004 Sample Final A, Problem 16

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

Solution:

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