Quadratic equations

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)^{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}\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 

and 

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

Hence, the roots of the equation are

 

QUESTION 16

21 x^{2}-28 x+10=0

Sol :

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

Hence, the roots of the equation are

 

QUESTION 17

8 x^{2}-9 x+3=0

Sol :

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

Hence, the roots of the equation are

 

QUESTION 18

13 x^{2}+7 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

Hence, the roots of the equation are

 

QUESTION 19

2 x^{2}+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

Hence, the roots of the equation are

 

QUESTION 20

\sqrt{3} x^{2}-\sqrt{2} x+3 \sqrt{3}=0

Sol :

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

Hence, the roots of the equation are

 

QUESTION 21

\sqrt{2} x^{2}+x+\sqrt{2}=0

Sol :

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

Hence, the roots of the equation are

 

QUESTION 22

x^{2}+x+\frac{1}{\sqrt{2}}=0

Sol :

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

Hence, the roots of the equation are

 

QUESTION 23

x^{2}+\frac{x}{\sqrt{2}}+1=0

Sol :

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

Hence, the roots of the equation are

 

QUESTION 24

\sqrt{5} x^{2}+x+\sqrt{5}=0

Sol :

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

Hence, the roots of the equation are

 

QUESTION 25

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

Sol :

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

Hence, the roots of the equation are

 

QUESTION 26

x^{2}-2 x+\frac{3}{2}=0

Sol :

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

Hence, the roots of the equation are

 

QUESTION 27

3 x^{2}-4 x+\frac{20}{3}=0

Sol :

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

Hence, the roots of the equation are

Leave a Reply

Your email address will not be published. Required fields are marked *