Q. 8.116 - Intermediate Algebra

Question

Solve: $\sqrt{y-5}+5=y$.

Step-by-Step Solution

Answer

The solutions are, $y=6$ and $y=5$.

Step 1 Given equation is,

$\sqrt{y-5}+5=y$.

The product of Binomial squares pattern is given by,

${\left(a-b\right)}^2=a^2-2ab+b^2$.

The objective is, we need to find the solution for the given equation.

To isolate the radical, subtract $5$ on both sides.

$$\begin{gathered} \sqrt{y-5}+5-5=y-5 \\ \sqrt{y-5}=y-5 \end{gathered}$$

Step 2 Now square on both sides.

$y-5={\left(y-5\right)}^2$

Now use the product of binomial square pattern formula on the right side of the equation.

$$\begin{gathered} y-5=y^2-2\left(y\right)\left(5\right)+{\left(5\right)}^2 \\ y-5=y^2-10y+25 \end{gathered}$$

Simplify the obtained equation.

$$\begin{gathered} y^2-10y+25-y+5=0 \\ {y}^2-11y+30=0 \end{gathered}$$

Now factorize it.

$\left(y-6\right)\left(y-5\right)=0$

Now use the zero product property.

$$\begin{gathered} y-6=0 \\ y=6 \\ y-5=0 \\ y=5 \end{gathered}$$

Step 3 Let us check the answer.

Substitute $y=6$ into the equation $\sqrt{y-5}+5=y$.

$$\begin{gathered} \sqrt{6-5}+5=6 \\ \sqrt{1}+5=6 \\ 1+5=6 \\ 6=6 \end{gathered}$$

which is true.

Substitute $y=5$ into the equation $\sqrt{y-5}+5=y$.

$$\begin{gathered} \sqrt{5-5}+5=5 \\ \sqrt{0}+5=5 \\ 5=5 \end{gathered}$$

which is true.