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Everyday events like determining the area of a square garden, estimating a car's speed, or solving business word problems that can be expressed using quadratic equations. Therefore, what is a quadratic equation? A one-variable second-degree polynomial equation is a quadratic equation. It shows students the standard form of a quadratic equation, techniques for solving quadratic equations like factorization and the quadratic formula, and the nature of roots using the discriminant. It also enables students to grasp actual uses of quadratic equations in several sectors, including physics, engineering, and economics. Students are encouraged to study the NCERT Class 10th Maths Syllabus thoroughly.
These NCERT Exemplar Class 10 Maths chapter 4 solutions are created by our experts in 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 Class 10 Maths chapter 4 exemplar 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.
Class 10 Maths Chapter 4 Solutions Exercise: 4.1 Page number: 36-38 Total questions: 11 |
Answer:
A quadratic equation in x is an equation that can be written in the standard form. $a x^2+b x+c=0$ where $\mathrm{a}, \mathrm{b}$ and c are real number with $a \neq 0$
$x^2+2 x+1=(4-x)^2+3$
$⇒x^2+2 x+1=16-8x+x^2+3$
$⇒10x=18$
$⇒10x-18=0$
$10 x-18=0$ is not in the form of $a x^2+b x+c=0$
Hence, it is not a quadratic equation.
(B)
$\begin{aligned} & -2 x^2=(5-x)\left(2 x-\frac{2}{5}\right) \\ & -2 x^2=10 x-2-2 x^2+\frac{2}{5} x \\ & 10 x+\frac{2}{5} x-2=0 \\ & \frac{50 x+2 x-10}{5}=0 \\ & 5 x-10=0\end{aligned}$
This is not in the form of $a x^2+b x+c=0$
Hence, it is not a quadratic equation
$(C)(k+1) x^2+\frac{3}{2} x=7$ where $k=-1$
Put $k=-1$
$\begin{aligned} & ((-1)+1) x^2+\frac{3}{2} x=7 \\ & \frac{3}{2} x-7=0 \\ & \frac{3 x-14}{2}=0 \\ & 3 x-14=0\end{aligned}$
This is not in the form of $a x^2+b x+c=0$
Hence, this is not a quadratic equation.
(D)
$\begin{aligned} & x^3-x^2=(x-1)^3 \\ & x^3-x^2=(x)^3-(1)^3(1)+3(x)(1)^2 \\ & x^3-x^2-x^3+1+3 x^2-3 x=0 \\ & 2 x^2-3 x+1=0\end{aligned}$
This is in the form of $a x^2+b x+c=0$
Hence, $x^3-x^2=(x-1)^3$ is a quadratic equation.
Question 2: Which of the following is not a quadratic equation?
(C) $(\sqrt{2} x+\sqrt{3})^2+x^2=3 x^2-5 x$
$(D)\left(x^2+2 x\right)^2=x^4+3+4 x^3$
Answer:
Quadratic equation: A quadratic equation in x is an equation that can be written in the standard form $a x^2+b x+c=0$, $a \neq 0$
Where $\mathrm{a}, \mathrm{b}$ and c are real numbers
$\begin{aligned} & 2(x-1)^2=4 x^2-2 x+1 \\ & 2\left(x^2+1-2 x\right)=4 x^2-2 x+1\end{aligned}$
Using $(a-b)^2=a^2+b^2-2 a b$
$\begin{aligned} & 2 x^2+2-4 x-4 x^2+2 x-1=0 \\ & -2 x^2-2 x+1=0\end{aligned}$
It is a quadratic equation because it is in the form $a x^2+b x+c=0$
(B)
$\begin{aligned} & 2 x-x^2=x^2+5 \\ & 2 x-x^2-x^2-5=0 \\ & -2 x^2+2 x-5=0 \\ & -2 x^2+2 x-5=0\end{aligned}$
It is a quadratic equation because it is in the form $a x^2+b x+c=0$
(C)
$\begin{aligned} & (\sqrt{2} x+\sqrt{3})^3+x^2=3 x^2-5 x \\ & (\sqrt{2} x)^2+(\sqrt{3})^2+2(\sqrt{2} x)(\sqrt{3})+x^2=x^2-5 x \\ & \quad 2 x^2+3+2 \sqrt{6} x+x^2-3 x^2+5 x=0 \\ & (2 \sqrt{6}+5) x+3=0\end{aligned}$
It is not in the form of $a x^2+b x+c=0$
Hence, it is not a quadratic equation.
(D)
$\begin{aligned} & \quad\left(x^2+2 x\right)^2=x^4+3+4 x^3 \\ & \left(x^2\right)^2+(2 x)^2+2\left(x^2\right)(2 x)=x^4+3+4 x^3 \\ & x^4+4 x^2+4 x^3-x^4-3-4 x^3=0 \\ & 4 x^2-3=0\end{aligned}$
It is in the form of $a x^2+b x+c=0$
Hence, it is a quadratic equation
Only option (C) is not in the form of $a x^2+b x+c=0$
Hence, $(C)$ is not a quadratic equation.
Question 3: Which of the following equations has 2 as a root?
Answer:
(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 satisfied when x = 2.
Hence, option C has 2 as a root.
Question 4: If $\frac{1}{2}$ is a root of the equation $x^2+k x-\frac{5}{4}=0$, then the value of k is
Answer:
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, the value of k is 2.
Hence, the correct answer is (A).
Question 5: Which of the following equations has the sum of its roots as 3?
(C) $\sqrt{2} x^2-\frac{3}{\sqrt{2}} x+1=0$
Answer:
(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 a sum of its roots 3
Question 6: Values of k for which the quadratic equation $2 x^2-k x+k=0$ has equal roots is
Answer:
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.
Answer:
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
$\left (3x+ \frac{1}{8} \right )^{2}=\frac{1+64\times\sqrt{2}}{64}$ $\text{ Hence}\left (\frac{1}{8} \right )^{2}=\frac{1}{64}$ $\text{ should be added and subtracted in the equation}$ $9x^{2}+\frac{3}{4}x-\sqrt{2}=0$
$\text{ for solving by completing the square method. }$
Hence, the correct answer is option (B).
Question 8: The quadratic equation $2 x^2-\sqrt{5} x+1=0$ has
Answer:
(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) $2 x^2-3 \sqrt{2} x+\frac{9}{4}=0$
Answer:
We know that if roots are distinct and real, then $b^{2}-4ac>0$
$(A)2x^{2}-3\sqrt{2}x+\frac{9}{4}=0$
$\text{compare with }$ $ax^{2}+bx+c=0$ $\text{where}$ $a \neq 0$
$a=2, b=-3\sqrt{2},c=\frac{9}{4}$
$\\b^{2}-4ac=\left ( -\frac{3}{\sqrt{2}} \right )^{2}-4(2)\left ( \frac{9}{4} \right )\\ =9 \times 2-18\\ =18-18=0\\ b^{2}-4ac=0\text{(two equal real roots)}$
$(B)\ x^{2}+x-5=0$
$b^{2}-4ac=(1)^{2}-4(1)(-5)\\ =1+20 =21\\ b^{2}-4ac>0\text{(two distinct real roots) }$
$(C)\ x^{2}+3x+2\sqrt{2}=0$
$\begin{aligned} & b^2-4 a c=(3)^2-4(1)(2 \sqrt{2}) \\ & =9-8 \sqrt{2} \\ & =9-11.3=-2.3 \\ & b^2-4 a c<0(\text { no real roots })\end{aligned}$
$(D)\ 5x^{2}-3x+1=0$
$\\b^{2}-4ac=\left (-3 \right )^{2}-4(5)\left ( 1 \right )\\ =9 -20\\ =-11\\ b^{2}-4ac<0\text{(no real roots)}$
Here only option (B) has two distinct real roots.
Question 11: $\left(x^2+1\right)^2-x^2=0$ has
Answer:
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.
Hence, the correct answer is option (C).
Class 10 Maths chapter 4 solutions Exercise: 4.2 Page number: 38-39 Total questions: 7 |
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 $ax^{2}+bx+c=0$ 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 has 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 equation 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, $ax^{2} + bx + c = 0$
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 that 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 that 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 that 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 that 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 that satisfy equation (1) are the roots of the equation.
As we know, 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 are at most 2 values of x that satisfy the equation.
Hence, the given statement that 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 that 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 that 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 have have the same sign, and the 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: A quadratic equation with an integral coefficient has integral roots. Justify your answer
Answer:
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 that satisfy equation (1) are the roots of the equation.
Let us take a quadratic equation with an 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 the integral.
Hence, the given statement is false.
Answer:
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.
Answer:
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
Answer:
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 that 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 does not satisfy equation (2)
Hence,e 0.2 is not a root of the equation $x^{2}-0.4=0$
Hence, the given statement is false.
Answer:
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 that 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.
Class 10 Maths chapter 4 solutions Exercise: 4.3 Page number: 40 Total questions: 2 |
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 $ax^{2}+bx+c=0$ 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 factorization 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
$\\\frac{2}{5}x^{2}-x-\frac{3}{5}=0\\ \frac{2x^{2}-5x-3}{5}=0\\ 2x^{2}-5x-3=0\\ 2x^{2}-6x+x-3=0\\ 2x(x-3)+1(x-3)=0\\ (x-3)(2x+1)=0\\ x-3=0\;\;\;\;\;\;\;\;\;\;\;\;\; 2x+1=0\\ x=3\;\;\;\;\;\;\;\;\;\;\;\;\; 2x=-1 \;\;\;\;x=\frac{-1}{2}\\$
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 $21x^{2}-2x+\frac{1}{21}=0$
Class 10 Maths chapter 4 solutions Exercise: 4.4 Page number: 42-43 Total questions: 8 |
Question 1:
Find whether the following equations have real roots. If real roots exist, find them.
$(i)8x^{2}+2x-3=0$
$(ii)-2x^{2}+3x+2=0$
$(iii)5x^{2}-2x-10=0$
$(iv)\frac{1}{2x-3}+\frac{1}{x-5}=1,x\neq \frac{3}{2},5$
$(v)x^{2}+5\sqrt{5}x-70=0$
Answer:
(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 2: Find a natural number whose square diminished by 84 is equal to thrice 8 more than the given number
Answer:
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 3: A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.
Answer:
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.
Answer:
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 the 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 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 the original speed of the train is 45 km per hour.
Answer:
Let Zeba's current age = x
According to the 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 than 1 then it is equal to – 4 and age cannot be negative.
Hence, x = 14 i.e., the age of Zeba is 14 years
Answer:
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 the 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.
Answer:
Let the distance between the pond and the loan boundary be x.
Hence, the length of the pond =50-x-x
Then the breadth of the pond =40-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
Breadth of pond =40-2(8)
=40-16=24m
Answer:
At t minutes past 2 pm, the time needed by the minute hand to show 3 pm = 60 – t
According to the 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
For other chapters, you can refer here
Exemplar Class 10 Maths Solutions Chapter 4 pdf downloads are available through online tools for the students to access this content in an offline version, so that no breaks in continuity are faced while practicing Exemplar Class 10 Maths Chapter 4 solution.
These Class 10 Maths NCERT exemplar chapter 4 solutions provide a basic knowledge of quadratic equations, which has great importance in higher classes.
The questions based on quadratic equations can be practiced in a better way, along with these solutions.
The NCERT exemplar Class 10 Maths chapter 4 solution quadratic equations has a good amount of problems for practice and is sufficient for a student to solve the questions of other books.
Students can check the following links for more in-depth learning.
Students can check the following links for more in-depth learning.
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.
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.
The chapter of Quadratic Equations is important for Board examinations as it holds around 4-5% weightage of the whole paper.
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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