NCERT Exemplar Class 10 Maths Solutions Chapter 4 Quadratic Equations

Question:1

State whether the following quadratic equations have two distinct real roots. Justify your answer.
$(i)x^2-3x+4=0$
$(ii)2x^2+x-1=0$
$(iii)2x^2-6x+\frac{9}{2}=0$
$(iv)3x^2-4x+1=0$
$(v)(x+4{)}^2-8x=0$
$(vi)(x-\sqrt{2}{)}^2-2(x+1)=0$
$(vii)\sqrt{2}{x}^2-\frac{3}{\sqrt{2}}x+\frac{1}{2}=0$
$(viii)x(1-x)-2=0$
$(ix)(x-1)(x+2)+2=0$
$(x)(x+1)(x-2)+x=0$

Answer:

(i) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $a{x}^2+bx+c=0$ Where a, b and c are real numbers with $a\ne 0$
The given equation is $x^2-3x+4=0$
Here a=1,b=-3,c=4
$$\begin{gathered} D=b^2-4ac \\ =(-3{)}^2-4(1)(4) \\ =9-16=-7 \\ {b}^2-4ac<0 \end{gathered}$$
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 $a{x}^2+bx+c=0$ Where a, b and c are real numbers with $a\ne 0$
The given equation is $2x^2+x-1=0$
Here a=2,b=-1,c=-1
$$\begin{gathered} D=b^2-4ac \\ =(1{)}^2-4(2)(-1) \\ =1+8=9 \\ {b}^2-4ac>0 \end{gathered}$$
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 $a{x}^2+bx+c=0$ Where a, b and c are real numbers with $a\ne 0$
The given equation is $2x^2-6x+\frac{9}{2}=0$
Here $a=2,b=-6,c=\frac{9}{2}$
$$\begin{gathered} D=b^2-4ac \\ =(-6{)}^2-4(2)(\frac{9}{4}) \\ =36-36=0 \\ {b}^2-4ac=0 \end{gathered}$$
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 $a{x}^2+bx+c=0$ Where a, b and c are real numbers with $a\ne 0$
The given equation is $3x^2-4x+1=0$
Here a=3,b=-4,c=1
$$\begin{gathered} D=b^2-4ac \\ =(-4{)}^2-4(3)(1) \\ =16-12=4 \\ {b}^2-4ac>0 \end{gathered}$$
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 $a{x}^2+bx+c=0$ Where a, b and c are real numbers with $a\ne 0$
The given equation is $(x+4{)}^2-8x=0$
$(x{)}^2+(4{)}^2+2(x)(4)-8x=0$ $(using(a+b{)}^2=(a{)}^2+(b{)}^2+2ab)$
$$\begin{gathered} (x{)}^2+16+8x-8x=0 \\ {x}^2+16=0 \end{gathered}$$
$$\begin{gathered} a=1,b=0,c=16 \\ =(0{)}^2-4(1)(16) \\ =-64 \\ {b}^2-4ac<0 \end{gathered}$$
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 $a{x}^2+bx+c=0$ Where a, b and c are real numbers with $a\ne 0$
The given equation is $(x-\sqrt{2}{)}^2-2(x+1)=0$
$(x{)}^2+(\sqrt{2}{)}^2+2(x)(\sqrt{2})-2x-2=0$ $(using(a+b{)}^2=(a{)}^2+(b{)}^2+2ab)$
$$\begin{gathered} (x{)}^2+2-2\sqrt{2}x-2x-2=0 \\ {x}^2-(2\sqrt{2}+2)x=0 \end{gathered}$$
Compare with $a{x}^2+bx+c=0$ where $a\ne 0$
$$\begin{gathered} a=1,b=-(2\sqrt{2}+2),c=0 \\ {b}^2-4ac=(-(2\sqrt{2}+2){)}^2-4(1)(0) \\ =(2\sqrt{2}+2{)}^2-0 \end{gathered}$$
$=(2\sqrt{2}{)}^2+(2{)}^2+2(2\sqrt{2})(2)$ $(using(a+b{)}^2=(a{)}^2+(b{)}^2+2ab)$
$$\begin{gathered} =8+4+8\sqrt{2} \\ =12+8\sqrt{2} \end{gathered}$$
$b^2-4ac>0$
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 $a{x}^2+bx+c=0$ Where a, b and c are real numbers with $a\ne 0$
The given equation is $\sqrt{2}{x}^2-\frac{3}{\sqrt{2}}x+\frac{1}{2}=0$
Here $a=\sqrt{2},b=\frac{3}{\sqrt{2}},c=\frac{1}{2}$
$$\begin{gathered} b^2-4ac={(-\frac{3}{\sqrt{2}})}^2-4(\sqrt{2})\left(\frac{1}{2}\right) \\ =\frac{9}{2}-2\sqrt{2} \\ =4.5-3.8=1.7 \\ {b}^2-4ac>0 \end{gathered}$$
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 $a{x}^2+bx+c=0$ Where a, b and c are real numbers with $a\ne 0$
The given equation is $x(1-x)-2=0$
$$\begin{gathered} x-(x{)}^2-2=0 \\ -x^2+x-2=0 \end{gathered}$$
compare with $a{x}^2+bx+c=0$ where $a\ne 0$
$$\begin{gathered} a=-1,b=1,c=-2 \\ {b}^2-4ac=(1{)}^2-4(-1)(-2) \\ 1-8=-7 \\ {b}^2-4ac<0 \end{gathered}$$
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 $a{x}^2+bx+c=0$ Where a, b and c are real numbers with $a\ne 0$
The given equation is $(x-1)(x+2)+2=0$
$$\begin{gathered} x(x+2)-1(x+2)+2=0 \\ {x}^2+2x-x-2+2=0 \\ {x}^2+x=0 \end{gathered}$$
$$\begin{gathered} a=1,b=1,c=0 \\ {b}^2-4ac=(1{)}^2-4(1)(0) \\ =1 \end{gathered}$$
$b^2-4ac>0$
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 $a{x}^2+bx+c=0$ Where a, b and c are real numbers with $a\ne 0$
The given equation is $(x+1)(x-2)+x=0$
$$\begin{gathered} x(x-2)+1(x-2)+x=0 \\ {x}^2-2x+x-2+x=0 \\ {x}^2-2x+2x-2=0 \\ {x}^2-2=0 \end{gathered}$$
Compare with $a{x}^2+bx+c=0$ where $a\ne 0$
$$\begin{gathered} a=1,b=0,c=-2 \\ {b}^2-4ac=(0{)}^2-4(1)(-2) \\ =8 \end{gathered}$$
$b^2-4ac>0$
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.

Answer:

(i) False
Solution
Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, $a{x}^2+bx+c=0$
Where a, b and c are real numbers with $a\ne 0$
Root: If $a{x}^2+bx+c=0$ …..(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 $x^2-4=0$
$$\begin{gathered} {x}^2-4=0 \\ {x}^2=4 \\ x=\pm \sqrt{4} \\ x=2x=-2 \end{gathered}$$
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, $a{x}^2+bx+c=0$
Where a, b and c are real numbers with $a\ne 0$
Root: If $a{x}^2+bx+c=0$ …..(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 $a{x}^2+bx+c=0$ where $a\ne 0$
$$\begin{gathered} a=1,b=-1,c=2 \\ {b}^2-4ac=(-1{)}^2-4(1)(2) \\ =1-8 \\ =-7 \\ {b}^2-4ac<0 \end{gathered}$$
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, $a{x}^2+bx+c=0$
Where a, b and c are real numbers with $a\ne 0$
Root: If $a{x}^2+bx+c=0$ …..(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 $x^2-4x+4=0$
$$\begin{gathered} {x}^2-4x+4=0 \\ {x}^2+(2{)}^2-2(x)(2) \\ (x-2{)}^2=0(using(a-b{)}^2=a^2+b^2-2ab) \\ (x-2)=0 \\ x=2 \end{gathered}$$
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, $a{x}^2+bx+c=0$
Where a, b and c are real numbers with $a\ne 0$
Root: If $a{x}^2+bx+c=0$ …..(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 $a{x}^2+bx+c=0$
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, $a{x}^2+bx+c=0$
Where a, b and c are real numbers with $a\ne 0$
Root: If $a{x}^2+bx+c=0$ …..(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 $a{x}^2+bx+c=0$ where $a\ne 0$
$$\begin{gathered} a=1,b=1,c=-2 \\ {b}^2-4ac=(1{)}^2-4(1)(-2) \\ =1+8 \\ =9 \\ {b}^2-4ac>0 \end{gathered}$$
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, $a{x}^2+bx+c=0$
Where a, b and c are real numbers with $a\ne 0$
Root: If $a{x}^2+bx+c=0$ …..(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 $a{x}^2+bx+c=0$ where $a\ne 0$
$$\begin{gathered} a=1,b=0,c=4 \\ {b}^2-4ac=(0{)}^2-4(1)(4) \\ =16 \\ {b}^2-4ac<0 \end{gathered}$$
Here $b^2-4ac<0$
Hence these types of equations have no real roots.

Question:3

A quadratic equation with integral coefficient has integral roots. Justify your answer

Answer:

False
Solution
Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, $a{x}^2+bx+c=0$
Where a, b and c are real numbers with $a\ne 0$
Root: If $a{x}^2+bx+c=0$ …..(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 $a{x}^2+bx+c=0$ where $a\ne 0$
(a=2,b=-3,c=-5)
let us find the roots of the equation
$$\begin{gathered} x=\frac{-b+\sqrt{b^2-4ac}}{2a} \\ x=\frac{3\pm \sqrt{9-4(2)(-5)}}{2\times 2} \\ x=\frac{3\pm \sqrt{9+40}}{4} \\ x=\frac{3+\sqrt{49}}{4} \\ x=\frac{3+7}{4}=\frac{5}{7}x=\frac{3-7}{4}=-1 \end{gathered}$$
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.

Answer:

True
Let us suppose a quadratic equation with rational coefficients.
$x^2-2x+4=0$
compare with $a{x}^2+bx+c=0$ where $a\ne 0$
a=-1,b=-2,c=4
let us find the roots of the equation
$$\begin{gathered} x=\frac{-b+\sqrt{b^2-4ac}}{2a} \\ x=\frac{2\pm \sqrt{4-4(-1)(4)}}{2(-1)} \\ x=\frac{2\pm \sqrt{20}}{-2} \\ x=\frac{2\pm 2\sqrt{5}}{-2} \\ x=\frac{2+2\sqrt{5}}{-2}x=\frac{2-2\sqrt{5}}{-2} \end{gathered}$$
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?

Answer:

True
Let us consider an equation with distinct irrational numbers
$\sqrt{2}{x}^2-5\sqrt{2}x+4\sqrt{2}=0$
compare with $a{x}^2+bx+c=0$ where $a\ne 0$
$a=\sqrt{2},b=-5\sqrt{2},c=4\sqrt{2}$
let us find the roots of the equation
$$\begin{gathered} x=\frac{-b+\sqrt{b^2-4ac}}{2a} \\ x=\frac{5\sqrt{2}\pm \sqrt{50-4(\sqrt{2})(4\sqrt{2})}}{2\sqrt{2}} \\ x=\frac{5\sqrt{2}\pm \sqrt{50-32}}{2\sqrt{2}} \\ x=\frac{5\sqrt{2}\pm \sqrt{18}}{2\sqrt{2}} \\ x=\frac{\sqrt{2}(5\pm 3)}{2\sqrt{2}}=\frac{5\pm 3}{2} \\ x=\frac{5+3}{2}=4x=\frac{5-3}{2}=1 \end{gathered}$$
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

Answer:

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 $\ne$ 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)
$$\begin{gathered} (0.2{)}^2-0.4=0 \\ \left(\frac{2}{10}\right)-0.4=0 \\ \frac{4}{100}-0.4=0 \\ 0.04-0.4=0 \\ -0.36\ne 0 \end{gathered}$$
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.

Answer:

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 $\ne$ 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)
$$\begin{gathered} {x}^2+0(x)-y=0 \\ {x}^2=y \\ x=\pm \sqrt{y} \\ x=+\sqrt{y}x=-\sqrt{y} \end{gathered}$$
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:
$(i)2x^2-3x-5=0$
$(ii)5x^2+13x+8=0$
$(iii)-3x^2+5x+12=0$
$(iv)-x^2+7x-10=0$
$(v)x^2+2\sqrt{2}x-6=0$
$(vi)x^2+3\sqrt{5}x+10=0$
$(vii)\frac{1}{2}{x}^2-\sqrt{11}x+1=0$

Answer:

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

(ii) $-1,\frac{-8}{5}$
$5x^2+13x+8=0$
Compare with $a{x}^2+bx+c=0$ where $a\ne 0$
Here a=5, b=13, c=8
$$\begin{gathered} x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ x=\frac{-13\pm \sqrt{(13{)}^2-4(5)(8)}}{2\times 5} \\ x=\frac{-13\pm \sqrt{169-160}}{10} \\ x=\frac{-13\pm 3}{10} \\ x=\frac{-13+3}{10}=\frac{-10}{10}=-1x=\frac{-13-3}{10}=\frac{-16}{10}=\frac{-8}{5} \end{gathered}$$
$x=-1,\frac{-8}{5}$
$(-1,\frac{-8}{5})$ are the roots of the equation $5x^2+13x+8=0$
(iii) $\frac{-4}{3},3$
$-3x^2+5x+12=0$
Compare with $a{x}^2+bx+c=0$ where $a\ne 0$
Here a=-3, b=5, c=12
$$\begin{gathered} x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ x=\frac{-5\pm \sqrt{(5{)}^2-4(-3)(12)}}{2\times (-3)} \\ x=\frac{-5\pm \sqrt{25+144}}{-6} \\ x=\frac{-5\pm 13}{-6} \\ x=\frac{-5+13}{-6}=\frac{8}{-6}=\frac{-4}{3}x=\frac{-5-13}{-6}=\frac{-18}{-6}=3 \end{gathered}$$
$x=\frac{-4}{3},3$
$(\frac{-4}{3},3)$ $\text{are the roots of the equation}$ $-3x^2+5x+12=0$
(iv) 5,2
$-x^2+7x-10=0$
Compare with $a{x}^2+bx+c=0$ where $a\ne 0$
Here a=-1, b=7, c=-10
$$\begin{gathered} x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ x=\frac{-7\pm \sqrt{49-4(-1)(-10)}}{2\times (-1)} \\ x=\frac{-7\pm \sqrt{9}}{-2} \\ x=\frac{-7\pm 3}{-2} \\ x=\frac{-7+3}{-2}=\frac{-4}{-2}=2x=\frac{-7-3}{-2}=\frac{-10}{-2}=5 \end{gathered}$$
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 $a{x}^2+bx+c=0$ where $a\ne 0$
Here $a=1,b=2\sqrt{2},c=-6$
$$\begin{gathered} x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ x=\frac{-2\sqrt{2}\pm \sqrt{(2\sqrt{2}{)}^2-4(1)(-6)}}{2\times 1} \\ x=\frac{-2\sqrt{2}\pm \sqrt{8+24}}{2} \\ x=\frac{2\sqrt{2}\pm 4\sqrt{2}}{2} \\ x=\frac{2(\sqrt{2}\pm 2\sqrt{2})}{2} \\ x=-\sqrt{2}\pm 2\sqrt{2} \\ x=-\sqrt{2}+2\sqrt{2}=\sqrt{2}x=-\sqrt{2}-2\sqrt{2}=-3\sqrt{2} \end{gathered}$$
$x=\sqrt{2},-3\sqrt{2}$
$\sqrt{2},-3\sqrt{2}$ 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 $a{x}^2+bx+c=0$ where $a\ne 0$
$a=1,b=3\sqrt{5},c=10$
$$\begin{gathered} x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ x=\frac{3\sqrt{5}\pm \sqrt{(3\sqrt{5}{)}^2-4(1)(10)}}{2\times 1} \\ x=\frac{3\sqrt{5}\pm \sqrt{45-40}}{2} \\ x=\frac{3\sqrt{5}\pm \sqrt{5}}{2} \\ x=\frac{3\sqrt{5}+\sqrt{5}}{2}=\frac{4\sqrt{5}}{2}=2\sqrt{5}x=\frac{3\sqrt{5}-\sqrt{5}}{2}=\frac{2\sqrt{5}}{2}=\sqrt{5} \end{gathered}$$
$x=2\sqrt{5},\sqrt{5}$
$2\sqrt{5},\sqrt{5}$ 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 $a{x}^2+bx+c=0$ where $a\ne 0$
Here $a=\frac{1}{2},b=-\sqrt{11},c=1$
$$\begin{gathered} x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ x=\frac{\sqrt{11}\pm \sqrt{(-\sqrt{11}{)}^2-4\left(\frac{1}{2}\right)(1)}}{2\times \frac{1}{2}} \\ x=\frac{\sqrt{11}\pm \sqrt{11-2}}{1} \\ x=\sqrt{11}\pm 3 \\ x=\sqrt{11}+3x=\sqrt{11}-3 \end{gathered}$$
$x=(\sqrt{11}+3),(\sqrt{11}-3)$
hence $(\sqrt{11}+3),(\sqrt{11}-3)$ 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
$(i)2x^2+\frac{5}{3}x-2=0$
$(ii)\frac{2}{5}{x}^2-x-\frac{3}{5}=0$
$(iii)3\sqrt{2}{x}^2-5x-\sqrt{2}=0$
$(iv)3x^2+5\sqrt{5}x-10=0$
$(v)21x^2-2x+\frac{1}{21}=0$

Answer:

(ii) $3,-\frac{1}{2}$
Solution
$$\begin{gathered} \frac{2}{5}{x}^2-x-\frac{3}{5}=0 \\ \frac{2x^2-5x-3}{5}=0 \\ 2x^2-5x-3=0 \\ 2x^2-6x+x-3=0 \\ 2x(x-3)+1(x-3)=0 \\ (x-3)(2x+1)=0 \\ x-3=02x+1=0 \\ x=32x=-1x=\frac{-1}{2} \end{gathered}$$
Hence $\frac{-3}{2},\frac{2}{3}$ are the roots of the equation.
(i) $\frac{-3}{2},\frac{2}{3}$
Solution
$$\begin{gathered} 2x^2+\frac{5}{3}x-2=0 \\ \frac{6x^2+5x-6}{3}=0 \\ 6x^2+5x-6=0 \\ 6x^2+9x-4x-6=0 \\ 3x(2x+3)-2(2x+3)=0 \\ (2x+3)(3x-2)=0 \\ 2x+3=03x-2=0 \\ 2x=-33x=2 \\ x=\frac{-3}{2}x=\frac{2}{3} \end{gathered}$$
Hence $3,-\frac{1}{2}$ are the roots of the equation.
(iii) $\sqrt{2},-\frac{1}{3\sqrt{2}}$
Solution
$$\begin{gathered} 3\sqrt{2}{x}^2-5x-\sqrt{2}=0 \\ 3\sqrt{2}{x}^2-6x+x-\sqrt{2}=0 \\ 3\sqrt{2}x(x-\sqrt{2})+1(x-\sqrt{2})=0 \\ (x-\sqrt{2})(3\sqrt{2}x+1)=0 \\ 3\sqrt{2}x+1=0x-\sqrt{2}=0 \\ 3\sqrt{2}x=-1x=\sqrt{2} \\ x=\frac{-1}{3\sqrt{2}} \end{gathered}$$
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
$$\begin{gathered} 3x^2+5\sqrt{5}x-10=0 \\ 3x^2+6\sqrt{5}x-\sqrt{5}x-10=0 \\ 3x(x+2\sqrt{5})-\sqrt{5}(x+2\sqrt{5})=0 \\ (3x-\sqrt{5})(x+2\sqrt{5})=0 \\ 3x-\sqrt{5}=5x+2\sqrt{5}=0 \\ 3x=\sqrt{5}x=-2\sqrt{5} \\ x=\frac{\sqrt{5}}{3} \end{gathered}$$
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
$$\begin{gathered} 21x^2-2x+\frac{1}{21}=0 \\ 21x^2-x-x+\frac{1}{21}=0 \\ x(21x-1)-\frac{1}{21}(21x-1)=0 \\ (21x-1)(x-\frac{1}{21})=0 \\ 21-x=0x-\frac{1}{21}=0 \\ 21x=1x=\frac{1}{21} \\ x=\frac{1}{21} \end{gathered}$$
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.

Answer: 14

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 $\frac{t^2}{4}$
$$\begin{gathered} 60-t=\frac{t^2}{4}-3 \\ 60-t=\frac{t^2-12}{4} \\ {t}^2-12=4(60-t) \\ {t}^2-12=240-4t \\ {t}^2+4t-12-240=0 \\ {t}^2+4t-252=0 \\ {t}^2+18t-14t-252=0 \\ t(t+18)-14(t+18)=0 \\ (t+18)(t-14)=0 \\ t=-18,14 \end{gathered}$$
Time can’t equal to negative term.
Hence t = 14 minutes