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

i) x^2 + 10 i x - 21 = 0      ii) x^2 + ( 1 - 2i) x - 2i = 0

iii) x^2 - ( 2\sqrt{3} + 3i) x + 6 \sqrt{3}i = 0      iv) 6x^2 - 17i x - 12 = 0

Answer:

i)      x^2 + 10 i x - 21 = 0

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

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

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

\Rightarrow x = - 7i  \text{ or } -3i

So the roots of given quadratic equation are -7i and -3i

ii)     x^2 + ( 1 - 2i) x - 2i = 0

\Rightarrow x^2 + x - 2i x - 2i  = 0

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

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

\Rightarrow x = - 1  \text{ or } 2i

So the roots of given quadratic equation are -1 and 2i

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

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

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

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

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

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

iv)    6x^2 - 17i x - 12 = 0

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

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

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

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

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

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Question 2: Solve the following quadratic equations:

i) x^2 - ( 3 \sqrt{2} + 2i)x + 6\sqrt{2} i = 0       ii) x^2 - ( 5-i) x + ( 18+i) = 0

iii) (2+i) x^2 - ( 5-i)x + 2( 1 - i) = 0      iv) x^2 - ( 2+i) x - ( 1 - 7i) = 0

v) ix^2 - 4x - 4i = 0       vi) x^2 + 4ix - 4 = 0      vii) 2x^2 + \sqrt{15} ix - i = 0

viii) x^2 - x + ( 1 + i) = 0      ix) ix^2 - x + 12 i = 0

x) x^2 - ( 3\sqrt{2}- 2i) x - \sqrt{2} i = 0       xi) x^2 - ( \sqrt{2} + i)x + \sqrt{2} i = 0

xii) 2x^2 - ( 3 + 7i) x + ( 9i-3) = 0

Answer:

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

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

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

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

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

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

ii)      x^2 - ( 5-i) x + ( 18+i) = 0

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

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

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

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

So the roots of given quadratic equation are 2+3i and ( 3 - 4i)

iii)     (2+i) x^2 - ( 5-i)x + 2( 1 - i) = 0

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

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

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

iv)     x^2 - ( 2+i) x - ( 1 - 7i) = 0

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

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

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

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

v)      ix^2 - 4x - 4i = 0 

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

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

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

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

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

\Rightarrow x = - 2i

vi)     x^2 + 4ix - 4 = 0

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

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

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

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

\Rightarrow x = - 2i

vii)   2x^2 + \sqrt{15} ix - i = 0

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

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

Substituting these values in

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

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

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

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

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

Squaring both sides

x^2 - y^2 + 2 xy i = 8i - 15

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

and 2xy = 8    … … … … … ii)

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

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

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

Adding i) and iii) we get

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

Substituting in i) we get

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

Since xy is positive, x and y will have the same sign.

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

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

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

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

viii)   x^2 - x + ( 1 + i) = 0

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

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

Substituting these values in

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

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

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

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

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

Squaring both sides

x^2 - y^2 + 2 xy i = -3-4i

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

and 2xy = -4    … … … … … ii)

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

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

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

Adding i) and iii) we get

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

Substituting in i) we get

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

Since xy is negative, x and y will have opposite sign.

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

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

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

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

ix)     ix^2 - x + 12 i = 0

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

\Rightarrow x^2 + ix + 12 = 0

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

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

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

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

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

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

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

Substituting these values in

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

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

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

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

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

Squaring both sides

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

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

and 2xy = -8\sqrt{2}    … … … … … ii)

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

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

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

Adding i) and iii) we get

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

Substituting in i) we get

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

Since xy is negative, x and y will have opposite sign.

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

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

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

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

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

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

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

Substituting these values in

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

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

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

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

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

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

xii)    2x^2 - ( 3 + 7i) x + ( 9i-3) = 0

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

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

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

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

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

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