# 004 Sample Final A, Problem 16

Solve. ${\displaystyle {\sqrt {x-3}}+5=x}$

Foundations
1) How do you solve for ${\displaystyle x}$ in the equation ${\displaystyle {\sqrt {x}}=5}$?
2) How do you find the zeros of ${\displaystyle f(x)=x^{2}+x-6}$?
1) You square both sides of the equation to get ${\displaystyle x=25}$.
2) You factor ${\displaystyle f(x)=0}$ to get ${\displaystyle (x+3)(x-2)=0}$. From here, we solve to get ${\displaystyle x=-3}$ or ${\displaystyle x=2}$.

Solution:

Step 1:
First, we get the square root by itself. Subtracting 5 from both sides, we get ${\displaystyle {\sqrt {x-3}}=x-5}$.
Step 2:
Now, to get rid of the square root, we square both sides of the equation.
So, we get ${\displaystyle x-3=(x-5)^{2}}$.
Step 3:
We multiply out the right hand side to get ${\displaystyle x-3=x^{2}-10x+25}$.
Step 4:
Getting all the terms on one side, we have ${\displaystyle 0=x^{2}-11x+28}$.
To solve, we can factor to get ${\displaystyle 0=(x-7)(x-4)}$.
Step 5:
The two possible solutions are ${\displaystyle x=7}$ and ${\displaystyle x=4}$.
But, plugging in ${\displaystyle x=4}$ into the problem, gives us ${\displaystyle 6={\sqrt {4-3}}+5=4}$, which is not true.
Thus, the only solution is ${\displaystyle x=7}$.
${\displaystyle x=7}$