### Home > CCA2 > Chapter 4 > Lesson 4.2.3 > Problem4-95

4-95.

Solve the equations below.

1. $\sqrt { x + 15 } = 5 + \sqrt { x }$

Square both sides.

$\left(\sqrt{x+15}\right)^2=\left(5+\sqrt{x}\right)^2$

$x+15=\left(5+\sqrt{x}\right)\left(5+\sqrt{x}\right)$

$x+15=25+10\sqrt{x}+x$

Isolate the square root of $x$ on one side of the equation.

$-10=10\sqrt{x}$

Divide both sides by $10$.

$-1=\sqrt{x}$

What can you take the square root of and get $−1$?

No real solutions.

1. $(y−6)^2+10=3y$

Expand the $(y−6)^2$.

Rearrange the equation so that it equals zero.

Solve by factoring and using the Zero Product Property, or use the Quadratic Formula.