Answer
$\{4\}$.
Work Step by Step
The given function is
$f(x)=x+\sqrt{x+5}$
Replace $f(x)$ with $7$.
$\Rightarrow 7=x+\sqrt{x+5}$
Subtract $x$ from both sides.
$\Rightarrow 7-x=x+\sqrt{x+5}-x$
Simplify.
$\Rightarrow 7-x=\sqrt{x+5}$
Square both sides.
$\Rightarrow (7-x)^2=(\sqrt{x+5})^2$
Use the special formula $(A-B)^2=A^2-2AB+B^2$
We have $A=7$ and $B=x$
$\Rightarrow (7)^2-2(7)( x)+(x)^2=x+5$
Simplify.
$\Rightarrow 49-14x+x^2=x+5$
Add $-x-5$ to both sides.
$\Rightarrow 49-14x+x^2-x-5=x+5-x-5$
Add like terms.
$\Rightarrow x^2-15x+44=0$
Rewrite the middle term $-15x$ as $-11x-4x$.
$\Rightarrow x^2-11x-4x+44=0$
Group terms.
$\Rightarrow (x^2-11x)+(-4x+44)=0$
Factor each group.
$\Rightarrow x(x-11)-4(x-11)=0$
Factor out $(x-11)$.
$\Rightarrow (x-11)(x-4)=0$
Set each factor equal to zero.
$\Rightarrow x-11=0$ or $ x-4=0$
Isolate $x$.
$\Rightarrow x=11$ or $ x=4$
Check $x=11$.
$f(11)=11+\sqrt{11+5}$
$f(11)=11+\sqrt{16}$
$f(11)=11+4$
$f(11)=15$ false.
Check $x=4$.
$f(4)=4+\sqrt{4+5}$
$f(4)=4+\sqrt{9}$
$f(4)=4+3$
$f(4)=7$ false.
The solution set is $\{4\}$.
