Question

Find the value of x.

Answer 1

$x^2 + (x + 3)^2 = (\sqrt{117})^2$

$x^2 + x^2 + 6x + 9 = 117$

$2x^2 + 6x -108 = 0$

$x^2 + 3x - 54 = 0$

$(x + 9)(x - 6) = 0$

$x + 9 = 0 ~~or~~ x - 6 = 0$

$x = -9 ~~ or~~ x = 6$

Since the side of a triangle cannot have a negative length, we discard the solution x = -9.

Answer: x = 6

Answer 2

${x}^{2} + {(x + 3)}^{2} = {( \sqrt{117} )}^{2} \\ 2 {x}^{2} + 6x + 9 = 117 \\ 2 {x}^{2} + 6x - 108 = 0\\ {x }^{2} + 3x - 54 = 0 \\ (x + 9)(x - 6) = 0 \\ x = - 9 \: or \: 6 \\ but \: x > 0 \\ so \: the \: answer \: is \: x = 6$
hope this would help you