NCERT Exemplar Class 10 Maths Solutions Chapter 4 - Quadratic Equations

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.

NCERT Class 10 Maths Exemplar Solutions Chapter 4

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

Answer:

A quadratic equation in x is an equation that can be written in the standard form. $a{x}^2+bx+c=0$ where $\mathrm{a},\mathrm{b}$ and c are real number with $a\ne 0$

$x^2+2x+1=(4-x{)}^2+3$
$\Rightarrow x^2+2x+1=16-8x+x^2+3$
$\Rightarrow 10x=18$
$\Rightarrow 10x-18=0$

$10x-18=0$ is not in the form of $a{x}^2+bx+c=0$

Hence, it is not a quadratic equation.

(B)

$\begin{array}{rl}& -2x^2=(5-x)(2x-\frac{2}{5})\\ & -2x^2=10x-2-2x^2+\frac{2}{5}x\\ & 10x+\frac{2}{5}x-2=0\\ & \frac{50x+2x-10}{5}=0\\ & 5x-10=0\end{array}$

This is not in the form of $a{x}^2+bx+c=0$

Hence, it is not a quadratic equation

$(C)(k+1)x^2+\frac{3}{2}x=7$ where $k=-1$

Put $k=-1$

$\begin{array}{rl}& ((-1)+1)x^2+\frac{3}{2}x=7\\ & \frac{3}{2}x-7=0\\ & \frac{3x-14}{2}=0\\ & 3x-14=0\end{array}$

This is not in the form of $a{x}^2+bx+c=0$

Hence, this is not a quadratic equation.

(D)

$\begin{array}{rl}& x^3-x^2=(x-1{)}^3\\ & x^3-x^2=(x{)}^3-(1{)}^3(1)+3(x)(1{)}^2\\ & x^3-x^2-x^3+1+3x^2-3x=0\\ & 2x^2-3x+1=0\end{array}$

This is in the form of $a{x}^2+bx+c=0$

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) $(\sqrt{2}x+\sqrt{3}{)}^2+x^2=3x^2-5x$

$(D){(x^2+2x)}^2=x^4+3+4x^3$

Answer:

Quadratic equation: A quadratic equation in x is an equation that can be written in the standard form $a{x}^2+bx+c=0$, $a\ne 0$

Where $\mathrm{a},\mathrm{b}$ and c are real numbers

$\begin{array}{rl}& 2(x-1{)}^2=4x^2-2x+1\\ & 2(x^2+1-2x)=4x^2-2x+1\end{array}$

Using $(a-b{)}^2=a^2+b^2-2ab$

$\begin{array}{rl}& 2x^2+2-4x-4x^2+2x-1=0\\ & -2x^2-2x+1=0\end{array}$

It is a quadratic equation because it is in the form $a{x}^2+bx+c=0$

(B)

$\begin{array}{rl}& 2x-x^2=x^2+5\\ & 2x-x^2-x^2-5=0\\ & -2x^2+2x-5=0\\ & -2x^2+2x-5=0\end{array}$

It is a quadratic equation because it is in the form $a{x}^2+bx+c=0$

(C)

$\begin{array}{rl}& (\sqrt{2}x+\sqrt{3}{)}^3+x^2=3x^2-5x\\ & (\sqrt{2}x{)}^2+(\sqrt{3}{)}^2+2(\sqrt{2}x)(\sqrt{3})+x^2=x^2-5x\\ & 2x^2+3+2\sqrt{6}x+x^2-3x^2+5x=0\\ & (2\sqrt{6}+5)x+3=0\end{array}$

It is not in the form of $a{x}^2+bx+c=0$

Hence, it is not a quadratic equation.

(D)

$\begin{array}{rl}& {(x^2+2x)}^2=x^4+3+4x^3\\ & {\left(x^2\right)}^2+(2x{)}^2+2\left(x^2\right)(2x)=x^4+3+4x^3\\ & x^4+4x^2+4x^3-x^4-3-4x^3=0\\ & 4x^2-3=0\end{array}$

It is in the form of $a{x}^2+bx+c=0$

Hence, it is a quadratic equation

Only option (C) is not in the form of $a{x}^2+bx+c=0$

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) $3x^2-6x-2=0$

Answer:

(A) $x^2-4x+5=0$

Put x = 2

$$\begin{gathered} (2{)}^2-4(2)+5=0 \\ 4-8+5=0 \\ 9-8=0 \\ 1\ne 0 \end{gathered}$$

(B)

$$\begin{gathered} x^2+3x-12=0 \\ putx=2 \\ (2{)}^2+3(2)-12=0 \\ 4+6-12=0 \\ 10-12=0 \\ -2\ne 0 \end{gathered}$$

(C)

$$\begin{gathered} 2x^2-7x+6=0 \\ putx=2 \\ 2(2{)}^2-7(2)+6=0 \\ 8-14+6=0 \\ 14-14=0 \\ 0=0 \end{gathered}$$

(D)

$$\begin{gathered} 3x^2-6x-2=0 \\ putx=2 \\ 3(2{)}^2-6(2)-2=0 \\ 12-12-2=0 \\ -2\ne 0 \end{gathered}$$

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 $x^2+kx-\frac{5}{4}=0$, then the value of k is

(A) 2

(B) -2

(C) $\frac{1}{4}$

(D) $\frac{1}{2}$

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 $x=\frac{1}{2}$

$$\begin{gathered} {\left(\frac{1}{2}\right)}^2+k\left(\frac{1}{2}\right)-\frac{5}{4}=0 \\ \frac{1}{4}+\frac{k}{2}-\frac{5}{4}=0 \\ \frac{1+2k-5}{4}=0 \\ 2k-4=0 \\ 2k=4 \\ k=\frac{4}{2}=2 \\ k=2 \end{gathered}$$

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?

(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) $3x^2-3x+3=0$

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 $2x^2-kx+k=0$ has equal roots is

(A) 0 only

(B) 4

(C) 8 only

(D) 0,8

Answer:

Here the given equation is $2x^2-kx+k=0$

Compare with $a{x}^2+bx+c=0$ where $a\ne 0$

a=2,b=-k,c=k

It is given that roots are equal.

Hence

$$\begin{gathered} {b}^2-4ac=0 \\ (-k{)}^2-4(2)(k)=0 \\ {k}^2-8k=0 \\ k(k-8)=0 \\ k-8=0k=0 \\ k=8 \\ k=0,8 \end{gathered}$$

Hence for the value of k = 0, 8 the equation $a{x}^2+bx+c=0$ 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) $\frac{9}{64}$

Answer:

Here, the given quadratic equation is

$9x^2+\frac{3}{4}x-\sqrt{2}=0$

$$\begin{gathered} (3{)}^2+\frac{1}{4}(3x)-\sqrt{2}=0 \\ (3x{)}^2+2\times \left(\frac{1}{8}\right)\times (3x)-\sqrt{2}=0 \end{gathered}$$

$\text{[add and subtract}{\left(\frac{1}{8}\right)}^2]$

$$\begin{gathered} (3x{)}^2+2\times \left(\frac{1}{8}\right)\times (3x)+{\left(\frac{1}{8}\right)}^2-{\left(\frac{1}{8}\right)}^2-\sqrt{2}=0 \\ (3x{)}^2+2\times \left(\frac{1}{8}\right)\times (3x)+{\left(\frac{1}{8}\right)}^2={\left(\frac{1}{8}\right)}^2+\sqrt{2} \\ \because a^2+b^2+2ab=(a+b{)}^2 \end{gathered}$$

So

${(3x+\frac{1}{8})}^2=\frac{1+64\times \sqrt{2}}{64}$ $\text{ Hence}{\left(\frac{1}{8}\right)}^2=\frac{1}{64}$ $\text{ should be added and subtracted in the equation}$ $9x^2+\frac{3}{4}x-\sqrt{2}=0$

$\text{ for solving by completing the square method. }$

Hence, the correct answer is option (B).

Question 8: The quadratic equation $2x^2-\sqrt{5}x+1=0$ has

(A) two distinct real roots

(B) two equal real roots

(C) no real roots

(D) more than 2 real roots

Answer:

(C) no real roots

$\text{Here the given quadratic equation is }2x^2-\sqrt{5}x+1=0$

$\text{ Compare with }a{x}^2+bx+c=0$ $\text{ where }$ $a\ne 0$

$$\begin{gathered} a=2,b=\sqrt{5},c=1 \\ {b}^2-4ac=(-\sqrt{5}{)}^2-4(2)(1 \\ =5-8=-3 \end{gathered}$$

$b^2-4ac<0$

$\text{Hence the equation }2x^2-\sqrt{5}x+1=0\text{ has no real roots.}$

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) $5x^2-3x+1=0$

Answer:

We know that if roots are distinct and real, then $b^2-4ac>0$

$(A)2x^2-3\sqrt{2}x+\frac{9}{4}=0$

$\text{compare with }$ $a{x}^2+bx+c=0$ $\text{where}$ $a\ne 0$

$a=2,b=-3\sqrt{2},c=\frac{9}{4}$

$$\begin{gathered} {b}^2-4ac={(-\frac{3}{\sqrt{2}})}^2-4(2)\left(\frac{9}{4}\right) \\ =9\times 2-18 \\ =18-18=0 \\ {b}^2-4ac=0\text{(two equal real roots)} \end{gathered}$$

$(B)x^2+x-5=0$

$$\begin{gathered} b^2-4ac=(1{)}^2-4(1)(-5) \\ =1+20=21 \\ {b}^2-4ac>0\text{(two distinct real roots) } \end{gathered}$$

$(C)x^2+3x+2\sqrt{2}=0$

$\begin{array}{rl}& b^2-4ac=(3{)}^2-4(1)(2\sqrt{2})\\ & =9-8\sqrt{2}\\ & =9-11.3=-2.3\\ & b^2-4ac<0(\text{ no real roots })\end{array}$

$(D)5x^2-3x+1=0$

$$\begin{gathered} {b}^2-4ac={(-3)}^2-4(5)\left(1\right) \\ =9-20 \\ =-11 \\ {b}^2-4ac<0\text{(no real roots)} \end{gathered}$$

Here only option (B) has 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.

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? ]

$$\begin{gathered} x^4+1+2x^2-x^2=0 \\ {x}^4+x^2+1=0 \end{gathered}$$

Put $x^2=z$

$z^2+z+1=0$

Compare with $a{z}^2+bz+c=0$ where $a\ne 0$

Here a=1,b=1,c=1

$$\begin{gathered} b^2-4ac=(1{)}^2-4(1)(1) \\ =1-4=-3 \\ {b}^2-4ac<0 \end{gathered}$$

Here, the given equation has no real roots.

Hence, the correct answer is option (C).

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 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 $a{x}^2+bx+c=0$, 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 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, $a{x}^2+bx+c=0$

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 that 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 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 $a{x}^2+bx+c=0$ 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 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, $a{x}^2+bx+c=0$

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 that 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 that satisfy equation (1) are the roots of the equation.

As we know, 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 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, $a{x}^2+bx+c=0$

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 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.