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

Last Updated : 13 Jan, 2021

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 – (2âˆš3 + 3i)x + 6âˆš3i = 0

Solution:

x2 – 2âˆš3x – 3ix + 6âˆš3i = 0

x(x – 2âˆš3) – 3i(x – 2âˆš3) = 0

x = 3i, 2âˆš3

iv) 6x2 – 17ix + 12i2 = 0

Solution:

6x2 – 9ix – 8ix + 12i2 = 0

3x(2x – 3i) – 4i(2x – 3i) = 0

x = 4/3i, 3/2i

i) x2 – (3âˆš2 + 2i)x + 6âˆš2i = 0

Solution:

x2 – 3âˆš2x – 2ix + âˆš2i = 0

x(x – 2i)(x – 3âˆš2) = 0

(x – 2i)(x – 3âˆš2) = 0

x = 2i, 3âˆš2

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 + âˆš15ix – i = 0

Solution:

Here we use general form of quadratic equation

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

a = 2, b = âˆš15i, c = -i; substitute these value in formula

r1 = (-b + âˆšb2-4ac)/2a; r2 = (-b – âˆšb2-4ac)/2a

r1 = (- âˆš15i + âˆš-15 + 8i) / 4; r2 = (-âˆš15i – âˆš-15 + 8i) / 4

let âˆš-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, âˆš-15 + 8i = 1 + 4i, -1 – 4i

when âˆš-15 + 8i = 1 + 4i

r1 = (-âˆš15i + 1 + 4i) / 4 = 1 + (4 – âˆš15) i / 4

r2 = ( -âˆš15i -(1 + 4i) ) / 4 = -1 – (4 + âˆš15)i / 4

when âˆš-15 + 8i = -1 – 4i

r1 = (-âˆš15i – 1 – 4i) / 4 = -1 – (4 + âˆš15)i / 4

r2 = (-âˆš15i – (-1 – 4i) ) / 4 = 1 + (4 – âˆš15)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 Â±âˆšb2– 4ac) / 2a

x = -(-1) Â± (âˆš(-1)2 – 4i * 12i) / 2i

=1 Â±âˆš1 + 48/ 2i

= 1 Â± âˆš49 / 2i

=1 Â± 7 / 2i

= 8 / 2i, -6 / 2i

= 4 / i, -3 / i

= -4i, 3i

x) x2 – (3âˆš2 – 2i)x – âˆš2i = 0

Solution:

Applying discriminate rule

x = (-b + âˆšb2– 4ac) / 2a

x = ((3âˆš2 – 2i) Â± âˆš[-(3âˆš2 – 2i)]2 – 4 (-âˆš2i)) / 2

= ((3âˆš2 – 2i) Â± âˆš(3âˆš2 – 2i)2 + 4âˆš2i) / 2

= (3âˆš2 – 2i) / 2  Â± (4âˆš2i) / 2

xi) x2 – (âˆš2 + i)x + âˆš2i = 0

Solution:

x2 – âˆš2x – ix + âˆš2i = 0

x(x – âˆš2) – i(x – âˆš2) = 0

(x – i)(x – âˆš2) = 0

x = i, âˆš2

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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