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) $x^2+2 x+1=(4-x)^2+3$
$(B)-2 x^2=(5-x)\left(2 x-\frac{2}{5}\right)$
(C) $(k+1) x^2+\frac{3}{2} x=7$ where $k=-1$
(D) $x^3-x^2=(x-1)^3$

Answer:

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?

(A) $2(x-1)^2=4 x^2-2 x+1$

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

(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?

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

(B) $x^2+3 x-12=0$

(C) $2 x^2-7 x+6=0$

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

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

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

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

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?

(A) $2 x^2-3 x+6=0$

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

(C) $\sqrt{2} x^2-\frac{3}{\sqrt{2}} x+1=0$

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

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

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

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

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

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

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

(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

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

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

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

$b^{2}-4ac<0$

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

Question 9: Which of the following equations has two distinct real roots?

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

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

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

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

Here only option (B) has two distinct real roots.

Question 11: $\left(x^2+1\right)^2-x^2=0$ has

(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? ]

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

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).

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.

Question 4: Does there exist a quadratic equation whose coefficients are rational but both of its roots are irrational? Justify your answer.

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.

Question 5: Does there exist a quadratic equation whose coefficients are all distinct irrationals but both the roots are rational? Why?

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.

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:

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.

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 $21x^{2}-2x+\frac{1}{21}=0$

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.

Question 4: A train, traveling at a uniform speed of 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.

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.

Question 5: If Zeba were younger by 5 years than she 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?

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

Question 6: At present Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age. When Nisha grows to her mother’s present age, Asha’s age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha.

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.

Question 7: In the center of a rectangular lawn of dimensions 50 m × 40 m, a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be 1184 m2 [see Fig. 4.1]. Find the length and breadth of the pond.

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

Question 8: At t minutes past 2 pm, the time needed by the minute's hand of a clock to show 3 pm was found to be 3 minutes less than $\frac{t^{2}}{4}$ minutes. Find t.

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

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