Class 11 RD Sharma Solutions- Chapter 14 Quadratic Equations – Exercise 14.2

Question 1. Solve the following Quadratic Equations:

i) x² + 10ix – 21 = 0

Solution:

x² + 10ix – 21 = 0            

 x² + 7ix + 3ix – 21 = 0      

 x(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) x² + (1 – 2x)x – 2i = 0

Solution:

 x² + x – 2ix – 2i = 0

 x(x + 1) – 2i(x + 1) = 0

 (x + 1)(x – 2i) = 0

 x = 2i, -1

iii) x² – (2√3 + 3i)x + 6√3i = 0

Solution:

x² – 2√3x – 3ix + 6√3i = 0

x(x – 2√3) – 3i(x – 2√3) = 0

x = 3i, 2√3

iv) 6x² – 17ix + 12i² = 0

Solution:

6x² – 9ix – 8ix + 12i² = 0

3x(2x – 3i) – 4i(2x – 3i) = 0

x = 4/3i, 3/2i

Question 2. Solve the following quadratic equations:

i) x² – (3√2 + 2i)x + 6√2i = 0

Solution:

x² – 3√2x – 2ix + √2i = 0

x(x – 2i)(x – 3√2) = 0

(x – 2i)(x – 3√2) = 0

x = 2i, 3√2

ii) x² – (5 – i)x + (18 + i) = 0

Solution:

x² – 5x – ix +18 + i = 0

x² – (3 – 4i)x – (2 + 3i)x + (18 + i) = 0

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

(x – (2 + 3i)) (x – (3 – 4i)) = 0 

x = 2 + 3i,  3 – 4i

iii) (2 + i)x² – (5 – i)x + 2(1 – i) = 0

Solution:

(2 + i)x² – 2x – (3 – i)x + 2(1 – i) = 0

x[(2 + i)x – 2] – (1 – i)[(2 + i)x – 2] = 0                  (we have, i² = -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) x² – (2 + i)x – (1 – 7i) = 0

Solution:

x² – (2 + i)x – (1 – 7i) = 0

x² – (3 – i)x + (1 – 2i)x – (1 – 7i) = 0

x(x – (3 – i)) + (1 – 2i)(x – (3 – i)) = 0

[x + (1 – 2i)] [x – (3 – i)] = 0

x = -1 + 2i, 3 – i

v) ix² – 4x – 4i

Solution:

ix² + 4i² + 4i³ = 0         [ i² = -1;   (4i³) = 4i(i²) = 4i(-1) = -4i ]

i(x^2 + 2ix + 2ix + 4i²) = 0                       

x^2 + 2ix + 2ix + 4i² = 0

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

x = -2i, -2i

vi) x² + 4ix – 4 = 0

Solution:

x² + 4ix + 4i² = 0    [i² = -1]

x² + 2ix + 2ix + 4i² = 0

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

x = -2i, -2i

vii) 2x² + √15ix – i = 0

Solution:

Here we use general form of quadratic equation

Comparing with general form of quadratic equation ax² + bx + c

a = 2, b = √15i, c = -i; substitute these value in formula

r1 = (-b + √b²-4ac)/2a; r2 = (-b – √b²-4ac)/2a

r1 = (- √15i + √-15 + 8i) / 4; r2 = (-√15i – √-15 + 8i) / 4

let √-15 + 8i = a + bi

=> -15 + 8i = (a + bi)²

=> -15 + 8i = a² – b² + 2abi

=> a² – b² = – 15 and 2abi = 8i

we have, (a² + b²)² = ((a² – b²)²) + 4 * a² * b²

               (a² + b²)² = (-15)² + 64 = 289

                a² + b² = 17

Solving  a² – b² = -15 and   a² + b² = 17 we get

a² = 1 and b² = 16

a = 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) x² – x + (1 + i) = 0

Solution:

x² – x + (1 + i) = 0

x² – ix – (1 – i)x + i(1 – i) = 0

(x – i)(x – (1 – i)) = 0

x = i,1 – i

ix) ix² – x + 12i = 0

Solution:

Applying discriminate rule

x = (-b ±√b²– 4ac) / 2a

x = -(-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) x² – (3√2 – 2i)x – √2i = 0

Solution:

Applying discriminate rule

x = (-b + √b²– 4ac) / 2a

x = ((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) x² – (√2 + i)x + √2i = 0

Solution:

x² – √2x – ix + √2i = 0

x(x – √2) – i(x – √2) = 0

(x – i)(x – √2) = 0

x = i, √2

xii) 2x² – (3 + 7i)x + (9i – 3) = 0

Solution:

2x² – 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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