# Find real values of x and y for which(x4 + 2xi) – (3x2 + iy) = (3 – 5i) + (1 + 2iy).

We have, (x4 + 2xi) – (3x2 + iy) = (3 – 5i) + (1 + 2iy).

x4 + 2xi - 3x2 + iy = 3 – 5i + 1 + 2iy

(x4 - 3x2) + i(2x - y) = 4 + i(2y - 5)

On equating real and imaginary parts, we get

x4 - 3x2 = 4 and 2x - y = 2y - 5

x4 - 3x2 - 4 = 0 eq(i) and 2x - y - 2y + 5 = 0 eq(ii)

Now from eq (i), x4 - 3x2 - 4 = 0

x4 - 4x2 + x2 - 4 = 0

x2 (x2 - 4) + 1(x2 - 4) = 0

(x2 - 4)(x2 + 1) = 0

x2 - 4 = 0 and x2 + 1 = 0

x = ±2 and x = √ - 1

Real value of x = ±2

Putting x = 2 in eq (ii), we get

2x - 3y + 5 = 0

2×2 - 3y + 5 = 0

4 - 3y + 5 = 0 = 9 - 3y = 0

y = 3

Putting x = - 2 in eq (ii), we get

2x - 3y + 5 = 0

2× - 2 - 3y + 5 = 0

- 4 - 3y + 5 = 0 = 1 - 3y = 0 Rate this question :

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Evaluate:

(i) (iii) .

RS Aggarwal - Mathematics