Question

Solve the equation by completing square method.

$x^2-20x=21$

Short answer

The solutions are$x=-1,21$.

Step-by-step solution

Given Information.   

The Given equation is $x^2-20x=21$.

Isolate the variable terms on one side and the constant terms on the other. 

The variable terms are on the left side.

$x^2-20x=21$

Here$b=-20$.

Find 12b2 and add to the both sides of equation. 

$$\begin{gathered} ={\left(\frac{1}{2}(-20)\right)}^2 \\ ={\left(-10\right)}^2 \\ =100 \end{gathered}$$

Add $100$both sides.

$$\begin{gathered} x^2-20x+100=21+100 \\ {x}^2-20x+100=121 \end{gathered}$$

Factor the perfect square trinomial as a binomial square. 

$(x-10{)}^2=121$

Use the square root property.

$$\begin{gathered} x-10=\pm \sqrt{121} \\ x-10=\pm 11 \\ x=10\pm 11 \end{gathered}$$

Rewrite to show two solutions.

$$\begin{gathered} x=10+11=21 \\ x=10-11=-1 \\ x=-1,21 \end{gathered}$$

Check the solutions. 

Substitute $x=-1$

$$\begin{gathered} x^2-20x=21 \\ (-1{)}^2-20(-1)\stackrel{?}{=}21 \\ 1+20\stackrel{?}{=}21 \\ 21=21True \end{gathered}$$

Substitute $x=21$

$$\begin{gathered} x^2-20x=21 \\ (21{)}^2-20(21)\stackrel{?}{=}21 \\ 441-420\stackrel{?}{=}21 \\ 21=21True \end{gathered}$$

Thus, the solutions are$x=-1,21$.