Q. 64.0( 4 Votes )

Find real values of x and y for which

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

Answer :

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



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