How do you solve for x in $\sqrt{42 - x} + x = 13$ $sqrt(42-x)+x =13$ ?

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Answer 1 · 2015-02-07T09:03:58

Okay start by

Rewriting equation as $\sqrt{42 - x} = 13 - x$ $sqrt(42- x ) = 13 - x$

Now square both sides

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

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

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

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