrt{2}, c=-6$
$\\x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\\ x=\frac{-2\sqrt{2}\pm \sqrt{(2\sqrt{2})^{2}-4(1)(-6)}}{2 \times 1}\\ x=\frac{-2\sqrt{2} \pm \sqrt{8+24}}{2}\\ x=\frac{2\sqrt{2} \pm 4\sqrt{2}}{2}\\ x=\frac{2\left (\sqrt{2} \pm 2 \sqrt{2} \right )}{2}\\ x=-\sqrt{2}\pm2\sqrt{2}\\ x=-\sqrt{2}+2\sqrt{2}=\sqrt{2}\;\; \; \;\;\; x=-\sqrt{2}-2\sqrt{2}=-3\sqrt{2}$
$x=\sqrt{2},-3\sqrt{2}$
$\sqrt{2},-3\sqrt{2}$ are the roots of the equation $x^{2}+2\sqrt{2}x-6=0$
(vi) $2\sqrt{5},\sqrt{5}$
$x^{2}+3\sqrt{5}x+10=0$
Compare with $ax^{2}+bx+c=0$ where $a \neq 0$
$a=1, b=3\sqrt{5}, c=10$
$\\x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\\ x=\frac{3\sqrt{5}\pm \sqrt{(3\sqrt{5})^{2}-4(1)(10)}}{2 \times 1}\\ x=\frac{3\sqrt{5} \pm \sqrt{45-40}}{2}\\ x=\frac{3\sqrt{5} \pm \sqrt{5}}{2}\\ x=\frac{3\sqrt{5} + \sqrt{5}}{2}=\frac{4\sqrt{5}}{2}=2\sqrt{5}\;\; \; \;\;\; x=\frac{3\sqrt{5} - \sqrt{5}}{2}=\frac{2\sqrt{5}}{2}=\sqrt{5}$
$x=2\sqrt{5},\sqrt{5}$
$2\sqrt{5},\sqrt{5}$ are the roots of the equation $x^{2}+3\sqrt{5}x+10=0$
(vii) $\sqrt{11}+3,\sqrt{11}-3$
$\frac{1}{2}x^{2}-\sqrt{11}x+1=0$
Compare with $ax^{2}+bx+c=0$ where $a \neq 0$
Here $a=\frac{1}{2},b=-\sqrt{11},c=1$
$\\x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\\ x=\frac{\sqrt{11}\pm \sqrt{(-\sqrt{11})^{2}-4\left ( \frac{1}{2} \right )(1)}}{2 \times \frac{1}{2}}\\ x=\frac{\sqrt{11}\pm\sqrt{11-2}}{1}\\ x=\sqrt{11}\pm 3\\ x=\sqrt{11}+ 3\;\;\;\;\;\;\;x=\sqrt{11}- 3$
$x=\left (\sqrt{11}+ 3 \right ),\left (\sqrt{11}- 3 \right )$
hence $\left (\sqrt{11}+ 3 \right ),\left (\sqrt{11}- 3 \right )$ are the root of the equation $\frac{1}{2}x^{2}-\sqrt{11}x+1=0$

Question:2

Find the roots of the following quadratic equations by the factorisation method
$(i)2x^{2}+\frac{5}{3}x-2=0$
$(ii)\frac{2}{5}x^{2}-x-\frac{3}{5}=0$
$(iii)3 \sqrt{2}x^{2}-5x-\sqrt{2}=0$
$(iv)3x^{2}+5\sqrt{5}x-10=0$
$(v)21x^{2}-2x+\frac{1}{21}=0$

(ii) $3,-\frac{1}{2}$
Solution
$\\\frac{2}{5}x^{2}-x-\frac{3}{5}=0\\ \frac{2x^{2}-5x-3}{5}=0\\ 2x^{2}-5x-3=0\\ 2x^{2}-6x+x-3=0\\ 2x(x-3)+1(x-3)=0\\ (x-3)(2x+1)=0\\ x-3=0\;\;\;\;\;\;\;\;\;\;\;\;\; 2x+1=0\\ x=3\;\;\;\;\;\;\;\;\;\;\;\;\; 2x=-1 \;\;\;\;x=\frac{-1}{2}\\$
Hence $\frac{-3}{2},\frac{2}{3}$ are the roots of the equation.
(i) $\frac{-3}{2},\frac{2}{3}$
Solution
$2x^{2}+\frac{5}{3}x-2=0\\ \frac{6x^{2}+5x-6}{3}=0\\ 6x^{2}+5x-6=0\\ 6x^{2}+9x-4x-6=0\\ 3x(2x+3)-2(2x+3)=0\\ (2x+3)(3x-2)=0\\ 2x+3=0\;\;\;\;\;\;\;\;\;\;\;\;\; 3x-2=0\\ 2x=-3\;\;\;\;\;\;\;\;\;\;\;\;\; 3x=2\\ x=\frac{-3}{2}\;\;\;\;\;\;\;\;\;\;\;\;\; x=\frac{2}{3}\\$
Hence $3,-\frac{1}{2}$ are the roots of the equation.
(iii) $\sqrt{2},-\frac{1}{3\sqrt{2}}$
Solution
$3\sqrt{2}x^{2}-5x-\sqrt{2}=0\\ 3\sqrt{2}x^{2}-6x+x-\sqrt{2}=0\\ 3\sqrt{2}x(x-\sqrt{2})+1(x-\sqrt{2})=0\\ (x-\sqrt{2})(3\sqrt{2}x+1)=0\\ 3\sqrt{2}x+1=0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x-\sqrt{2}=0\\ 3\sqrt{2}x=-1 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x=\sqrt{2}\\ x=\frac{-1}{3\sqrt{2}}$
Hence $\sqrt{2},-\frac{1}{3\sqrt{2}}$ are root of the equation $3\sqrt{2}x^{2}-5x-\sqrt{2}=0\\$
(iv) $-2\sqrt{5},-\frac{\sqrt{5}}{3}$
Solution
$3x^{2}+5\sqrt{5}x-10=0\\ 3x^{2}+6\sqrt{5}x-\sqrt{5}x-10=0\\ 3x(x+2\sqrt{5})-\sqrt{5}(x+2\sqrt{5})=0\\ (3x-\sqrt{5})(x+2\sqrt{5})=0\\ 3x-\sqrt{5}=5 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x+2\sqrt{5}=0\\ 3x=\sqrt{5} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x=-2\sqrt{5}\\ x=\frac{\sqrt{5}}{3}$
Hence $-2\sqrt{5},-\frac{\sqrt{5}}{3}$ are root of the equation $3x^{2}+5\sqrt{5}x-10=0$
(v) $\frac{1}{21}$
solution
$21x^{2}-2x+\frac{1}{21}=0\\ 21x^{2}-x-x+\frac{1}{21}=0\\ x(21x-1)-\frac{1}{21}(21x-1)=0\\ (21x-1)\left ( x-\frac{1}{21} \right )=0\\ 21-x=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x-\frac{1}{21}=0\\ 21x=1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x=\frac{1}{21}\\ x=\frac{1}{21}$
Hence $\frac{1}{21}$ are the roots of the equation $21x^{2}-2x+\frac{1}{21}=0$

Question:8

At t minutes past 2 pm, the time needed by the minutes hand of a clock to show 3 pm was found to be 3 minutes less than $\frac{t^{2}}{4}$ minutes. Find t.

At t minutes past 2 pm, the time need by minute hand to show 3 pm = 60 – t
According to question, it is given that (60 – t) is equal to 3 minutes less than $\frac{t^{2}}{4}$
$60-t=\frac{t^{2}}{4}-3\\ 60-t=\frac{t^{2}-12}{4}\\ t^{2}-12=4(60-t)\\ t^{2}-12=240-4t\\ t^{2}+4t-12-240=0\\ t^{2}+4t-252=0\\ t^{2}+18t-14t-252=0\\ t(t+18)-14(t+18)=0\\ (t+18)(t-14)=0\\ t=-18,14\\$
Time can’t equal to negative term.
Hence t = 14 minutes

Question:7

length = 34m, breadth = 24m
Solution
Let the distance between the pond and loan boundary is x.
Hence the length of pond =50-x-x
From the figure
Area of lawn = Area of pond + Area of grass
50 × 40=(50-2x)(40-2x)+1184 (using Area of rectangle = l × B)
2000-1184=2000-100x-80x+4x2
4x2-180x+2000-2000+1184=0
4x2-180x+1184=0
Divide by 4
x2-45x+296=0
x2 - 37x - 8x + 296 = 0
x(x-37)-8(x-37)=0
(x-37)(x-8)=0
x = 37, 8
If x = 37
Length of pond =50-2(37)
=50-74
=-24
Which is not possible because length can’t be negative.
If x = 8
Length of pond =50-2(8)
=50-16=34m
=40-16=24m

Question:6

5, 27
Solution
Let Nisha’s present age = x
At present Asha is 2 more than the square of Nisha’s age.
Hence, Asha present age = x2 + 2
According to question
x2 + 2 - x = 10x – 1 - ( x2 + 2 )
x2 + 2 – x = 10x – 1 – x2 - 2
x2 + 2 – x - 10x + x2 + 3=0
2x2-11x+5=0
2x(x-5)-1(x-5)=0
(x-5)(2x-1)=0
$x=5,\frac{1}{2}$?
If Nisha’s age is 5
Asha age = (5)2+2=27
If Nisha’s age is 1/2 then Asha’s age is 2.25 which is not possible.

Question:5

If Zeba were younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than five times her actual age. What is her age now?

Let zeba’s current age = x
According to question
$\\(x-5)^{2}=11+5x\\ x^{2}+25-10=11+5x\;\;\;\;\;\;\;\(using \;\;(a-b)^{2}=a^{2}+b^{2}-2ab)\\ x^{2}+25-10x-11-5x=0\\ x^{2}-15x+14=0\\ x^{2}-14x-x+14=0\\ x(x-14)-1(x-14)=0\\ (x-14)(x-1)\\$
x = 14
we can’t take 1 because if we take 5 years younger of 1 then it is equal to – 4 and age cannot be negative.
Hence x = 14 i.e., the age of zeba is 14 years

Question:4

A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/h more. Find the original speed of the train.

45 km per hour.
Solution
Let the original speed of the train = x km per hour.
Total distance = 360 km
$\text{Time taken by train to reach at final point =}\left ( \frac{360}{x} \right )\left ( \because speed=\frac{distance}{time} \right )$
If train’s speed increased by 5 then speed = (x + 5) km per hour. Time taken by train to cover 360 km at a speed of (x + 5) km per hour=
$\left ( \frac{360}{x+5} \right )\left ( \because speed=\frac{distance}{time} \right )$
Now it is given that the difference between the time taken at a speed of x and at a speed of x + 5 is 48 min.
$\frac{360}{x} - \frac{360}{x+5} =\frac{48}{60}(hour)\\ \frac{360(x+5)-360x}{x(x+5)}=\frac{48}{60}(hour)\\ 5(360x+1800-360x)=4x(x+5)\\ 9000=4x^{2}+20x\\ 4x^{2}+20x-9000=0$
Or
$x^{2}+5x-2250=0$ (divide by 4)
$x^{2}+5x-2250=0\\ x^{2}+50x-45x-2250=0\\ x(x+50)-45(x+50)=0\\ x=-50,45$
We can’t take x = – 50 because speed is always positive.
Hence original speed of train is 45 km per hour.

Question:3

A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.

Let the natural number = x
It is given that x is increased by 12, equal to 160 times its reciprocal
$x+12=\frac{160}{x}\\ x(x+12)=160\\ x^{2}+12x-160=0\\ x^{2}+20x-8x-160=0\\ x(x+20)-8(x+20)=0\\ (x+20)(x-8)=0\\ x+20=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x-8=0\\ x=-20\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x=8$
We can’t take x = – 20 because it is given that x is a natural number.
Hence x = 8 is the required natural number.

Question:2

Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number

12
Let the natural number = x
It is given that when the square of x is diminished by 84 then it is equal to thrice of 8 more than x
$x^{2}-84=3(x+8)\\ x^{2}-84=3x+24\\ x^{2}-3x-84-24=0\\ x^{2}-3x-108=0\\ x^{2}-12x+9x-108=0\\ x(x-12)+9(x-12)=0\\ (x-12)(x+9)=0$
x = 12, – 9
We can’t take – 9 because it is given that x is a natural number.
Hence x = 12 is the required natural number.

Question:1

Find whether the following equations have real roots. If real roots exist, find them.
$(i)8x^{2}+2x-3=0$
$(ii)-2x^{2}+3x+2=0$
$(iii)5x^{2}-2x-10=0$
$(iv)\frac{1}{2x-3}+\frac{1}{x-5}=1,x\neq \frac{3}{2},5$
$(v)x^{2}+5\sqrt{5}x-70=0$

(i) We Know that b2-4ac > 0 in the case of real roots and b2 - 4ac < 0 in the case of imaginary roots.
For the equation $8x^{2}+2x-3=0$
Compare with $ax^{2}+bx+c=0$ where $a\neq0$
a=8,b=2,c=-3
$\\b^{2} - 4ac =(2)^{2} - 4(8)(-3)\\ =4+96\\ =100\\ b^{2}-4ac>0$
Hence the equation has real roots
$\\8x^{2}+2x-3=0\\ 8x^{2}+6x-4x-3=0\\ 2x(4x+3)-1(4x+3)=0\\ (4x+3)(2x-1)=0\\ 4x+3=0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 2x-1=0\\ 4x=-3 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 2x=1\\ x=\frac{-3}{4} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x= \frac{1}{2}\\$
Roots of the equation $8x^{2}+2x-3=0$ are $\frac{-3}{4}, \frac{1}{2}\\$
(ii) We Know that b2-4ac > 0 in the case of real roots and b2 - 4ac < 0 in the case of imaginary roots.
For the equation $-2x^{2}+3x+2=0$
Compare with $ax^{2}+bx+c=0$ where $a\neq0$
a=-2,b=3,c=2
$\\b^{2} - 4ac =(3)^{2} - 4(-2)(2)\\ =9+16\\ =25\\ b^{2}-4ac>0$
Hence the equation has real roots
$\\-2x^{2}+3x+2=0\\ -2x^{2}+4x-x+2=0\\-2x(x-2)-1(x-2)=0\\(x-2)(-2x-1)=0\\ -2x-1=0 \;\;\;\;\;\;\;\;\;\;\;\;\; x-2=0\\ -2x=1 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x=2\\ x=\frac{-1}{2}$
Roots of the equation $-2x^{2}+3x+2=0$ are $2, \frac{-1}{2}\\$
(iii) We Know that b2-4ac > 0 in the case of real roots and b2 - 4ac < 0 in the case of imaginary roots.
For the equation $5x^{2}-2x-10=0$
Compare with $ax^{2}+bx+c=0$ where $a\neq0$
a=5,b=-2,c=-10
$\\b^{2} - 4ac =(-2)^{2} - 4(5)(-10)\\ =4+200\\ =204\\ b^{2}-4ac>0$
Hence the equation has real roots
$\\5x^{2}-2x-10=0\\ x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\\ x=\frac{2 \pm \sqrt{4-4(5)(-10)}}{2(5)}\\ x=\frac{2 \pm \sqrt{4+200}}{10}\\ x=\frac{2 \pm \sqrt{204}}{10}\\ x=\frac{2(1 \pm \sqrt{51})}{10}\\ x=\frac{1 \pm \sqrt{51}}{5}\\ x=\frac{1 + \sqrt{51}}{5}, x=\frac{1 - \sqrt{51}}{5}\\$
Roots of the equation $5x^{2}-2x-10=0$ are $\frac{1 + \sqrt{51}}{5}, \frac{1 - \sqrt{51}}{5}$
(iv) We Know that b2-4ac > 0 in the case of real roots and b2 - 4ac < 0 in the case of imaginary roots.
For the equation $\frac{1}{2x-3}+\frac{1}{x-5}=1$
$\\\frac{(x-5)+(2x-3)}{(2x-3)(x-5)}=1\\ x-5+2x-3=2x(x-5)3(x-5)\\ 3x-8=2x^{2}-10x-3x+15\\ 2x^{2}-13x+15-3x+8=0\\ 2x^{2}-16x+23=0\\$
Compare with $ax^{2}+bx+c=0$ where $a\neq0$
a=2,b=-16,c=23
$\\b^{2}-4ac=(-16)^{2}-4(2)(23)\\ =256-184\\ =72\\ b^{2}-4ac>0$
Hence the equation has real roots
$\\x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\\ x=\frac{16 \pm \sqrt{72}}{2(2)}\;\;\;\;\;\;\;\;\;\;(\because b^{2}-4ac=72)\\ x=\frac{16 \pm 6\sqrt{2}}{4}=\frac{8 \pm 3\sqrt{2}}{2}\\ x=\frac{8 + 3\sqrt{2}}{2},\frac{8 - 3\sqrt{2}}{2}$
Roots of the equation $\frac{1}{2x-3}+\frac{1}{x-5}=1$ are $\frac{8 + 3\sqrt{2}}{2},\frac{8 - 3\sqrt{2}}{2}$
(v) We Know that b2-4ac > 0 in the case of real roots and b2 - 4ac < 0 in the case of imaginary roots.
For the equation $x^{2}+5\sqrt{5}x-70=0$
Compare with $ax^{2}+bx+c=0$ where $a\neq0$
$\\a=1, b=5\sqrt{5},c=-70\\ b^{2} - 4ac =(5\sqrt{5})^{2} - 4(1)(-70)\\ =125+280\\ =405\\ b^{2}-4ac>0$
Hence the equation has real roots
$\\x^{2}+5\sqrt{5}x-70=0\\x^{2}+7\sqrt{5}x-2\sqrt{5}x-70=0\\x(x+7\sqrt{5})-2\sqrt{5}(x+7\sqrt{5})=0\\(x+7\sqrt{5})(x-2\sqrt{5}) \\ x+7\sqrt{5}=0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x+7\sqrt{5}=0\\ x=-7\sqrt{5} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x=2\sqrt{5}\\$
Roots of the equation $x^{2}+5\sqrt{5}x-70=0$ are $-7\sqrt{5}, 2\sqrt{5}\\$

Question:1

Choose the correct answer from the given four options in the following questions: Which of the following is a quadratic equation?
$\\(A) x^{2}+2x+1=(4-x)^{2}+3\\ (B)-2x^{2}=(5-x)\left ( 2x-\frac{2}{5} \right )\\ (C)(k+1)x^{2}+\frac{3}{2}x=7\; \; where\;\; k=-1\\ (D)x^{3}-x^{2}=(x-1)^{3}$

Answer(D)$x^{3} - x^{2} = (x - 1)^{3}$
Solution:
A quadratic equation in x is an equation that can be written in the standard form.
$ax^{2} + bx + c = 0$ where a, b and c are real number with $a \neq 0$
$\\(A) x^{2} + 2x + 1 = (4 - x)^{2} + 3\\ x^{2} + 2x + 1 = (4)^{2} + (x)^{2} - 2(4)(x) + 3\;\;\;\; (Using (a - b)^{2} = a^{2} + b^{2} - 2ab)\\ x^{2} + 2x + 1 = 16 + x^{2} - 8x + 3\\ x^{2} + 2x + 1 - 16 - x^{2} + 8x - 3 = 0\\ 10x - 18 = 0$
$10x - 18 = 0$ is not in the form of $ax^{2} + bx + c = 0$
Hence it is not a quadratic equation.
(B)
$\\-2x^{2}=(5-x)\left ( 2x-\frac{2}{5} \right )\\ -2x^{2}=10x-2-2x^{2}+\frac{2}{5}x\\ 10x+\frac{2}{5}x-2=0\\ \frac{50x+2x-10}{5}=0\\ 5x-10=0$
This is not in the form of $ax^{2} + bx + c = 0$
Hence it is not a quadratic equation
$(C)(k+1)x^{2}+\frac{3}{2}x=7\; \; where\;\; k=-1\\$
Put k = – 1
$\\((-1)+1)x^{2}+\frac{3}{2}x=7\\ \frac{3}{2}x-7=0\\ \frac{3x-14}{2}=0\\ 3x-14=0$
This is not in the form of $ax^{2} + bx + c = 0$
Hence this is not a quadratic equation.
(D)
$\\x^{3}-x^{2}=(x - 1)^{3}\\ x^{3}-x^{2} =(x)^{3} - (1)^{3} (1) + 3(x)(1)^{2}\;\;\;\; [using\; (a - b)^{3} = a^{3} - b^{3} - 3a^{2}b + 3ab^{2}]\\ x^{3} - x^{2} - x^{3} + 1 + 3x^{2} - 3x = 0\\ 2x^{2} - 3x + 1 = 0$
This is in the form of $ax^{2} + bx + c = 0$
Hence $x^{3}-x^{2}=(x-1)^{3}$ is a quadratic equation.

Question:2

Which of the following is not a quadratic equation?
$\\(A)2(x-1)^{2}=4x^{2}-2x+1\\ (B)2x-x^{2}=x^{2}+5\\ (C)(\sqrt{2}x+\sqrt{3})^{2}+x^{2}=3x^{2}-5x\\ (D)(x^{2}+2x)^{2}=x^{4}+3+4x^{3}$

$(C)(\sqrt{2}x+\sqrt{3})^{2}+x^{2}=3x^{2}-5x$
Quadratic equation: A quadratic equation in x is an equation that can be written in the standard form $ax^{2} + bx + c = 0$, $a \neq 0$
Where a, b and c are real numbers
$2(x-1)^{2}=4x^{2}-2x+1$
$2(x^{2}+1-2x)=4x^{2}-2x+1$
[using (a – b)2 = a2 + b2 – 2ab]
$\\2x^{2}+2-4x-4x^{2}+2x-1=0\\ -2x^{2}-2x+1=0$
It is a quadratic equation because it is in the form
$ax^{2} + bx + c = 0$
(B)
$\\2x-x^{2}=x^{2}+5\\ 2x-x^{2}-x^{2}-5=0\\ -2x^{2}+2x-5=0\\ -2x^{2}+2x-5=0$
It is a quadratic equation because it is in the form $ax^{2} + bx + c = 0$
(C)
$(\sqrt{2}x+\sqrt{3})^{3}+x^{2}=3x^{2}-5x\\ (\sqrt{2}x)^{2}+(\sqrt{3})^{2}+2(\sqrt{2}x)(\sqrt{3})+x^{2}=x^{2}-5x$ $\left [using \;(a+b)^{2}=a^{2}+b^{2}+2ab \right ]$
$2x^{2}+3+2\sqrt{6}x+x^{2}-3x^{2}+5x=0\\ (2\sqrt{6}+5)x+3=0$
It is not in the form of $ax^{2} + bx + c = 0$
Hence it is not a quadratic equation.
(D)
$(x^{2} + 2x)^{2} = x^{4} + 3 + 4x^{3}\\ (x^{2})^{2} + (2x)^{2} + 2(x^{2})(2x) = x^{4} + 3 + 4x^{3}$ $\left [using \;(a+b)^{2}=a^{2}+b^{2}+2ab \right ]$
$x^{4} + 4x^{2} + 4x^{3} - x^{4} - 3 - 4x^{3} = 0\\ 4x^{2}-3=0$
It is in the form of $ax^{2} + bx + c = 0$
Hence it is a quadratic equation
Only option (C) is not in the form of $ax^{2} + bx + c = 0$
Hence (C) is not a quadratic equation.

Question:3

Which of the following equations has 2 as a root ?
$\\(A)x^{2}-4x+5=0\\ (B)x^{2}+3x-12=0\\ (C)2x^{2}-7x+6=0\\ (D)3x^{2}-6x-2=0$

(C)$(C)2x^{2}-7x+6=0$
Solution
(A) $x^{2}-4x+5=0$
Put x = 2
$(2)^{2}-4(2)+5=0\\ 4-8+5=0\\ 9-8=0\\ 1 \neq 0$
(B)
$x^{2}+3x-12=0\\ put\;\;x=2\\ (2)^{2}+3(2)-12=0\\ 4+6-12=0\\ 10-12=0\\ -2\neq0$
(C)
$2x^{2}-7x+6=0\\ put\;\;x=2\\ 2(2)^{2}-7(2)+6=0\\ 8-14+6=0\\ 14-14=0\\ 0= 0$
(D)
$3x^{2}-6x-2=0\\ put\;\;x=2\\ 3(2)^{2}-6(2)-2=0\\ 12-12-2=0\\ -2\neq 0$
Hence option (C) is only satisfy x = 2.
Hence C has 2 as a root.

Question:4

If $\frac{1}{2}$ is a root of the equation $x^{2}+kx-\frac{5}{4}=0$ , then the value of k is
$(A)2\\ (B)-2\\ (C)\frac{1}{4}\\ (D)\frac{1}{2}$

As we know that if $\frac{1}{2}$ is a root of the equation $x^{2}+kx-\frac{5}{4}=0$ then $x=\frac{1}{2}$ should satisfy its equation.
Put $x=\frac{1}{2}$
$\\\left (\frac{1}{2} \right )^{2}+k\left (\frac{1}{2} \right )-\frac{5}{4}=0\\ \frac{1}{4}+\frac{k}{2}-\frac{5}{4}=0\\ \frac{1+2k-5}{4}=0\\ 2k-4=0\\ 2k=4\\ k=\frac{4}{2}=2\\ k=2$
Hence value of k is 2.

Question:5

Which of the following equations has the sum of its roots as 3?
$\\(A)2x^{2}-3x+6=0\\ (B)-x^{2}+3x-3=0\\ (C)\sqrt{2}x^{2}-\frac{3}{\sqrt{2}}x+1=0\\ (D)3x^{2}-3x+3=0$

(B) $(B)-x^{2}+3x-3=0$
If $(B)ax^{2}+bx+c=0$ is a quadratic equation the sum of its roots are $-\frac{b}{a}$
(A)
$2x^{2}-3x+6=0$
Here b=-3,a=2
Sum of its roots =$-\frac{b}{a}$ = $-\left ( \frac{-3}{2} \right )=\frac{3}{2}$
(B)
$-x^{2}+3x-3=0$
Here b=-3,a=3
Sum of its roots =$-\frac{b}{a}=\frac{-3}{-1}=3$
(C)
$\sqrt{2}x^{2}-\frac{3}{\sqrt{2}}x+1=0$
Here $b=\frac{-3}{\sqrt{2}}, a=\sqrt{2}$
Sum of its roots = $-\frac{b}{a}=-\left (\frac{-3}{\sqrt{2}} \right )\times \sqrt{2}=\frac{3}{2}$
(D)
$3x^{2}-3x+3=0$
Here b=-3,a=3
Sum of its roots $=-\frac{b}{a}=\frac{(-3)}{3}=\frac{3}{3}=1$
Hence only B has sum of its roots 3

Question:6

Values of k for which the quadratic equation $2x^{2}-kx+k=0$ has equal roots is
(A) 0 only (B) 4 (C) 8 only (D) 0, 8

(D) 0, 8
Here the given equation is $2x^{2}-kx+k=0$
Compare with $ax^{2}+bx+c=0$ where $a \neq 0$
a=2,b=-k,c=k
It is given that roots are equal.
Hence
$\\b^{2}-4ac=0\\ (-k)^{2}-4(2)(k)=0\\ k^{2}-8k=0\\ k(k-8)=0\\ k-8=0 \;\;\;\;\;\; k=0\\ k=8\\ k=0,8$
Hence for the value of k = 0, 8 the equation $ax^{2}+bx+c=0$ has equal roots.

Question:7

Which constant must be added and subtracted to solve the quadratic equation$9x^{2}+\frac{3}{4}x-\sqrt{2}=0$ by the method of completing the square?
$\\(A)\frac{1}{8}\\ (B)\frac{1}{64}\\ (C)\frac{1}{4}\\ (D)\frac{9}{64}$

Here the given quadratic equation is
$9x^{2}+\frac{3}{4}x-\sqrt{2}=0$
$\\(3)^{2}+\frac{1}{4}(3x)-\sqrt{2}=0\\ (3x)^{2}+2 \times \left ( \frac{1}{8} \right )\times(3x)-\sqrt{2}=0$
$\text{[add and subtract}\left ( \frac{1}{8} \right )^{2}]$
$\\(3x)^{2}+2 \times \left ( \frac{1}{8} \right )\times(3x)+\left ( \frac{1}{8} \right )^{2}-\left ( \frac{1}{8} \right )^{2}-\sqrt{2}=0\\ (3x)^{2}+2 \times \left ( \frac{1}{8} \right )\times(3x)+\left ( \frac{1}{8} \right )^{2}=\left ( \frac{1}{8} \right )^{2}+\sqrt{2}\\ \because a^{2}+b^{2}+2ab=(a+b)^{2}\\$
So
$\left (3x+ \frac{1}{8} \right )^{2}=\frac{1+64\times\sqrt{2}}{64}$ $\text{ Hence}\left (\frac{1}{8} \right )^{2}=\frac{1}{64}$ $\text{ should be added and subtracted in the equation}$ $9x^{2}+\frac{3}{4}x-\sqrt{2}=0$
$\text{ for solving by completing the square method. }$

Question:

The quadratic equation $2x^{2}-\sqrt{5}x+1=0$ has
(A) two distinct real roots (B) two equal real roots
(C) no real roots (D) more than 2 real roots

(C) no real roots
$\text{Here the given quadratic equation is }2x^{2}-\sqrt{5}x+1=0$
$\text{ Compare with }ax^{2}+bx+c=0$ $\text{ where }$ $a \neq 0$
$\\a=2, b=\sqrt{5},c=1\\ b^{2}-4ac=(-\sqrt{5})^{2}-4(2)(1\\ =5-8=-3$
$b^{2}-4ac<0$
$\text{Hence the equation }2x^{2}-\sqrt{5}x+1=0\text{ has no real roots.}$

Question:9

Which of the following equations has two distinct real roots?
$\\(A)2x^{2}-3\sqrt{2}x+\frac{9}{4}=0\\ (B)x^{2}+x-5=0\\ (C)x^{2}+3x+2\sqrt{2}=0\\ (D)5x^{2}-3x+1=0$

We know that if roots are distinct and real, then $b^{2}-4ac>0$
$(A)2x^{2}-3\sqrt{2}x+\frac{9}{4}=0$
$\text{compare with }$ $ax^{2}+bx+c=0$ $\text{where}$ $a \neq 0$
$a=2, b=-3\sqrt{2},c=\frac{9}{4}$
$\\b^{2}-4ac=\left ( -\frac{3}{\sqrt{2}} \right )^{2}-4(2)\left ( \frac{9}{4} \right )\\ =9 \times 2-18\\ =18-18=0\\ b^{2}-4ac=0\text{(two equal real roots)}$
$(B)\ x^{2}+x-5=0$
$b^{2}-4ac=(1)^{2}-4(1)(-5)\\ =1+20 =21\\ b^{2}-4ac>0\text{(two distinct real roots) }$
$(C)\ x^{2}+3x+2\sqrt{2}=0$

$(D)\ 5x^{2}-3x+1=0$
$\\b^{2}-4ac=\left (-3 \right )^{2}-4(5)\left ( 1 \right )\\ =9 -20\\ =-11\\ b^{2}-4ac<0\text{(no real roots)}$
Here only option (B) have two distinct real roots.

Question:11

$(x^{2}+1)^{2}-x^{2}=0$ has
(A) four real roots (B) two real roots
(C) no real roots (D) one real root.

(C) no real roots
Solution
Here the given equation is $(x^{2}+1)^{2}-x^{2}=0$
$(x^{2})^{2}+(1)^{2}+2(x^{2})(1)-x^{2}=0$ [using? ]
$x^{4}+1+2x^{2}-x^{2}=0\\ x^{4}+x^{2}+1=0$
Put $x^{2}=z$
$z^{2}+z+1=0$
Compare with $az^{2}+bz+c=0$ where $a \neq 0$
Here a=1,b=1,c=1
$b^{2}-4ac=(1)^{2}-4(1)(1)\\ =1-4=-3\\ b^{2}-4ac<0$
Here the given equation has no real roots.

NCERT Exemplar Solutions Class 10 Maths Chapter 4 Important Topics:

• How to judge the nature of roots by observing the quadratic equation.
• Students will learn different techniques to solve the quadratic equation such as factorisation or making perfect square etc.
• NCERT exemplar Class 10 Maths solutions chapter 4 discusses the methods to form a quadratic equation as mentioned in the CBSE Class 10 Maths Syllabus. For example: sum of a number and its reciprocal is equal to 5. Find out the number.” This information will turn into a quadratic equation of variable that represents the number.
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NCERT Class 10 Maths Exemplar Solutions for Other Chapters:

 Chapter 1 Real Numbers Chapter 2 Polynomials Chapter 3 Pair of Linear Equations in Two Variables Chapter 5 Arithmetic Progressions Chapter 6 Triangles Chapter 7 Coordinate Geometry Chapter 8 Introduction to Trigonometry & Its Equations Chapter 9 Circles Chapter 10 Constructions Chapter 11 Areas related to Circles Chapter 12 Surface Areas and Volumes Chapter 13 Statistics and Probability

Features of NCERT Exemplar Class 10 Maths Solutions Chapter 4:

These Class 10 Maths NCERT exemplar chapter 4 solutions provide a detailed knowledge of quadratic equations. The knowledge of quadratic equations will be useful even for students who do not pursue mathematics in higher classes. For students aspiring for engineering entrance examinations such as JEE Advanced, this chapter will be very important. Quadratic Equations practice problems of competitive level are also provided in the exemplar which can help the student to prepare well for all sorts of examinations.

The Class 10 Maths NCERT exemplar solutions chapter 4 Quadratic Equations cover a wide range of practice problems and solving them can help to excel in other books such as NCERT Class 10 Maths, RD Sharma Class 10 Maths, RS Aggarwal Class 10 Maths et cetera.

Students can easily download the pdf version of these solutions using NCERT exemplar Class 10 Maths solutions chapter 4 pdf download to review these detailed solutions while studying NCERT exemplar Class 10 Maths chapter 4.

Check Chapter-Wise Solutions of Book Questions

 Chapter No. Chapter Name Chapter 1 Real Numbers Chapter 2 Polynomials Chapter 3 Pair of Linear Equations in Two Variables Chapter 4 Quadratic Equations Chapter 5 Arithmetic Progressions Chapter 6 Triangles Chapter 7 Coordinate Geometry Chapter 8 Introduction to Trigonometry Chapter 9 Some Applications of Trigonometry Chapter 10 Circles Chapter 11 Constructions Chapter 12 Areas Related to Circles Chapter 13 Surface Areas and Volumes Chapter 14 Statistics Chapter 15 Probability

Also, Check NCERT Books and NCERT Syllabus here

1. Does every quadratic equation have two roots?

No, it is not necessary for any quadratic equation to have two roots. Quadratic equation can have a unique root as well as some quadratic equations cannot have any roots or zeros.

2. What is the importance of discriminant in quadratic equation?

The discriminant of quadratic equations is very important to judge the nature of roots.

By evaluating the value of discriminant, we can depict that the given quadratic equation will have any real root or not.

3. Is the chapter Quadratic Equations important for Board examinations

The chapter of Quadratic Equations is important for Board examinations as it holds around 4-5% weightage of the whole paper.

4. Is Quadratic equation chapter of extreme importance from a board examination perspective?

Quadratic Equation is one of the most important topics of Algebra of Class 10 and holds around 8-10% marks of the final paper. If a student practices and refers NCERT exemplar Class 10 Maths solutions chapter 4 he/she can score well in this vertical.

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Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

Yes, you can definitely apply for diploma courses after passing 10th CBSE. In fact, there are many diploma programs designed specifically for students who have completed their 10th grade.

Generally, passing 10th CBSE with a minimum percentage (often 50%) is the basic eligibility for diploma courses. Some institutes might have specific subject requirements depending on the diploma specialization.

There is a wide range of diploma courses available in various fields like engineering (e.g., mechanical, civil, computer science), computer applications, animation, fashion design, hospitality management, and many more.

You can pursue diplomas at various institutions like:

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9