## Intermediate Algebra for College Students (7th Edition)

$x=6$
Given: $\sqrt {2x-3}=x-3$ Take square on both sides. we have, $(\sqrt {2x-3})^2=(x-3)^2$ or, $2x-3=x^2-6x+9$ or, $x^2-8x+12=0$ Factoization as follows: $(x-6)(x-2)=0$ or, $x=6,2$ Plug $x=6$ in the expression $\sqrt {2x-3}=x-3 \implies \sqrt {2(6)-3}=6-3 \implies 3 =3$, we get the value of $x=6$ satisfies the expression.