Class 11: Quadratic Equations – Exercise 14.2 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Solve the following quadratic equations by factorization method:

$\displaystyle \text{i) } x^2 + 10 i x - 21 = 0 \hspace{1.0cm} \text{ii) } x^2 + ( 1 - 2i) x - 2i = 0$

$\displaystyle \text{iii) } x^2 - ( 2\sqrt{3} + 3i) x + 6 \sqrt{3}i = 0 \hspace{1.0cm} \text{iv) } 6x^2 - 17i x - 12 = 0$

Answer:

$\displaystyle \text{i) } x^2 + 10 i x - 21 = 0$

$\displaystyle \Rightarrow x^2 + 7ix + 3ix - 21 = 0$

$\displaystyle \Rightarrow x( x + 7i) + 3i ( x + 7i) = 0$

$\displaystyle \Rightarrow (x+7i)(x+3i) = 0$

$\displaystyle \Rightarrow x = - 7i \text{ or } -3i$

So the roots of given quadratic equation are $\displaystyle -7i \text{ and } -3i$

$\displaystyle \text{ii) } x^2 + ( 1 - 2i) x - 2i = 0$

$\displaystyle \Rightarrow x^2 + x - 2i x - 2i = 0$

$\displaystyle \Rightarrow x( x+1) - 2i ( x+1) = 0$

$\displaystyle \Rightarrow (x+1)(x-2i) = 0$

$\displaystyle \Rightarrow x = - 1 \text{ or } 2i$

So the roots of given quadratic equation are $\displaystyle -1 \text{ and } 2i$

$\displaystyle \text{iii) } x^2 - ( 2\sqrt{3} + 3i) x + 6 \sqrt{3}i = 0$

$\displaystyle \Rightarrow x^2 - 2\sqrt{3} x - 3x i + 6\sqrt{3} i = 0$

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

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

$\displaystyle \Rightarrow x = 2 \sqrt{3}$ or $\displaystyle x = 3i$

So the roots of given quadratic equation are $\displaystyle 2 \sqrt{3} \text{ and } 3i$

$\displaystyle \text{iv) } 6x^2 - 17i x - 12 = 0$

$\displaystyle \Rightarrow 6x^2 - 9ix - 8ix - 12 = 0$

$\displaystyle \Rightarrow 3x(2x-3i) - 4i ( 2x-3i) = 0$

$\displaystyle \Rightarrow (2x-3i)(3x-4i) = 0$

$\displaystyle \Rightarrow x = \frac{3}{2} i$ or $\displaystyle x = \frac{4}{3} i$

So the roots of given quadratic equation are $\displaystyle \frac{3}{2} i \text{ and } \frac{4}{3} i$

$\displaystyle \\$

Question 2: Solve the following quadratic equations:

$\displaystyle \text{i) } x^2 - ( 3 \sqrt{2} + 2i)x + 6\sqrt{2} i = 0 \hspace{1.0cm} \text{ii) } x^2 - ( 5-i) x + ( 18+i) = 0$

$\displaystyle \text{iii) } (2+i) x^2 - ( 5-i)x + 2( 1 - i) = 0 \hspace{1.0cm} \text{iv) } x^2 - ( 2+i) x - ( 1 - 7i) = 0$

$\displaystyle \text{v) } ix^2 - 4x - 4i = 0 \hspace{1.0cm} \text{vi) } x^2 + 4ix - 4 = 0 \hspace{1.0cm} \text{vii) } 2x^2 + \sqrt{15} ix - i = 0$

$\displaystyle \text{viii) } x^2 - x + ( 1 + i) = 0 \hspace{1.0cm} \text{ix) } ix^2 - x + 12 i = 0$

$\displaystyle \text{x) } x^2 - ( 3\sqrt{2}- 2i) x - \sqrt{2} i = 0 \hspace{1.0cm} \text{xi) } x^2 - ( \sqrt{2} + i)x + \sqrt{2} i = 0$

$\displaystyle \text{xii) } 2x^2 - ( 3 + 7i) x + ( 9i-3) = 0$

Answer:

$\displaystyle \text{i) } x^2 - ( 3 \sqrt{2} + 2i)x + 6\sqrt{2} i = 0$

$\displaystyle \Rightarrow x^2 - 3 \sqrt{2} x - 2 i x + 6\sqrt{2} i = 0$

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

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

$\displaystyle \Rightarrow x = 3\sqrt{2}, 2i$

So the roots of given quadratic equation are $\displaystyle 3\sqrt{2} \text{ and } 2i$

$\displaystyle \text{ii) } x^2 - ( 5-i) x + ( 18+i) = 0$

$\displaystyle \Rightarrow x^2 - 5x + i x + 18 + i = 0$

$\displaystyle \Rightarrow x^2 - ( 3-4i) x - ( 2+3i) x + ( 18 + i) = 0$

$\displaystyle \Rightarrow x [ x - ( 2 + 3i) ] - ( 2 + 3i) [x - ( 3 - 4i)] = 0$

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

So the roots of given quadratic equation are $\displaystyle 2+3i \text{ and } ( 3 - 4i)$

$\displaystyle \text{iii) } (2+i) x^2 - ( 5-i)x + 2( 1 - i) = 0$

$\displaystyle \Rightarrow (2+i)^2-2x-(3-i)x+2(1-i) = 0$

$\displaystyle \Rightarrow [x-(1-i)][(2+i)x - 2] = 0$

$\displaystyle \Rightarrow x = (1-i)$ or $\displaystyle x = \frac{2}{2+i} \times \frac{2-i}{2-i} = \frac{4-2i}{5}$

$\displaystyle \text{iv) } x^2 - ( 2+i) x - ( 1 - 7i) = 0$

$\displaystyle \Rightarrow x^2 - ( 3 - i )x + ( 1 - 2i)x - ( 1 - 7i) = 0$

$\displaystyle \Rightarrow x[x-(3-i)]+(1-2i)[x-(3-i)] = 0$

$\displaystyle \Rightarrow [x - ( 3 - i)][ x + 1 - 2i] = 0$

$\displaystyle \Rightarrow x = 3 - i$ or $\displaystyle x = -1 + 2i$

$\displaystyle \text{v) } ix^2 - 4x - 4i = 0$

$\displaystyle \Rightarrow ix^2 + 4i^2 x + 4i^3 = 0$

$\displaystyle \Rightarrow x^2 + 4ix + 4i^2 = 0$

$\displaystyle \Rightarrow x^2 + 2ix + 2ix + 4i^2 =0$

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

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

$\displaystyle \Rightarrow x = - 2i$

$\displaystyle \text{vi) } x^2 + 4ix - 4 = 0$

$\displaystyle \Rightarrow x^2 + 4ix + 4i^2 = 0$

$\displaystyle \Rightarrow x^2 + 2ix + 2ix + 4i^2 = 0$

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

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

$\displaystyle \Rightarrow x = - 2i$

$\displaystyle \text{vii) } 2x^2 + \sqrt{15} ix - i = 0$

Comparing the given equation with general form of quadratic equation $\displaystyle ax^2 + bx + c = 0$ we get,

$\displaystyle \Rightarrow a = 2, \ b= \sqrt{15}i \text{ and } c = -i$

Substituting these values in

$\displaystyle \alpha = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$

$\displaystyle \text{Therefore } \alpha = \frac{-\sqrt{15}i \pm \sqrt{15i^2 - 4 \times 2 \times (-1)} }{2 \times 2}$

$\displaystyle \Rightarrow \alpha = \frac{-\sqrt{15}i \pm \sqrt{8i-15} }{4}$

Now lets calculate the roots of $\displaystyle \sqrt{8i-15}$

Let $\displaystyle x + i y = \sqrt{8i-15}$

Squaring both sides

$\displaystyle x^2 - y^2 + 2 xy i = 8i - 15$

$\displaystyle \Rightarrow x^2 - y^2 = - 15$ … … … … … i)

$\displaystyle \text{and } 2xy = 8$ … … … … … ii)

We know $\displaystyle (x^2 + y^2)^2 = ( x^2 - y^2 )^2 + 4 x^2 y^2$

$\displaystyle \Rightarrow (x^2 +y^2)^2 = 225 + 64 = 289$

$\displaystyle \Rightarrow x^2 + y^2 = 17$ … … … … … iii)

Adding i) and iii) we get

$\displaystyle 2x^2 = 2 \Rightarrow x^2 =1 \Rightarrow x = \pm 1$

Substituting in i) we get

$\displaystyle y^2 = x^2 + 15 \Rightarrow y^2 = 16 \Rightarrow y = \pm 4$

Since $\displaystyle xy$ is positive, $\displaystyle x \text{ and } y$ will have the same sign.

$\displaystyle \therefore$ if $\displaystyle x = 1, y = 4$ and if $\displaystyle x = -1 , y = -4$

$\displaystyle \therefore \sqrt{8i-15} = 1+4i \text{ or } -1-4i$

$\displaystyle \therefore \alpha = \frac{-\sqrt{15}i + (1+4i)}{4} = \frac{1 + ( 4 - \sqrt{15})i}{4}$

$\displaystyle \text{and } \beta = \frac{-\sqrt{15}i - (1+4i)}{4} = \frac{-1-(4+\sqrt{15})i}{4}$

$\displaystyle \text{viii) } x^2 - x + ( 1 + i) = 0$

Comparing the given equation with general form of quadratic equation $\displaystyle ax^2 + bx + c = 0$ we get,

$\displaystyle \Rightarrow a = 1, \ b= -1 \text{ and } c = (1+i)$

Substituting these values in

$\displaystyle \alpha = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$

$\displaystyle \text{Therefore } \alpha = \frac{1 \pm \sqrt{(-1)^2 - 4 \times 1 \times (1+i)} }{2 \times 1}$

$\displaystyle \Rightarrow \alpha = \frac{1 \pm \sqrt{-3-4i} }{2}$

Now lets calculate the roots of $\displaystyle \sqrt{-3-4i}$

Let $\displaystyle x + i y = \sqrt{-3-4i}$

Squaring both sides

$\displaystyle x^2 - y^2 + 2 xy i = -3-4i$

$\displaystyle \Rightarrow x^2 - y^2 = -3$ … … … … … i)

$\displaystyle \text{and } 2xy = -4$ … … … … … ii)

We know $\displaystyle (x^2 + y^2)^2 = ( x^2 - y^2 )^2 + 4 x^2 y^2$

$\displaystyle \Rightarrow (x^2 +y^2)^2 = 9 + 16 = 25$

$\displaystyle \Rightarrow x^2 + y^2 = 5$ … … … … … iii)

Adding i) and iii) we get

$\displaystyle 2x^2 = 2 \Rightarrow x^2 =1 \Rightarrow x = \pm 1$

Substituting in i) we get

$\displaystyle y^2 = x^2 + 3 \Rightarrow y^2 = 4 \Rightarrow y = \pm 2$

Since $\displaystyle xy$ is negative, $\displaystyle x \text{ and } y$ will have opposite sign.

$\displaystyle \therefore$ if $\displaystyle x = 1, y = -2$ and if $\displaystyle x = -1 , y = 2$

$\displaystyle \therefore \sqrt{-3-4i} = (1-2i) \text{ or } (-1+2i)$

$\displaystyle \therefore \alpha = \frac{1 + (1-2i)}{2} = 1- i$

$\displaystyle \text{and } \beta = \frac{1 + (-1+2i)}{2} = i$

$\displaystyle \text{ix) } ix^2 - x + 12 i = 0$

$\displaystyle \Rightarrow i( x^2 + i x + 12) = 0$

$\displaystyle \Rightarrow x^2 + ix + 12 = 0$

$\displaystyle \Rightarrow x^2 + 4ix - 3i x + 12 = 0$

$\displaystyle \Rightarrow x( x+4i)- 3( x + 4i) = 0$

$\displaystyle \Rightarrow (x+ 4i) ( x - 3i) = 0$

$\displaystyle \Rightarrow x = -4 \text{ or } x = 3i$

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

Comparing the given equation with general form of quadratic equation $\displaystyle ax^2 + bx + c = 0$ we get,

$\displaystyle \Rightarrow a = 1, \ b= -( 3\sqrt{2} -2i) \text{ and } c = -\sqrt{2}i$

Substituting these values in

$\displaystyle \alpha = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$

$\displaystyle \text{Therefore } \alpha = \frac{( 3\sqrt{2} -2i) \pm \sqrt{(-( 3\sqrt{2} -2i))^2 - 4 \times 1 \times (-\sqrt{2}i)} }{2 \times 1}$

$\displaystyle \Rightarrow \alpha = \frac{( 3\sqrt{2} -2i) \pm \sqrt{14-8\sqrt{2} i} }{2}$

Now lets calculate the roots of $\displaystyle \sqrt{14-8\sqrt{2} i}$

Let $\displaystyle x + i y = \sqrt{14-8\sqrt{2} i}$

Squaring both sides

$\displaystyle x^2 - y^2 + 2 xy i = 14-8\sqrt{2} i$

$\displaystyle \Rightarrow x^2 - y^2 = 14$ … … … … … i)

$\displaystyle \text{and } 2xy = -8\sqrt{2}$ … … … … … ii)

We know $\displaystyle (x^2 + y^2)^2 = ( x^2 - y^2 )^2 + 4 x^2 y^2$

$\displaystyle \Rightarrow (x^2 +y^2)^2 = 196 + 128 = 324$

$\displaystyle \Rightarrow x^2 + y^2 = 18$ … … … … … iii)

Adding i) and iii) we get

$\displaystyle 2x^2 = 32 \Rightarrow x^2 =16 \Rightarrow x = \pm 4$

Substituting in i) we get

$\displaystyle y^2 = x^2 -14 \Rightarrow y^2 = 16-14= 2 \Rightarrow y = \pm \sqrt{2}$

Since $\displaystyle xy$ is negative, $\displaystyle x \text{ and } y$ will have opposite sign.

$\displaystyle \therefore$ if $\displaystyle x = 4, y = -\sqrt{2}$ and if $\displaystyle x = -4 , y = \sqrt{2}$

$\displaystyle \therefore \sqrt{14-8\sqrt{2} i} = (4-\sqrt{2}i) \text{ or } (-4+\sqrt{2} i)$

$\displaystyle \therefore \alpha = \frac{(3\sqrt{2}-2i) \pm (4-\sqrt{2} i)}{2}$

$\displaystyle \text{Hence the roots of the equation are } \frac{(3\sqrt{2}-2i) \pm (4-\sqrt{2} i)}{2}$

$\displaystyle \text{xi) } x^2 - ( \sqrt{2} + i)x + \sqrt{2} i = 0$

Comparing the given equation with general form of quadratic equation $\displaystyle ax^2 + bx + c = 0$ we get,

$\displaystyle \Rightarrow a = 1, \ b= -( \sqrt{2}+i) \text{ and } c = \sqrt{2}i$

Substituting these values in

$\displaystyle \alpha = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$

$\displaystyle \text{Therefore } \alpha = \frac{( \sqrt{2} +i) \pm \sqrt{(-(\sqrt{2} +i)^2 - 4 \times 1 \times (\sqrt{2}i)} }{2 \times 1}$

$\displaystyle \Rightarrow \alpha = \frac{(\sqrt{2}+i) \pm \sqrt{(\sqrt{2})^2 - 1^2 - 2 \sqrt{2} i} }{2}$

$\displaystyle \Rightarrow \alpha = \frac{(\sqrt{2}+i) \pm \sqrt{(\sqrt{2}-i)^2} }{2}$

$\displaystyle \Rightarrow \alpha = \frac{(\sqrt{2}+i) \pm (\sqrt{2}-i)}{2}$

$\displaystyle \Rightarrow \alpha = \sqrt{2} \text{ or } i$

$\displaystyle \text{xii) } 2x^2 - ( 3 + 7i) x + ( 9i-3) = 0$

$\displaystyle \Rightarrow 2x^2 - 3x - 7x i + 9 i - 3 = 0$

$\displaystyle \Rightarrow 2x^2 - 3 x - xi - 6xi + 9i - 3 = 0$

$\displaystyle \Rightarrow x ( 2x - 3 - i) - 3i ( 2x - 3 - i) = 0$

$\displaystyle \Rightarrow (2x-3-i)(x - 3i) = 0$

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

ICSE / ISC / CBSE Mathematics Portal for K12 Students

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Class 11: Quadratic Equations – Exercise 14.2

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