How do you solve for x in #sqrt(42-x)+x =13#?

1 Answer
Feb 7, 2015

Okay start by

Rewriting equation as #sqrt(42- x ) = 13 - x#

Now square both sides

So we are left with the equation # 42 - x = (13 - x)^2 = x^2− 26x +169 #

Next step bring all the terms to one side of the equation
SO # x^2 - 26 x + 169- 42 + x = 0 #

So using the quadratic formula i get that The values of x are = #x=17.908326913195985,7.091673086804016#