**Question 1. Solve the following Quadratic Equations:**

**i) x**^{2 }+ 10ix – 21 = 0

^{2 }+ 10ix – 21 = 0

**Solution:**

x

^{2 }+ 10ix – 21 = 0x

^{2 }+ 7ix + 3ix – 21 = 0x(x + 7i) + 3i(x + 7i) = 0

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

x + 7i = 0 ; x + 3i = 0;

x = -7i, -3i

Hence, the roots are -7i, -3i

**ii)**** ****x**^{2 }+ (1 – 2x)x – 2i = 0

^{2 }+ (1 – 2x)x – 2i = 0

**Solution:**

x

^{2 }+ x – 2ix – 2i = 0x(x + 1) – 2i(x + 1) = 0

(x + 1)(x – 2i) = 0

x = 2i, -1

**iii) x**^{2 }– (2√3 + 3i)x + 6√3i = 0

^{2 }– (2√3 + 3i)x + 6√3i = 0

**Solution:**

x

^{2 }– 2√3x – 3ix + 6√3i = 0x(x – 2√3) – 3i(x – 2√3) = 0

x = 3i, 2√3

**iv) 6x**^{2 }– 17ix + 12i^{2 }= 0

^{2 }– 17ix + 12i

^{2 }= 0

**Solution:**

6x

^{2 }– 9ix – 8ix + 12i^{2 }= 03x(2x – 3i) – 4i(2x – 3i) = 0

x = 4/3i, 3/2i

**Question 2. Solve the following quadratic equations:**

**i) x**^{2 }– (3√2 + 2i)x + 6√2i = 0

^{2 }– (3√2 + 2i)x + 6√2i = 0

**Solution:**

x

^{2 }– 3√2x – 2ix + √2i = 0x(x – 2i)(x – 3√2) = 0

(x – 2i)(x – 3√2) = 0

x = 2i, 3√2

**ii) x**^{2 }– (5 – i)x + (18 + i) = 0

^{2 }– (5 – i)x + (18 + i) = 0

**Solution:**

x

^{2 }– 5x – ix +18 + i = 0x

^{2 }– (3 – 4i)x – (2 + 3i)x + (18 + i) = 0x(x – (3 – 4i)) – (2 + 3i)x + (18 + i) = 0

(x – (2 + 3i)) (x – (3 – 4i)) = 0

x = 2 + 3i, 3 – 4i

**iii) (2 + i)x**^{2 }– (5 – i)x + 2(1 – i) = 0

^{2 }– (5 – i)x + 2(1 – i) = 0

**Solution:**

(2 + i)x

^{2 }– 2x – (3 – i)x + 2(1 – i) = 0x[(2 + i)x – 2] – (1 – i)[(2 + i)x – 2] = 0 (we have,

i^{2 }= -1)[x – (1 – i)] = 0 or [(2 + i)x- 2] = 0

x = 1 – i or x = 2 / 2 + i

further we can extend x = 2 / 2 + i

x = 2 / 2 + i * 2 – i / 2 – i

x = 4 – 2i / 4 + 1

x = 4 / 5 – 2 / 5i

**iv) x**^{2 }– (2 + i)x – (1 – 7i) = 0

^{2 }– (2 + i)x – (1 – 7i) = 0

**Solution:**

x

^{2 }– (2 + i)x – (1 – 7i) = 0x

^{2 }– (3 – i)x + (1 – 2i)x – (1 – 7i) = 0x(x – (3 – i)) + (1 – 2i)(x – (3 – i)) = 0

[x + (1 – 2i)] [x – (3 – i)] = 0

x = -1 + 2i, 3 – i

**v) ix**^{2 }– 4x – 4i

^{2 }– 4x – 4i

**Solution:**

ix

^{2 }+ 4i^{2 }+ 4i^{3 }= 0 [ i^{2}= -1; (4i^{3}) = 4i(i^{2}) = 4i(-1) = -4i ]i(x^2 + 2ix + 2ix + 4i

^{2}) = 0x^2 + 2ix + 2ix + 4i

^{2 }= 0x(x + 2i) + 2i(x + 2i) = 0

x = -2i, -2i

**vi) x**^{2 }+ 4ix – 4 = 0

^{2 }+ 4ix – 4 = 0

**Solution:**

x

^{2 }+ 4ix + 4i^{2 }= 0[i^{2 }= -1]x

^{2 }+ 2ix + 2ix + 4i^{2 }= 0x(x + 2i) + 2i(x + 2i) = 0

x = -2i, -2i

**vii) 2x**^{2 }+ √15ix – i = 0

^{2 }+ √15ix – i = 0

**Solution:**

Here we use general form of quadratic equation

Comparing with general form of quadratic equation ax

^{2}+ bx + ca = 2, b = √15i, c = -i; substitute these value in formula

r1 = (-b + √b^{2}-4ac)/2a; r2 = (-b – √b^{2}-4ac)/2ar1 = (- √15i + √-15 + 8i) / 4; r2 = (-√15i – √-15 + 8i) / 4

let √-15 + 8i = a + bi

=> -15 + 8i = (a + bi)

^{2}=> -15 + 8i = a

^{2 }– b^{2 }+ 2abi=> a

^{2 }– b^{2 }= – 15 and 2abi = 8iwe have, (a

^{2 }+ b^{2})^{2 }= ((a^{2 }– b^{2})^{2}) + 4 * a^{2 }* b^{2}(a

^{2 }+ b^{2})^{2 }= (-15)^{2 }+ 64 = 289a

^{2 }+ b^{2}= 17Solving a

^{2 }– b^{2 }= -15 and a^{2 }+ b^{2}= 17 we geta

^{2 }= 1 and b^{2 }= 16a = 1,b = 4 or a = -1,b = -4

so, √-15 + 8i = 1 + 4i, -1 – 4i

when √-15 + 8i = 1 + 4i

r1 = (-√15i + 1 + 4i) / 4 = 1 + (4 – √15) i / 4

r2 = ( -√15i -(1 + 4i) ) / 4 = -1 – (4 + √15)i / 4

when √-15 + 8i = -1 – 4i

r1 = (-√15i – 1 – 4i) / 4 = -1 – (4 + √15)i / 4

r2 = (-√15i – (-1 – 4i) ) / 4 = 1 + (4 – √15)i / 4

**viii) x**^{2 }– x + (1 + i) = 0

^{2 }– x + (1 + i) = 0

**Solution:**

x

^{2 }– x + (1 + i) = 0x

^{2 }– ix – (1 – i)x + i(1 – i) = 0(x – i)(x – (1 – i)) = 0

x = i,1 – i

**ix) ix**^{2 }– x + 12i = 0

^{2 }– x + 12i = 0

**Solution:**

Applying discriminate rule

x = (-b ±√b^{2}– 4ac) / 2ax = -(-1)

±(√(-1)^{2 }– 4i * 12i) / 2i=1

±√1 + 48/ 2i= 1

±√49 / 2i=1

±7 / 2i= 8 / 2i, -6 / 2i

= 4 / i, -3 / i

= -4i, 3i

**x) x**^{2 }– (3√2 – 2i)x – √2i = 0

^{2 }– (3√2 – 2i)x – √2i = 0

**Solution:**

Applying discriminate rule

x = (-b + √b^{2}– 4ac) / 2ax = ((3√2 – 2i)

±√[-(3√2 – 2i)]^{2 }– 4 (-√2i)) / 2= ((3√2 – 2i)

±√(3√2 – 2i)^{2}+ 4√2i) / 2= (3√2 – 2i) / 2

±(4√2i) / 2

**xi) x**^{2 }– (√2 + i)x + √2i = 0

^{2 }– (√2 + i)x + √2i = 0

**Solution:**

x

^{2 }– √2x – ix + √2i = 0x(x – √2) – i(x – √2) = 0

(x – i)(x – √2) = 0

x = i, √2

**xii) 2x**^{2 }– (3 + 7i)x + (9i – 3) = 0

^{2 }– (3 + 7i)x + (9i – 3) = 0

**Solution:**

2x

^{2 }– 3x – 7ix + (9i – 3) = 0(2x – 3 – i)(x – 3i) = 0

(x – 3 + i / 2) (x – 3i) = 0

x = 3 + i/2, 3i

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