Question

If x+y =12 & xy =27 find the value of x^3+y^2

Answer 1

Hey friend,

Here is the solution :-

$x + y = 12 \: \: \: \: \: \: \: xy = 27 \\ \\ from \: \: \: x + y = 12 \\ we \: get \: x = 12 - y \\ so \: xy = (12 - y)y = 27 \\ (12 - y)y = 27 \\ 12y - {y}^{2} = 27 \\ { - y}^{2} + 12y = 27 \\ {y}^{2} - 12y = - 27 \\ \\ now \: by \: completing \: square \: method \\ \\ {y}^{2} - 2 \times 6y + {6}^{2} = - 27 + {6}^{2} \\ \\ (y - 6)^{2} = - 27 + 36 \\ \\ {(y - 6)}^{2} = 9 \\ {(y - 6)} = \sqrt{9} \\ (y - 6) = + - 3 \\ \\ y = 3 + 6 = 9 \\ \\ y = - 3 + 6 = 3 \\ \\ when \: we \: take \: y = 9 \\ x + 9 = 12 \\ x = 12 - 9 \\ x = 3 \\ \\ now \: we \: have \: x = 3 \: and \: y = 9 \\ so \: \: \: {x}^{3} + {y}^{2} = {3}^{3} + {9}^{2} = 9 + 18 = 27$

I hope this will help you....