Q. 8.115 - Intermediate Algebra

Question

Solve: $\sqrt{x - 2 }+ 2 = x.$

Step-by-Step Solution

Answer

The solutions are,$x=3$ and $x=2$.

Step 1 Given equation is,

$\sqrt{x - 2 }+ 2 = x.$

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 $2$ on both sides.

$$\begin{gathered} \sqrt{x - 2 }+ 2-2= x-2 \\ \sqrt{x - 2 }=x-2 \end{gathered}$$

Step 2 Now square on both sides.

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

Now use the product of binomial squares pattern on the right side.

$$\begin{gathered} x-2=x^2-2\left(x\right)\left(2\right)+{\left(2\right)}^2 \\ x-2=x^2-4x+4 \end{gathered}$$

Now simplify the obtained equation.

$$\begin{gathered} x-2=x^2-4x+4 \\ {x}^2-4x+4-x+2=0 \\ {x}^2-5x+6=0 \end{gathered}$$

Now factorize the obtained equation.

$\left(x-3\right)\left(x-2\right)=0$

Now use the zero product property.

$$\begin{gathered} x-3=0 \\ x=3 \\ x-2=0 \\ x=2 \end{gathered}$$

Step 3 Let us check the answer.

Substitute $x=3$ into the equation.

$$\begin{gathered} \sqrt{3 - 2 }+ 2 = 3 \\ \sqrt{1}+2=3 \\ 1+2=3 \\ 3=3 \end{gathered}$$

which is true.

Substitute $x=2$ into the equation $\sqrt{x - 2 }+ 2 = x$.

$$\begin{gathered} \sqrt{2 - 2 }+ 2 = 2 \\ \sqrt{0}+2=2 \\ 0+2=2 \\ 2=2 \end{gathered}$$

which is true.