Exercise 14.1

QUESTION 1

$x^{2}+1=0$

Sol :

$\Rightarrow x^{2}-1 i^{2}=0$

$\Rightarrow(x+i)(x-i)=0$ $\left[\left(a^{2}-b^{2}\right)=(a+b)(a-b)\right]$

$\Rightarrow(x+i)=0 \text { or }(x-i)=0$

$\Rightarrow x=-i \text { or } x=i$

Hence , the roots of the equation are i and -i

QUESTION 2

$9 x^{2}+4=0$

Sol :

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

$\Rightarrow(3 x)^{2}-(2 i)^{2}=0$

$\Rightarrow(3 x+2 i)(3 x-2 i)=0$ $\left[\left(a^{2}-b^{2}\right)=(a+b)(a-b)\right]$

$\Rightarrow(3 x+2 i)=0 \text{ or }(3 x-2 i)=0$

$\Rightarrow 3 x=-2 i \text { or } 3 x=2 i$

$\Rightarrow x=-\frac{2 i}{3} \quad \text { or } x=\frac{2 i}{3}$

Hence, the roots of the equation are $\frac{2 i}{3} \text { and }-\frac{2 i}{3}$

QUESTION 3

$x^{2}+2 x+5=0$

Sol :

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

$\Rightarrow(x+1)^{2}-(2 i)^{2}=0$ $\left[(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$

$\Rightarrow(x+1+2 i)(x+1-2 i)=0$ $\left[a^{2}-b^{2}=(a+b)(a-b)\right]$

$\Rightarrow(x+1+2 i)=0 \text { or, }(x+1-2 i)=0$

$\Rightarrow x=-(1+2 i) \text { or, } x=-1+2 i$

Hence, the roots of the equation are $-1+2 i \text { and }-1-2 i$

QUESTION 4

$4 x^{2}-12 x+25=0$

Sol :

$\Rightarrow 4 x^{2}-12 x+9+16=0$

$\Rightarrow(2 x)^{2}+3^{2}-2 \times 2 x \times 3-(4 i)^{2}=0$

$\Rightarrow(2 x-3)^{2}-(4 i)^{2}=0$

$\Rightarrow(2 x-3+4 i)(2 x-3-4 i)=0$ $\left[a^{2}-b^{2}=(a+b)(a-b)\right]$

$\Rightarrow(2 x-3+4 i)=0 \text { or, }(2 x-3-4 i)=0$

$\Rightarrow 2 x=3-4 i \text { or, } 2 x=3+4 i$

$\Rightarrow x=\frac{3}{2}-2 i \text { or, } x=\frac{3}{2}+2 i$

Hence , the roots of the equation are

QUESTION 5

$x^{2}+x+1=0$

Sol :

$\Rightarrow x^{2}+x+\frac{1}{4}+\frac{3}{4}=0$

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

$\Rightarrow\left(x+\frac{1}{2}\right)^{2}-\left(\frac{\sqrt{3 i}}{2}\right)^{2}=0$

$\Rightarrow\left(x+\frac{1}{2}+\frac{\sqrt{3} i}{2}\right)\left(x+\frac{1}{2}-\frac{\sqrt{3} i}{2}\right)=0$

$\Rightarrow\left(x+\frac{1}{2}+\frac{\sqrt{3 i}}{2}\right)=0 \text { or }\left(x+\frac{1}{2}-\frac{\sqrt{3} i}{2}\right)=0$

$\Rightarrow x=-\frac{1}{2}-\frac{\sqrt{3} i}{2} \quad \text { or, } x=-\frac{1}{2}+\frac{\sqrt{3} i}{2}$

Hence , the roots of the equation are $-\frac{1}{2}-i \frac{\sqrt{3}}{2} \text { and }-\frac{1}{2}+i \frac{\sqrt{3}}{2}$

QUESTION 6

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

Sol :

$\Rightarrow(2 x)^{2}-i^{2}=0$

$\Rightarrow(2 x)^{2}-(i)^{2}=0$

$\Rightarrow(2 x+i)(2 x-i)=0$

$\Rightarrow(2 x+i)=0 \text { or }(2 x-i)=0$

$\Rightarrow 2 x=-i \text { or } 2 x=i$

$\Rightarrow x=-\frac{i}{2} \quad \text { or } x=\frac{i}{2}$

Hence, the roots of the equation are $\frac{1}{2} i \text { and }-\frac{1}{2} i$

QUESTION 7

$x^{2}-4 x+7=0$

Sol :

$\Rightarrow x^{2}-4 x+4+3=0$

$\Rightarrow x^{2}-2 \times x \times 2+2^{2}-(\sqrt{3} i)^{2}=0$

$\Rightarrow(x-2)^{2}-(\sqrt{3} i)^{2}=0$

$\Rightarrow(x-2+\sqrt{3} i)(x-2-\sqrt{3} i)=0$

$\Rightarrow(x-2+\sqrt{3 i})=0 \text { or }(x-2-\sqrt{3} i)=0$

$\Rightarrow x=2-\sqrt{3} i \quad \text { or, } \quad x=2+\sqrt{3} i$

Hence, the roots of the equation are $2 \pm i \sqrt{3}$

QUESTION 8

$x^{2}+2 x+2=0$

Sol :

$\Rightarrow x^{2}+2 x+1+1=0$

$\Rightarrow x^{2}+2 \times x \times 1+1^{2}-(i)^{2}=0$

$\Rightarrow(x+1)^{2}-(i)^{2}=0$

$\Rightarrow(x+1+i)(x+1-i)=0$

$\Rightarrow(x+1+i)=0 \text { or }(x+1-i)=0$

$\Rightarrow x=-1-i \text { or } x=-1+i$

Hence , the roots of the equation are $-1+i \text { and }-1-i$

QUESTION 9

$5 x^{2}-6 x+2=0$

Sol :

Comparing the given equation with general form of the quadratic equation $a x^{2}+b x+c=0$ , we get $a=5$ $b=-6 \text { and } c=2$

Substituting these values in $\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$ and $\beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$ we get

$\alpha=\frac{6+\sqrt{36-4 \times 5 \times 2}}{2 \times 5}$ and $\beta=\frac{6-\sqrt{36-4 \times 2 \times 5}}{2 \times 5}$

$\Rightarrow \alpha=\frac{6+\sqrt{-4}}{10}$ and $\beta=\frac{6-\sqrt{-4}}{10}$

$\Rightarrow \alpha=\frac{6+\sqrt{4 i^{2}}}{10}$ and $\beta=\frac{6-\sqrt{4 i^{2}}}{10}$

$\Rightarrow \alpha=\frac{6+2 i}{10}$ and $\beta=\frac{6-2 i}{10}$

$\Rightarrow \alpha=\frac{2(3+i)}{10}$ and $\beta=\frac{2(3-i)}{10}$

$\alpha=\frac{3}{5}+\frac{1}{5} i$ and $\beta=\frac{3}{5}-\frac{1}{5} i$

Hence , the roots of the equation are $\frac{3}{5} \pm \frac{1}{5} i$

QUESTION 10

$21 x^{2}+9 x+1=0$

Sol :

Comparing the given equation with general form of the quadratic equation $a x^{2}+b x+c=0$ , we get $a=21$ $b=9 \text { and } c=1$

Substituting these values in $\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$ and $\beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$ we get

$\Rightarrow \alpha=\frac{-9+\sqrt{81-4 \times 21 \times 1}}{2 \times 21}$ and $\beta=\frac{-9-\sqrt{81-4 \times 21 \times 1}}{2 \times 21}$

$\Rightarrow \alpha=\frac{-9+\sqrt{3} i}{42}$ and $\beta=\frac{-9-\sqrt{3} i}{42}$

$\Rightarrow \alpha=-\frac{9}{42}+\frac{\sqrt{3} i}{42}$ and $\beta=-\frac{9}{42}-\frac{\sqrt{3} i}{42}$

$\Rightarrow \alpha=-\frac{3}{14}+\frac{\sqrt{3} i}{42}$ and $\beta=-\frac{3}{14}-\frac{\sqrt{3} i}{42}$

Hence , the roots of the equation are $-\frac{3}{14} \pm \frac{i \sqrt{3}}{42}$

QUESTION 11

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

Sol :

$\Rightarrow x^{2}-x+\frac{1}{2}+\frac{3}{4}=0$

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

$\Rightarrow\left(x-\frac{1}{2}\right)^{2}-\left(\frac{i \sqrt{3}}{2}\right)^{2}=0$

$\Rightarrow\left(x-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)\left(x-\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)=0$

$\Rightarrow\left(x-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)=0$ or $\left(x-\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)=0$

$\Rightarrow x=\frac{1}{2}-\frac{i \sqrt{3}}{2}$ or $x=\frac{1}{2}+\frac{i \sqrt{3}}{2}$

Hence , the roots of the equation are $\frac{1}{2} \pm \frac{i \sqrt{3}}{2}$

QUESTION 12

$x^{2}+x+1=0$

Sol :

We have :

$\Rightarrow x^{2}+x+\frac{1}{4}+\frac{3}{4}=0$

$\Rightarrow\left(x+\frac{1}{2}\right)^{2}-\frac{3}{4} i^{2}=0$

$\Rightarrow\left(x+\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)\left(x+\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)=0$

$\Rightarrow\left(x+\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)=0$ or $\left(x+\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)=0$

$\Rightarrow x=-\frac{1}{2}-\frac{i \sqrt{3}}{2}$ or $x=-\frac{1}{2}+\frac{i \sqrt{3}}{2}$

Hence , the roots of the equation are $-\frac{1}{2} \pm \frac{i \sqrt{3}}{2}$

QUESTION 13

$17 x^{2}-8 x+1=0$

Sol :

Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$ , we get $a=17$ $b=-8 \text { and } c=1$

Substituting these values in $\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$ and $\beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$ we get

$\alpha=\frac{8+\sqrt{64-4 \times 17 \times 1}}{2 \times 17}$ and $\beta=\frac{8-\sqrt{64-4 \times 17 \times 1}}{2 \times 17}$

$\Rightarrow \alpha=\frac{8+\sqrt{64-68}}{34}$ and $\beta=\frac{8-\sqrt{64-68}}{34}$

$\Rightarrow \alpha=\frac{8+\sqrt{-4}}{34}$ and $\beta=\frac{8-\sqrt{-4}}{34}$

$\Rightarrow \alpha=\frac{8+\sqrt{4 i^{2}}}{34}$ and $\beta=\frac{8-\sqrt{4 i^{2}}}{34}$

$\Rightarrow \alpha=\frac{8+2 i}{34}$ and $\beta=\frac{8-2 i}{34}$

$\Rightarrow \alpha=\frac{4+i}{17}$ and $\beta=\frac{4-i}{17}$

$\Rightarrow \alpha=\frac{4}{17}+\frac{1}{17} i$ and $\beta=\frac{4}{17}-\frac{1}{17} i$

Hence, the roots of the equation are $\frac{4}{17} \pm \frac{1}{17} i$

QUESTION 14

$27 x^{2}-10+1=0$

Sol :

Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$ , we get $a=27$ $b=-10 \text { and } c=1$

Substituting these values in $\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$ and $\beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$

$\Rightarrow \alpha=\frac{10+\sqrt{100-4 \times 27 \times 1}}{2 \times 27}$ and $\beta=\frac{10-\sqrt{100-4 \times 27 \times 1}}{2 \times 27}$

$\Rightarrow \alpha=\frac{10+\sqrt{100-108}}{54}$ and $\beta=\frac{10-\sqrt{100-108}}{54}$

$\Rightarrow \quad \alpha=\frac{10+\sqrt{-8}}{54}$ and $\beta=\frac{10-\sqrt{-8}}{54}$

$\Rightarrow \alpha=\frac{10+\sqrt{8 i^{2}}}{54}$ and

and

QUESTION 15

$17 x^{2}+28 x+12=0$

Sol :

Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$ , we get