# How do you use the pythagorean theorem to solve a=x-5 b=x 2 c=13?

##### 1 Answer
Apr 22, 2016

So acceptable value of x is 10

#### Explanation:

Here sides are
$a = x - 5$
$b = x + 2$
$c = 13$
Taking c as larger side i.e. hypotenuse
we can apply Pythagorean theorem

${a}^{2} + {b}^{2} = {c}^{2}$
$\implies {\left(x - 5\right)}^{2} + {\left(x + 2\right)}^{2} = {13}^{2}$
$\implies \left({x}^{2} - 10 x + 25 + {x}^{2} + 4 x + 4\right) = 169$
$\implies \left(2 {x}^{2} - 6 x - 140\right) = 0$
$\implies {x}^{2} - 3 x - 70 = 0$
$\implies {x}^{2} - 10 x + 7 x - 70 = 0$
$\implies x \left(x - 10\right) + 7 \left(x - 10\right) = 0$
$\implies \left(x - 10\right) \left(x + 7\right) = 0$
So acceptable value of x is 10
otherwise a and b will be negative for x = -7