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#### Question 1:

Choose the correct answer from the given four options.
Which of the following is a quadratic equation?
(A) x2 + 2x + 1 = (4 – x)2 + 3
(B) $-2{x}^{2}=\left(5-x\right)\left(2x-\frac{2}{5}\right)$
(C)
(D) x3 x2 = (x – 1)3

An equation is of the form $a{x}^{2}+bx+c=0$$a\ne 0$ is called a quadratic equation. So, simplify each part of the question and check whether it is in the form of $a{x}^{2}+bx+c=0$ or not.

(a) Given that,
${x}^{2}+2x+1={\left(4-x\right)}^{2}+3$
$⇒{x}^{2}+2x+1=16+{x}^{2}-8x+3\phantom{\rule{0ex}{0ex}}$
$⇒10x-18=0$, which is not a quadratic equation.

(b) Given that,
$-2{x}^{2}=\left(5-x\right)\left(2x-\frac{2}{5}\right)$
$⇒-2{x}^{2}=10x-2{x}^{2}-2+\frac{2x}{5}$
$⇒50x+2x-10=0$
$⇒52x-10=0$, which is not a quadratic equation.

(c) Given that,
${x}^{2}\left(k+1\right)+\frac{3}{2}x=7$, where k = −1
$⇒{x}^{2}\left(-1+1\right)+\frac{3}{2}x=7$
$⇒3x-14=0$, which is not a quadratic equation.

(d) Given that,
${x}^{3}-{x}^{2}={\left(x-1\right)}^{3}$
$⇒{x}^{3}-{x}^{2}={x}^{3}-3{x}^{2}+3x-1\phantom{\rule{0ex}{0ex}}⇒-{x}^{2}+3{x}^{2}-3x+1=0$
$⇒2{x}^{2}-3x+1=0$, which is a quadratic equation.

Hence, the correct answer is option D.

#### Question 2:

Choose the correct answer from the given four options.
Which of the following is not a quadratic equation?
(A) 2(x – 1)2 = 4x2 – 2x + 1
(B) 2xx2 = x2 + 5
(C) ${\left(\sqrt{2}x+\sqrt{3}\right)}^{2}+{x}^{2}=3{x}^{2}-5x$
(D) (x2 + 2x)2 = x4 + 3 + 4x3

An equation is of the form $a{x}^{2}+bx+c=0$$a\ne 0$ is called a quadratic equation. So, simplify each part of the question and check whether it is in the form of $a{x}^{2}+bx+c=0$ or not.

(a) Given that,
$2{\left(x-1\right)}^{2}=4{x}^{2}-2x+1$
$⇒2\left({x}^{2}+1-2x\right)=4{x}^{2}-2x+1\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}+2-4x=4{x}^{2}-2x+1$
$⇒2{x}^{2}+2x-1=0$, which is a quadratic equation.

(b) Given that,
$2x-{x}^{2}={x}^{2}+5$
$⇒2{x}^{2}-2x+5=0$, which is a quadratic equation.

(c) Given that,
${\left(\sqrt{2}x+\sqrt{3}\right)}^{2}+{x}^{2}=3{x}^{2}-5x$
$⇒2{x}^{2}+3+2\sqrt{6}x+{x}^{2}=3{x}^{2}-5x$
$⇒5x+3+2\sqrt{6}x=0$, which is not a quadratic equation.

(d) Given that,
${\left({x}^{2}+2x\right)}^{2}={x}^{4}+3+4{x}^{3}$
$⇒{x}^{4}+4{x}^{2}+4{x}^{3}={x}^{4}+3+4{x}^{3}$
$⇒4{x}^{2}-3=0$, which is a quadratic equation.

Hence, the correct answer is option C.

#### Question 3:

Choose the correct answer from the given four options.
Which of the following equations has 2 as a root?
(A) x2 – 4x + 5 = 0
(B) x2 + 3x – 12 = 0
(C) 2x2 – 7x + 6 = 0
(D) 3x2 – 6x – 2 = 0

If α is one of the root of quadratic equation $a{x}^{2}+bx+c=0$, then x = α  satisfies the equation $a{\alpha }^{2}+b\alpha +c=0.$

(a) Given, ${x}^{2}-4x+5=0$
Substituting x = 2 in ${x}^{2}-4x+5$, we get
${2}^{2}-4\left(2\right)+5$
$⇒4-8+5=1\ne 0$, So we conclude, x = 2 is not a root of ${x}^{2}-4x+5=0$.

(b) Given, ${x}^{2}+3x-12=0$
Substituting x = 2 in ${x}^{2}+3x-12$, we get
${\left(2\right)}^{2}+3\left(2\right)-12$
$⇒4+6-12=-2\ne 0$, So we conclude, x = 2 is not a root of ${x}^{2}+3x-12=0$.

(c) Given, $2{x}^{2}-7x+6=0$
Substituting x = 2 in $2{x}^{2}-7x+6$, we get
$2{\left(2\right)}^{2}-7\left(2\right)+6$
$⇒8-14+6=14-14=0$, So we conclude, x = 2 is a root of $2{x}^{2}-7x+6=0$.

(d) Given, $3{x}^{2}-6x-2=0$
Substituting x = 2 in $3{x}^{2}-6x-2$, we get
$3{\left(2\right)}^{2}-6\left(2\right)-2$
$⇒12-12-2=-2\ne 0$, So we conclude, x = 2 is not a root of $3{x}^{2}-6x-2=0$.

Hence, the correct answer is option C.

#### Question 4:

Choose the correct answer from the given four options.
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)$\frac{1}{2}$

Since, $\frac{1}{2}$ is a root of given equation, then put $x=\frac{1}{2}$ in given equation and get the value of k.

Given,
$\frac{1}{2}$ is a root of the quadratic equation ${x}^{2}+kx-\frac{5}{4}=0.$
$⇒{\left(\frac{1}{2}\right)}^{2}+k\left(\frac{1}{2}\right)-\frac{5}{4}=0\phantom{\rule{0ex}{0ex}}⇒\frac{1}{4}+\frac{k}{2}-\frac{5}{4}=0\phantom{\rule{0ex}{0ex}}⇒\frac{1+2k-5}{4}=0$
$⇒2k-4=0\phantom{\rule{0ex}{0ex}}⇒2k=4⇒k=2$

Thus, we conclude the value of k = 2.
Hence, the correct answer is option A.

#### Question 5:

Choose the correct answer from the given four options.
Which of the following equations has the sum of its roots as 3?
(A) 2x2 – 3x + 6 = 0
(B) –x2 + 3x – 3 = 0
(C) $\sqrt{2}{x}^{2}-\frac{3}{\sqrt{2}}x+1=0$
(D) 3x2 – 3x + 3 = 0

If α and β are the roots of the quadratic equation $a{x}^{2}+bx+c=0$$a\ne 0$, then sum of roots = α + β = .

(a) Given, $2{x}^{2}-3x+6=0$.
Comparing with $a{x}^{2}+bx+c=0$, we get
a = 2, b = −3 and c = 6.
∴ Sum of the roots = $\frac{-b}{a}=\frac{-\left(-3\right)}{2}=\frac{3}{2}$.
So, we conclude sum of the roots of given quadratic equation is not 3.

(b) Given, $-{x}^{2}+3x-3=0$.
Comparing with $a{x}^{2}+bx+c=0$, we get
a = −1, b = 3 and c = −3.
∴Sum of the roots = $\frac{-b}{a}=\frac{-\left(3\right)}{-1}=3$.
So, we conclude sum of the roots of given quadratic equation is 3.

(c) Given, $\sqrt{2}{x}^{2}-\frac{3}{\sqrt{2}}x+1=0$.
Comparing with $a{x}^{2}+bx+c=0$, we get
a = 2, b = −3 and c = $\sqrt{2}$.
∴Sum of the roots = $\frac{-b}{a}=\frac{-\left(-3\right)}{2}=\frac{3}{2}$.
So, we conclude sum of the roots of given quadratic equation is not 3.

(d) Given, $3{x}^{2}-3x+3=0$.
Comparing with $a{x}^{2}+bx+c=0$, we get
a = 1, b = −1 and c = 1.
∴Sum of the roots = $\frac{-b}{a}=\frac{-\left(-1\right)}{1}=1$.
So, we conclude sum of the roots of given quadratic equation is not 3.

Hence, the correct answer is option B.

#### Question 6:

Choose the correct answer from the given four options.
Values of k for which the quadratic equation 2x2kx + k = 0 has equal roots is
(A) 0 only
(B) 4
(C) 8 only
(D) 0, 8

Given, $2{x}^{2}-kx+k=0$
Now, comparing with $a{x}^{2}+bx+c=0$, we get
a = 2, b = −k and c = k.

For equal roots, the discriminant must be zero.
i.e. we can say, $D={b}^{2}-4ac=0.$
$⇒{\left(-k\right)}^{2}-4\left(2\right)k=0\phantom{\rule{0ex}{0ex}}⇒{k}^{2}-8k=0\phantom{\rule{0ex}{0ex}}⇒k\left(k-8\right)=0$

k = 0, 8

Thus, the required values of k are 0 and 8.
Hence, the correct answer is option D.

#### Question 7:

Choose the correct answer from the given four options.
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) $\frac{9}{64}$

Given, equation $9{x}^{2}+\frac{3}{4}x-\sqrt{2}=0.$
$⇒{\left(3x\right)}^{2}+\frac{1}{4}\left(3x\right)-\sqrt{2}=0$
Substituting 3x = y in above equation, we get
${y}^{2}+\frac{1}{4}y-\sqrt{2}=0$
$⇒{y}^{2}+\frac{1}{4}y+{\left(\frac{1}{8}\right)}^{2}-{\left(\frac{1}{8}\right)}^{2}-\sqrt{2}=0\phantom{\rule{0ex}{0ex}}⇒{\left(y+\frac{1}{8}\right)}^{2}=\frac{1}{64}+\sqrt{2}\phantom{\rule{0ex}{0ex}}⇒{\left(y+\frac{1}{8}\right)}^{2}=\frac{1+64\sqrt{2}}{64}$

Thus, $\frac{1}{64}$ must be added and subtracted to solve the given equation.
Hence, the correct answer is option B.

#### Question 8:

Choose the correct answer from the given four options.
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

Given, equation $2{x}^{2}-\sqrt{5}x+1=0.$
Now, comparing with $a{x}^{2}+bx+c=0$, we get
a = 2, $-\sqrt{5}$ and c = 1.
$⇒D={b}^{2}-4ac={\left(-\sqrt{5}\right)}^{2}-4×2×1=5-8$, where D is discriminant.
$⇒D=-3$, which is less than 0.

Thus, we conclude given quadratic equation has no real roots since discriminant is negative.
Hence, the correct answer is option C.

#### Question 9:

Choose the correct answer from the given four options.
Which of the following equations has two distinct real roots?
(A) $2{x}^{2}-3\sqrt{2}x+\frac{9}{4}=0$
(B) x2 + x – 5 = 0
(C) ${x}^{2}+3x+2\sqrt{2}=0$
(D) 5x2 – 3x + 1 = 0

If a quadratic equation is in the form of $a{x}^{2}+bx+c=0$$a\ne 0$ then
(i) If $D={b}^{2}-4ac>0$, then its roots are distinct and real.
(ii) If $D={b}^{2}-4ac=0$, then its roots are real and equal.
(iii) If $D={b}^{2}-4ac<0$, then its roots are not real or imaginary roots.

(a) Given, $2{x}^{2}-3\sqrt{2}x+\frac{9}{4}=0.$
Now, comparing with $a{x}^{2}+bx+c=0$, we get
a = 2, $-3\sqrt{2}$ and c = $\frac{9}{4}$.
$⇒D={b}^{2}-4ac={\left(-3\sqrt{2}\right)}^{2}-4×2×\frac{9}{4}=18-18=0$, where D is discriminant.
$⇒D=0$.

Thus, equation has real and equal roots.

(b) Given ${x}^{2}+x-5=0.$
Now, comparing with $a{x}^{2}+bx+c=0$, we get
a = 1, = 1 and c = −5.
$⇒D={b}^{2}-4ac={1}^{2}-4×1×\left(-5\right)=1+20=21$, where D is discriminant.
$⇒D=21$, which is greater than zero.

Thus, equation has real and distinct roots.

(c) Given ${x}^{2}+3x+2\sqrt{2}=0.$
Now, comparing with $a{x}^{2}+bx+c=0$, we get
a = 1, = 3 and c = $2\sqrt{2}$
$⇒D={b}^{2}-4ac={\left(3\right)}^{2}-4×1×\left(2\sqrt{2}\right)=9-8\sqrt{2}<0$, where D is discriminant.
$⇒D<0$

Thus, equation has no real roots.

(d) Given $5{x}^{2}-3x+1=0.$
Now, comparing with $a{x}^{2}+bx+c=0$, we get
a = 5, = −3 and c = $1$
$⇒D={b}^{2}-4ac={\left(-3\right)}^{2}-4×5×1=9-20<0$, where D is discriminant.
$⇒D<0$

Thus, equation has no real roots.

Hence, the correct answer is option B.

#### Question 10:

Choose the correct answer from the given four options.
Which of the following equations has no real roots?
(A) ${x}^{2}-4x+3\sqrt{2}=0$
(B) ${x}^{2}+4x-3\sqrt{2}=0$
(C) ${x}^{2}-4x-3\sqrt{2}=0$
(D) $3{x}^{2}+4\sqrt{3}x+4=0$

If a quadratic equation is in the form of $a{x}^{2}+bx+c=0$$a\ne 0$ then
(i) If $D={b}^{2}-4ac>0$, then its roots are distinct and real.
(ii) If $D={b}^{2}-4ac=0$, then its roots are real and equal.
(iii) If $D={b}^{2}-4ac<0$, then its roots are not real or imaginary roots.

(a) Given ${x}^{2}-4x+3\sqrt{2}=0.$
Now, comparing with $a{x}^{2}+bx+c=0$, we get
a = 1, = −4 and c = $3\sqrt{2}$.
$⇒D={b}^{2}-4ac={\left(-4\right)}^{2}-4×1×3\sqrt{2}=16-12\sqrt{2}<0\phantom{\rule{0ex}{0ex}}$, where D is discriminant.
$⇒D<0$.

Thus, equation has no real roots.

(b) Given ${x}^{2}+4x-3\sqrt{2}=0.$
Now, comparing with $a{x}^{2}+bx+c=0$, we get
a = 1, = 4 and c = $-3\sqrt{2}$.
$⇒D={b}^{2}-4ac={\left(4\right)}^{2}-4×1×\left(-3\sqrt{2}\right)=16+12\sqrt{2}>0\phantom{\rule{0ex}{0ex}}$, where D is discriminant.
$⇒D>0$.

Thus, equation has real and distinct roots.

(c) Given ${x}^{2}-4x-3\sqrt{2}=0.$
Now, comparing with $a{x}^{2}+bx+c=0$, we get
a = 1, = −4 and c = $-3\sqrt{2}$.
$⇒D={b}^{2}-4ac={\left(-4\right)}^{2}-4×1×\left(-3\sqrt{2}\right)=16+12\sqrt{2}>0\phantom{\rule{0ex}{0ex}}$, where D is discriminant.
$⇒D>0$.

Thus, equation has real and distinct roots.

(d) Given $3{x}^{2}+4\sqrt{3}x+4=0.$
Now, comparing with $a{x}^{2}+bx+c=0$, we get
a = 3, $4\sqrt{3}$ and c = 4.
$⇒D={b}^{2}-4ac={\left(4\sqrt{3}\right)}^{2}-4×3×4=48-48=0\phantom{\rule{0ex}{0ex}}$, where D is discriminant.
$⇒D=0$.

Thus, equation has real and equal roots.

Hence, the correct answer is option A.

#### Question 11:

Choose the correct answer from the given four options.
(x2 + 1)2x2 = 0 has
(A) four real roots
(B) two real roots
(C) no real roots
(D) one real root.

Given equation, ${\left({x}^{2}+1\right)}^{2}-{x}^{2}=0.$
$⇒{x}^{4}+1+2{x}^{2}-{x}^{2}=0\phantom{\rule{0ex}{0ex}}⇒{x}^{4}+{x}^{2}+1=0$

Let ${x}^{2}=y$
∴ ${\left({x}^{2}\right)}^{2}+{x}^{2}+1=0$
$⇒{y}^{2}+y+1=0$

Now, comparing with $a{y}^{2}+by+c=0$, we get.
a = 1, b = 1 and c = 1
Thus, we can say $D={b}^{2}-4ac$, where D is discriminant.
$={\left(1\right)}^{2}-4\left(1\right)\left(1\right)\phantom{\rule{0ex}{0ex}}=1-4=-3\phantom{\rule{0ex}{0ex}}$

Since, $D<0$.
Thus, we conclude given equation has no real roots.
Hence, the correct answer is option C.

#### Question 1:

(i) x2 – 3x + 4 = 0
(ii) 2x2 + x – 1 = 0
(iii) $2{x}^{2}-6x+\frac{9}{2}=0$
(iv) 3x2 – 4x + 1 = 0
(v) (x + 4)2 – 8x = 0
(vi) ${\left(x-\sqrt{2}\right)}^{2}-2\left(x+1\right)=0$
(vii) $\sqrt{2}{x}^{2}-\frac{3}{\sqrt{2}}x+\frac{1}{\sqrt{2}}=0$
(viii) x(1 – x) – 2 = 0
(ix) (x – 1) (x + 2) + 2 = 0
(x) (x + 1) (x – 2) + x = 0

If a quadratic equation is in the form of $a{x}^{2}+bx+c=0$$a\ne 0$ then
(i) If $D={b}^{2}-4ac>0$, then its roots are distinct and real.
(ii) If $D={b}^{2}-4ac=0$, then its roots are real and equal.
(iii) If $D={b}^{2}-4ac<0$, then its roots are not real or imaginary roots.

(i) Given ${x}^{2}-3x+4=0.$
Now, comparing with $a{x}^{2}+bx+c=0$, we get
a = 1, b = −3 and c = 4
$⇒D={b}^{2}-4ac={\left(-3\right)}^{2}-4\left(1\right)\left(4\right)=9-16=-7$, where D is discriminant.
$⇒D<0$.
Thus, equation has no real roots.

(ii) Given $2{x}^{2}+x-1=0$.
Now, comparing with $a{x}^{2}+bx+c=0$, we get
= 2, b = 1 and c = −1
$⇒D={b}^{2}-4ac={\left(1\right)}^{2}-4\left(2\right)\left(-1\right)=1+8=9$, where D is discriminant.
$⇒D>0$.
​Thus, equation has two distinct and real roots.

(iii) Given $2{x}^{2}-6x+\frac{9}{2}=0$.
Now, comparing with $a{x}^{2}+bx+c=0$, we get
= 2, b = −6 and c = $\frac{9}{2}$
$⇒D={b}^{2}-4ac={\left(-6\right)}^{2}-4\left(2\right)\left(\frac{9}{2}\right)=36-36=0$, where D is discriminant.
$⇒D=0$.
​Thus, equation has equal and real roots.

(iv) Given $3{x}^{2}-4x+1=0.$
Now, comparing with $a{x}^{2}+bx+c=0$, we get
= 3, b = −4 and c = 1
$⇒D={b}^{2}-4ac={\left(-4\right)}^{2}-4\left(3\right)\left(1\right)=16-12=4$, where D is discriminant.
$⇒D>0$.
Thus, equation has two distinct real roots.

(v) Given ${\left(x+4\right)}^{2}-8x=0$.
$⇒{x}^{2}+16+8x-8x=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+16=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+0x+16=0$
Now, comparing with $a{x}^{2}+bx+c=0$, we get
= 1, b = 0 and c = 16
$⇒D={b}^{2}-4ac={\left(0\right)}^{2}-4\left(1\right)\left(16\right)=-64$, where D is discriminant.
$⇒D<0$.
Thus, equation has no real roots.

(vi) Given  ${\left(x-\sqrt{2}\right)}^{2}-2\left(x+1\right)=0$.
$⇒{x}^{2}+{\left(\sqrt{2}\right)}^{2}-2x\sqrt{2}-2x-2=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+2-2\sqrt{2}x-2x-2=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-x\left(2\sqrt{2}+2\right)=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-x\left(2\sqrt{2}+2\right)+0=0$
Now, comparing with $a{x}^{2}+bx+c=0$, we get
= 1, b = $-\left(2\sqrt{2}+2\right)$ and c = 0
$⇒D={b}^{2}-4ac={\left(-\left(2\sqrt{2}+2\right)\right)}^{2}-4\left(1\right)\left(0\right)\phantom{\rule{0ex}{0ex}}$, where D is discriminant.
$⇒D={\left(2\sqrt{2}+2\right)}^{2}$
$⇒D>0$.
Thus, equation has two distinct real roots.

(vii) Given equation, $\sqrt{2}{x}^{2}-\frac{3}{\sqrt{2}}x+\frac{1}{\sqrt{2}}=0$
Now, comparing with $a{x}^{2}+bx+c=0$, we get
= $\sqrt{2}$b = $-\frac{3}{\sqrt{2}}$ and c = $\frac{1}{\sqrt{2}}$
$⇒D={b}^{2}-4ac={\left(-\frac{3}{\sqrt{2}}\right)}^{2}-4\left(\sqrt{2}\right)\left(\frac{1}{\sqrt{2}}\right)=\frac{9}{2}-4=\frac{1}{2}$, where D is discriminant.
$⇒D>0$.
Thus, equation has two distinct real roots.

(viii) Given equation $x\left(1-x\right)-2=0$.
$⇒x-{x}^{2}-2=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-x+2=0$
Now, comparing with $a{x}^{2}+bx+c=0$, we get
= 1, b = −1 and c = 2
$⇒D={b}^{2}-4ac={\left(-1\right)}^{2}-4\left(1\right)\left(2\right)=1-8=-7$, where D is discriminant.
$⇒D<0$.
Thus, equation has no real roots.

(ix) Given equation $\left(x-1\right)\left(x+2\right)+2=0$.
$⇒{x}^{2}+x-2+2=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+x+0=0$
Now, comparing with $a{x}^{2}+bx+c=0$, we get
= 1, b = 1 and c = 0
$⇒D={b}^{2}-4ac={\left(1\right)}^{2}-4\left(1\right)\left(0\right)=1$, where D is discriminant.
$⇒D>0$.
Thus, equation has two distinct and real roots.

(x) Given equation $\left(x+1\right)\left(x-2\right)+x=0$.
$⇒{x}^{2}+x-2x-2+x=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-2=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+0x-2=0$
Now, comparing with $a{x}^{2}+bx+c=0$, we get
= 1, b = 0 and c = −2
$⇒D={b}^{2}-4ac={\left(0\right)}^{2}-4\left(1\right)\left(-2\right)=0+8=8$, where D is discriminant.
$⇒D>0$.
Thus, equation has two distinct real roots.

#### Question 2:

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

(i) False, since a quadratic equation has exactly two roots.
(ii) False, since consider equation ${x}^{2}+4=0$, has no real roots.
(iii) False, since a quadratic has two and only two roots.
(iv) False, every quadratic equation has exactlytwo roots.
(v) True
Consider, $D={b}^{2}-4ac$

Thus, discriminant will always be positive.
(vi) True
Consider, $D={b}^{2}-4ac$
Now, if b = 0, then ${b}^{2}-4ac=-4ac$ and ac > 0.
Thus, discriminant will always be negative.

#### Question 3:

No
Consider the quadratic equation $5{x}^{2}-3x-8=0$ with integral coefficient.
Now lets solve the quadratic equation.
$⇒5{x}^{2}-3x-8=0.$
$⇒5{x}^{2}-8x+5x-8=0\phantom{\rule{0ex}{0ex}}⇒x\left(5x-8\right)+1\left(5x-8\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(5x-8\right)\left(x+1\right)=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{8}{5},-1$
Thus, the roots of quadratic equation are not integers.
Hence, we conclude roots can be integral, but not always integral.

#### Question 4:

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

Yes
Consider the quadratic equation $2{x}^{2}-3x-15=0$, where coefficients are rational.
Now lets solve the quadratic equation using quadratic formula $\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$, where a, b and c coefficients of the quadratic equation $a{x}^{2}+bx+c=0$.
Comparing the equation, we get
a = 2, b = −3 and c = −15
$⇒D={b}^{2}-4ac={\left(-3\right)}^{2}-4\left(2\right)\left(-15\right)=9+120=129$, where D is discriminant.
Thus, roots are given
$x=\frac{-\left(-3\right)±\sqrt{129}}{2×2}$
$⇒x=\frac{3±\sqrt{129}}{4}$
Thus, we conclude roots are irrational since $\sqrt{129}$ is irrational.

#### Question 5:

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

Yes
Consider the quadratic equation $\sqrt{3}{x}^{2}-7\sqrt{3}x+12\sqrt{3}=0$, where coefficients are distinct irrational.
Now lets solve the quadratic equation using quadratic formula $\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$, where ab and c  are coefficients of the quadratic equation $a{x}^{2}+bx+c=0$.
Comparing the equation, we get
a = $\sqrt{3}$b = $-7\sqrt{3}$ and c $12\sqrt{3}$
$⇒D={b}^{2}-4ac={\left(-7\sqrt{3}\right)}^{2}-4\left(\sqrt{3}\right)\left(12\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}$
$⇒D=49×3-4×12×3=3\left(49-48\right)$
$⇒D=3$, where D is discriminant.

Thus, roots are given
$x=\frac{7\sqrt{3}±\sqrt{3}}{2\sqrt{3}}$
$⇒x=\frac{7\sqrt{3}+\sqrt{3}}{2\sqrt{3}}$ or $\frac{7\sqrt{3}-\sqrt{3}}{2\sqrt{3}}$
$⇒x=\frac{\sqrt{3}\left(7+1\right)}{2\sqrt{3}}$or$\frac{\sqrt{3}\left(7-1\right)}{2\sqrt{3}}$
Solving further, we get

Thus, we conclude roots are rational.

#### Question 6:

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

Given equation, ${x}^{2}-0.4=0$
Now, substitute the value 0.2 in the given equation, we get
${\left(0.2\right)}^{2}-0.4=0.04-0.4\ne 0$

Thus, we conclude 0.2 is not a root of the given equation.

#### 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.

Given, b = 0 and c < 0, for the quadratic equation ${x}^{2}+bx+c=0.$
Now, substitute b = 0 in the given equation, we get
${x}^{2}+0+c=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+c=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=-c$

∴ $x=±\sqrt{-c}$

Thus, we conclude the roots of the given quadratic equation are numerically equal and opposite in sign.

#### Question 1:

Find the roots of the quadratic equations by using the quadratic formula in each of the following:
(i) 2x2 – 3x – 5 = 0
(ii) 5x2 + 13x + 8 = 0
(iii) –3x2 + 5x + 12 = 0
(iv) –x2 + 7x – 10 = 0
(v) ${x}^{2}+2\sqrt{2}x-6=0$
(vi) ${x}^{2}-3\sqrt{5}x+10=0$
(vii) $\frac{1}{2}{x}^{2}-\sqrt{11}x+1=0$

(i) Given equation, $2{x}^{2}-3x-5=0$.
Now, comparing the equation with $a{x}^{2}+bx+c=0$, we get
a = 2, b = −3 and c = −5

This implies that using quadratic formula $x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$ , we get
$=\frac{-\left(-3\right)±\sqrt{{\left(-3\right)}^{2}-4\left(2\right)\left(-5\right)}}{2\left(2\right)}$
$=\frac{3±\sqrt{9+40}}{4}=\frac{3±\sqrt{49}}{4}$
$=\frac{3±7}{4}=\frac{10}{4},\frac{-4}{4}\phantom{\rule{0ex}{0ex}}=\frac{5}{2},-1$

Thus, the roots of the given quadratic equation are $\frac{5}{2}$ and $-1.$

(ii) Given equation, $5{x}^{2}+13x+8=0$.
Now, comparing the equation with $a{x}^{2}+bx+c=0$, we get
a = 5, b = 13 and = 8

This implies that using quadratic formula $x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$ , we get
$=\frac{-\left(13\right)±\sqrt{{\left(13\right)}^{2}-4\left(5\right)\left(8\right)}}{2\left(5\right)}$
$=\frac{-13±\sqrt{169-160}}{10}=\frac{-13±\sqrt{9}}{10}$
$=\frac{-13±3}{10}=-\frac{10}{10},-\frac{16}{10}\phantom{\rule{0ex}{0ex}}=-1,-\frac{8}{5}$

Thus, the roots of the given quadratic equation are $-\frac{8}{5}$ and $-1.$

(iii) Given equation, $-3{x}^{2}+5x+12=0$.
Now, comparing the equation with $a{x}^{2}+bx+c=0$, we get
a = −3, b = 5 and = 12

This implies that using quadratic formula $x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$ , we get
$=\frac{-\left(5\right)±\sqrt{{\left(5\right)}^{2}-4\left(-3\right)\left(12\right)}}{2\left(-3\right)}$
$=\frac{-5±\sqrt{25+144}}{-6}=\frac{-5±\sqrt{169}}{-6}$
$=\frac{-5±13}{-6}=\frac{8}{-6},\frac{-18}{-6}\phantom{\rule{0ex}{0ex}}=-\frac{4}{3},3$

Thus, the roots of the given quadratic equation are $-\frac{4}{3}$ and $3.$

(iv) Given equation, $-{x}^{2}+7x-10=0$.
Now, comparing the equation with $a{x}^{2}+bx+c=0$, we get
a = −1, b = 7 and = −10

This implies that using quadratic formula $x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$ , we get
$=\frac{-\left(7\right)±\sqrt{{\left(7\right)}^{2}-4\left(-1\right)\left(-10\right)}}{2\left(-1\right)}$
$=\frac{-7±\sqrt{49-40}}{-2}=\frac{-7±\sqrt{9}}{-2}$
$=\frac{-7±\sqrt{9}}{-2}=\frac{-4}{-2},\frac{-10}{-2}\phantom{\rule{0ex}{0ex}}=2,5$

Thus, the roots of the given quadratic equation are $2$ and $5.$

(v) Given equation, ${x}^{2}+2\sqrt{2}x-6=0$.
Now, comparing the equation with $a{x}^{2}+bx+c=0$, we get
a = 1, b = $2\sqrt{2}$ and = −6

This implies that using quadratic formula $x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$ , we get
$=\frac{-\left(2\sqrt{2}\right)±\sqrt{{\left(2\sqrt{2}\right)}^{2}-4\left(1\right)\left(-6\right)}}{2\left(1\right)}$
$=\frac{-2\sqrt{2}±\sqrt{8+24}}{2}=\frac{-2\sqrt{2}±\sqrt{32}}{2}$
$=\sqrt{2},-3\sqrt{2}$

Thus, the roots of the given quadratic equation are $\sqrt{2}$ and $-3\sqrt{2}.$

(vi) Given equation, ${x}^{2}-3\sqrt{5}x+10=0$.
Now, comparing the equation with $a{x}^{2}+bx+c=0$, we get
a = 1, b = $-3\sqrt{5}$ and = 10

This implies that using quadratic formula $x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$ , we get
$=\frac{-\left(-3\sqrt{5}\right)±\sqrt{{\left(-3\sqrt{5}\right)}^{2}-4\left(1\right)\left(10\right)}}{2\left(1\right)}$
$=\frac{3\sqrt{5}±\sqrt{45-40}}{2}=\frac{3\sqrt{5}±\sqrt{5}}{2}$
$=2\sqrt{5},\sqrt{5}$

Thus, the roots of the given quadratic equation are $2\sqrt{5}$ and $\sqrt{5}.$

(vii) Given equation, $\frac{1}{2}{x}^{2}-\sqrt{11}x+1=0$.
Now, comparing the equation with $a{x}^{2}+bx+c=0$, we get
a = $\frac{1}{2}$b = $-\sqrt{11}$ and = 1

This implies that using quadratic formula $x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$ , we get
$=\frac{-\left(-\sqrt{11}\right)±\sqrt{{\left(-\sqrt{11}\right)}^{2}-4\left(\frac{1}{2}\right)\left(1\right)}}{2\left(\frac{1}{2}\right)}$
$=\frac{\sqrt{11}±\sqrt{11-2}}{1}=\sqrt{11}±\sqrt{9}$

Thus, the roots of the given quadratic equation are $3+\sqrt{11}$ and $\sqrt{11}-3$.

#### Question 2:

Find the roots of the following quadratic equations by the factorisation method:
(i) $2{x}^{2}+\frac{5}{3}x-2=0$

(ii) $\frac{2}{5}{x}^{2}-x-\frac{3}{5}=0$

(iii) $3\sqrt{2}{x}^{2}-5x-\sqrt{2}=0$

(iv) $3{x}^{2}+5\sqrt{5}x-10=0$

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

(i) Given $2{x}^{2}+\frac{5}{3}x-2=0.$
Multiplying both sides by 3, we get
$6{x}^{2}+5x-6=0\phantom{\rule{0ex}{0ex}}⇒6{x}^{2}+\left(9x-4x\right)-6=0\phantom{\rule{0ex}{0ex}}⇒6{x}^{2}+9x-4x-6=0\phantom{\rule{0ex}{0ex}}⇒3x\left(2x+3\right)-2\left(2x+3\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(2x+3\right)\left(3x-2\right)=0\phantom{\rule{0ex}{0ex}}$

Now, $2x+3=0$
$⇒x=-\frac{3}{2}$
and $3x-2=0$
$⇒x=\frac{2}{3}$
Hence, the roots of the equation are .

(ii) Given $\frac{2}{5}{x}^{2}-x-\frac{3}{5}=0.$
Multiplying both sides by 5, we get
$2{x}^{2}-5x-3=0$
$⇒2{x}^{2}-\left(6x-x\right)-3=0\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-6x+x-3=0\phantom{\rule{0ex}{0ex}}⇒2x\left(x-3\right)+1\left(x-3\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-3\right)\left(2x+1\right)=0$

Now, $x-3=0$
$⇒x=3$
and $2x+1=0$
$⇒x=-\frac{1}{2}$
Hence, the roots of the given equation are

(iii) Given $3\sqrt{2}{x}^{2}-5x-\sqrt{2}=0$.
$⇒3\sqrt{2}{x}^{2}-\left(6x-x\right)-\sqrt{2}=0\phantom{\rule{0ex}{0ex}}⇒3\sqrt{2}{x}^{2}-6x+x-\sqrt{2}=0\phantom{\rule{0ex}{0ex}}⇒3\sqrt{2}{x}^{2}-\left(3\sqrt{2}\right)\left(\sqrt{2}x\right)+x-\sqrt{2}=0$
$⇒3\sqrt{2}x\left(x-\sqrt{2}\right)+1\left(x-\sqrt{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-\sqrt{2}\right)\left(3\sqrt{2}x+1\right)=0\phantom{\rule{0ex}{0ex}}$

Now, $x-\sqrt{2}=0\phantom{\rule{0ex}{0ex}}$
$⇒x=\sqrt{2}$
and $3\sqrt{2}x+1=0$
$⇒x=-\frac{1}{3\sqrt{2}}=\frac{-\sqrt{2}}{6}$

Hence, the roots of the equation are $\frac{-\sqrt{2}}{6}$ and $\sqrt{2}.$

(iv) Given  $3{x}^{2}+5\sqrt{5}x-10=0.$
$⇒3{x}^{2}+6\sqrt{5}x-\sqrt{5}x-10=0$
$⇒3{x}^{2}+6\sqrt{5}x-\sqrt{5}x-2\left(\sqrt{5}\right)\left(\sqrt{5}\right)=0$
$⇒3x\left(x+2\sqrt{5}\right)-\sqrt{5}\left(x+2\sqrt{5}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+2\sqrt{5}\right)\left(3x-\sqrt{5}\right)=0$

Now, $x+2\sqrt{5}=0\phantom{\rule{0ex}{0ex}}$
$⇒x=-2\sqrt{5}$
and $3x-\sqrt{5}=0$
$⇒x=\frac{\sqrt{5}}{3}$

Hence, the roots of the equation are .

(v) Given $21{x}^{2}-2x+\frac{1}{21}=0.$
Multiplying both sides by 21, we get
$441{x}^{2}-42x+1=0\phantom{\rule{0ex}{0ex}}⇒441{x}^{2}-\left(21x+21x\right)+1=0\phantom{\rule{0ex}{0ex}}⇒441{x}^{2}-21x-21x+1=0\phantom{\rule{0ex}{0ex}}⇒21x\left(21x-1\right)-1\left(21x-1\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(21x-1\right)\left(21x-1\right)=0$

Now, $21x-1=0$
$⇒x=\frac{1}{21}$
and $21x-1=0$
$⇒x=\frac{1}{21}$

Hence, the roots of the quadratic equation are $\frac{1}{21}$ and $\frac{1}{21}$.

#### Question 1:

Find whether the following equations have real roots. If real roots exist, find them.
(i) 8x2 + 2x – 3 = 0

(ii) –2x2 + 3x + 2 = 0

(iii) 5x2 – 2x – 10 = 0

(iv)

(v) ${x}^{2}+5\sqrt{5}x-70=0$

(a) Firstly we will check the quadratic equation has real roots or not, for this we will verify the discriminant
If discriminant, $D={b}^{2}-4ac\ge 0$ then roots are real.
(b) If roots are real, then further we can factorise the equation or use the quadratic formula to obtain the roots of the equation.

(i) Given $8{x}^{2}+2x-3=0.$
Now, comparing with $a{x}^{2}+bx+c=0$, we get
a = 8, b = 2 and c = −3
$⇒D={b}^{2}-4ac={\left(2\right)}^{2}-4\left(8\right)\left(-3\right)$, where D is discriminant.
$=4+96=100>0$
Thus, the given equation has distinct and real toots.

Now, using quadratic formula $x=\frac{-b±\sqrt{D}}{2a}$, we get
$x=\frac{-2±\sqrt{100}}{16}=\frac{-2±10}{16}$
$=\frac{-2+10}{16},\frac{-2-10}{16}\phantom{\rule{0ex}{0ex}}=\frac{1}{2},-\frac{3}{4}$
Hence, the roots of the given equation are $\frac{1}{2}$ and $-\frac{3}{4}$.

(ii) Given $-2{x}^{2}+3x+2=0.$
Now, comparing with $a{x}^{2}+bx+c=0$, we get
a = −2, b = 3 and c = 2
$⇒D={b}^{2}-4ac={\left(3\right)}^{2}-4\left(-2\right)\left(2\right)$, where D is discriminant.
$=9+16=25>0$
Thus, the given equation has distinct and real toots.

Now, using quadratic formula $x=\frac{-b±\sqrt{D}}{2a}$, we get
$x=\frac{-3±\sqrt{25}}{2\left(-2\right)}=\frac{-3±5}{-4}$
$=\frac{-3+5}{-4},\frac{-3-5}{-4}\phantom{\rule{0ex}{0ex}}=-\frac{1}{2},2$
Hence, the roots of the given equation are $-\frac{1}{2}$ and $2$.

(iii) Given $5{x}^{2}-2x-10=0.$
Now, comparing with $a{x}^{2}+bx+c=0$, we get
a = 5, b = −2 and c = −10
$⇒D={b}^{2}-4ac={\left(-2\right)}^{2}-4\left(5\right)\left(-10\right)$, where D is discriminant.
$=4+200=204>0$
Thus, the given equation has distinct and real toots.

Now, using quadratic formula $x=\frac{-b±\sqrt{D}}{2a}$, we get
$x=\frac{-\left(-2\right)±\sqrt{204}}{2×5}=\frac{2±2\sqrt{51}}{10}$
$=\frac{1+\sqrt{51}}{5},\frac{1-\sqrt{51}}{5}$
Hence, the roots of the given equation are $\frac{1+\sqrt{51}}{5}$ and $\frac{1-\sqrt{51}}{5}$.

(iv) Given
$⇒\frac{x-5+2x-3}{\left(2x-5\right)\left(x-5\right)}=1\phantom{\rule{0ex}{0ex}}⇒\frac{3x-8}{2{x}^{2}-5x-10x+25}=1\phantom{\rule{0ex}{0ex}}⇒\frac{3x-8}{2{x}^{2}-15x+25}=1\phantom{\rule{0ex}{0ex}}⇒3x-8=2{x}^{2}-15x+25$
$⇒2{x}^{2}-18x+33=0$
Now, comparing with $a{x}^{2}+bx+c=0$, we get
a = 2, b = −18 and c = 33
$⇒D={b}^{2}-4ac={\left(-18\right)}^{2}-4\left(2\right)\left(33\right)$, where D is discriminant.
$=324-264=60>0$
Thus, the given equation has distinct and real toots.

Now, using quadratic formula $x=\frac{-b±\sqrt{D}}{2a}$, we get
$x=\frac{-\left(-18\right)±\sqrt{60}}{2\left(2\right)}=\frac{18±2\sqrt{15}}{4}$
$=\frac{18±2\sqrt{15}}{4}\phantom{\rule{0ex}{0ex}}=\frac{9+\sqrt{15}}{2},\frac{9-\sqrt{15}}{2}$
Hence, the roots of the given equation are $\frac{9+\sqrt{15}}{2}$ and $\frac{9-\sqrt{15}}{2}$.

(v) Given ${x}^{2}+5\sqrt{5}x-70=0.$
Now, comparing with $a{x}^{2}+bx+c=0$, we get
a = 1, b = $5\sqrt{5}$ and c = −70
$⇒D={b}^{2}-4ac={\left(5\sqrt{5}\right)}^{2}-4\left(1\right)\left(-70\right)$, where D is discriminant.
$=125+280=405>0$
Thus, the given equation has distinct and real toots.

Now, using quadratic formula $x=\frac{-b±\sqrt{D}}{2a}$, we get
$x=\frac{-5\sqrt{5}±\sqrt{405}}{2\left(1\right)}=\frac{-5\sqrt{5}±9\sqrt{5}}{2}$
$=\frac{-5\sqrt{5}+9\sqrt{5}}{2},\frac{-5\sqrt{5}-9\sqrt{5}}{2}\phantom{\rule{0ex}{0ex}}=2\sqrt{5},-7\sqrt{5}$
Hence, the roots of the given equation are $2\sqrt{5}$ and $-7\sqrt{5}$.

#### Question 2:

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

Let n be a required natural number.
Square of a natural number diminished by 84 = n2 − 84, and
thrice of 8 more than the natural number = 3(n+8).
Now, by given condition
${n}^{2}-84=3\left(n+8\right)\phantom{\rule{0ex}{0ex}}⇒{n}^{2}-84=3n+24\phantom{\rule{0ex}{0ex}}⇒{n}^{2}-3n-108=0\phantom{\rule{0ex}{0ex}}⇒{n}^{2}-12n+9n-108=0\phantom{\rule{0ex}{0ex}}⇒n\left(n-12\right)+9\left(n-12\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(n-12\right)\left(n+9\right)=0\phantom{\rule{0ex}{0ex}}$
or $n=-9$
Since, -9 is not a natural number, thus $n\ne -9$.
Hence, the required natural number is 12.

#### Question 3:

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

Let x be required natural number.
Natural number increased by 12 = x + 12, and 160 times its reciprocal = $\frac{160}{x}$.
Now, by given condition.
$x+12=\frac{160}{x}$
Multiplying both sides by x, we get
${x}^{2}+12x-160=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+\left(20x-8x\right)-160=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+20x-8x-160=0\phantom{\rule{0ex}{0ex}}$
$⇒x\left(x+20\right)-8\left(x+20\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+20\right)\left(x-8\right)=0$

Now, $x+20=0⇒x=-20$, which is not a natural number, and
$x-8=0⇒x=8.$
Hence, the required natural number is 8.

#### Question 4:

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

Let the original speed of the train km/h.
Thus, the increased speed of the train = $\left(x+5\right)$ km/h and distance = 360 km
Now, by given condition
$\frac{360}{x}-\frac{360}{x+5}=\frac{4}{5}\phantom{\rule{0ex}{0ex}}⇒\frac{360\left(x+5\right)-360x}{x\left(x+5\right)}=\frac{4}{5}\phantom{\rule{0ex}{0ex}}⇒\frac{360x+1800-360x}{{x}^{2}+5x}=\frac{4}{5}$
$⇒\frac{1800}{{x}^{2}+5x}=\frac{4}{5}$
$⇒{x}^{2}+5x=\frac{1800×5}{4}=2250$
$⇒{x}^{2}+5x-2250=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+50x-45x-2250=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+50\right)-45\left(x+50\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+50\right)\left(x-45\right)=0$

Now, $x+50=0⇒x=-50$, which is not possible since speed cannot be negative and
$x-45=0⇒x=45$.
Hence, the original speed of the train = 45 km/h

#### Question 5:

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

Let the actual age of zeba = $x$ years.
Five years ago, her age was  = $\left(x-5\right)$ years.
Now, by given condition
Square of her age = 11 more than five times her actual age.
∴ ${\left(x-5\right)}^{2}=5×x+11$
$⇒{\left(x-5\right)}^{2}=5x+11\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+25-10x=5x+11\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-15x+14=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-14x-x+14=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-14\right)-1\left(x-14\right)$
$⇒\left(x-1\right)\left(x-14\right)=0$

Now, $x=1$ not possible because her age five years ago is $x-5=1-5=-4$ i.e. age cannot be negative
Hence, the required zeba's age now 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.

Let Nisha's present age be $x$ years.
Then, Asha's present age = ${x}^{2}+2$
Now, when Nisha grows to her mother's present age i.e. after ${x}^{2}+2-x$ years. Then, Asha's age also increased by ${x}^{2}+2-x$ years.
Again by given condition.
Age of Asha = one year less than 10 times the present age of Nisha.
$\left({x}^{2}+2\right)+\left({x}^{2}+2\right)-x=10x-1$
$⇒2{x}^{2}-x+4=10x-1\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-11x+5=0\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-10x-x+5=0\phantom{\rule{0ex}{0ex}}⇒2x\left(x-5\right)-1\left(x-5\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-5\right)\left(2x-1\right)=0$
∴ $x=5$ or $x=\frac{1}{2}$
Here, $x=\frac{1}{2}$ is not possible since at $x=\frac{1}{2}$, Asha's age is $2\frac{1}{4}$ years which is not possible.
Hence, present age of Nisha is 5 years and age of Asha = ${x}^{2}+2={\left(5\right)}^{2}+2=25+2=27$ years.

#### Question 7:

In the centre 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.

Given that a rectangular pond has to be constructed in the centre of a rectangular lawn of dimensions 50 m ⨯ 40 m. So, the distance between pond and lawn would be same around the pond, say $x$ m.

Now, length of rectangular lawn ${l}_{1}=50$ m and breadth of rectangular lawn ${b}_{1}=40$ m
∴ Length of rectangular pond ${l}_{2}=50-\left(x+x\right)=50-2x$ and breadth of rectangular pond ${b}_{2}=40-\left(x+x\right)=40-2x$
Also, area of the grass surrounding the pond = 1184 m2

∴ Area of rectangular lawn − Area of rectangular pond = Area of grass surrounding the pond.
${l}_{1}×{b}_{1}-{l}_{2}×{b}_{2}=1184$
$⇒50×40-\left(50-2x\right)\left(40-2x\right)=1184\phantom{\rule{0ex}{0ex}}⇒2000-\left(2000-80x-100x+4{x}^{2}\right)=1184\phantom{\rule{0ex}{0ex}}⇒80x+100x-4{x}^{2}=1184\phantom{\rule{0ex}{0ex}}⇒4{x}^{2}-180x+1184=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-45x+296=0\phantom{\rule{0ex}{0ex}}$
$⇒{x}^{2}-37x-8x+296=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-37\right)-8\left(x-37\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-37\right)\left(x-8\right)=0$
∴ $x=8$ or $37$

Now, at $x=37$, length and breadth of pond are $-24$ and $-34$, respectively but length and breadth cannot be negative.
So, $x=37$ is not possible.

∴ Length of pond = $50-2x=50-2\left(8\right)=50-16=34$ m.
and breadth of pond = $40-2x=40-2\left(8\right)=40-16=24$ m.

Hence, required length and breadth of pond are 34 m and 24 m, respectively.

#### Question 8:

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

Given that, at $t$ min past 2 pm, the time needed by the min hand of a clock to show 3 pm was found to be 3 min less than $\frac{{t}^{2}}{4}$ min i.e.
$t+\left(\frac{{t}^{2}}{4}-3\right)=60$
$⇒4t+{t}^{2}-12=240\phantom{\rule{0ex}{0ex}}⇒{t}^{2}+4t-252=0\phantom{\rule{0ex}{0ex}}⇒{t}^{2}+18t-14t-252=0\phantom{\rule{0ex}{0ex}}⇒t\left(t+18\right)-14\left(t+18\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(t+18\right)\left(t-14\right)=0\phantom{\rule{0ex}{0ex}}$
∴ $t=14$ or $-18$
Since, time cannot be negative, so $t\ne -18$
Hence, the required value of $t$ is 14 min.