Question

Find a number(x) whose cube is equal to the square of a number (y) such that y=x+4

Answer 1

Answer:

x^3 = y^2

Also, y = x + 4

Put this value in the eqn:

x^3 = (x + 4)^2

=> x^3 = x^2 + 16 + 8x

=> x^3 - x^2 - 8x - 16 = 0

Let x be = 4:

=> 64 - 16 - 32 - 16

=> 0

Thus, (x - 4) is a factor.

On dividing (x - 4) by x^3 - x^2 - 8x - 16 we get:

x^2 + 3x + 4

roots of this eqn are not real.

Thus, there is only one real value of x, that it 4

Number is 4.

y = x^3/2

y = 8

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