Solve. ${\sqrt {x3}}+5=x$
Foundations

1) How do you solve for $x$ in the equation ${\sqrt {x}}=5$?

2) How do you find the zeros of $f(x)=x^{2}+x6$?

Answer:

1) You square both sides of the equation to get $x=25$.

2) You factor $f(x)=0$ to get $(x+3)(x2)=0$. 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 ${\sqrt {x3}}=x5$.


Step 2:

Now, to get rid of the square root, we square both sides of the equation.

So, we get $x3=(x5)^{2}$.

Step 3:

We multiply out the right hand side to get $x3=x^{2}10x+25$.

Step 4:

Getting all the terms on one side, we have $0=x^{2}11x+28$.

To solve, we can factor to get $0=(x7)(x4)$.

Step 5:

The two possible solutions are $x=7$ and $x=4$.

But, plugging in $x=4$ into the problem, gives us $6={\sqrt {43}}+5=4$, which is not true.

Thus, the only solution is $x=7$.

Final Answer:

$x=7$

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