Chapter 9, Problem 20MCQ

To determine

Solve the equation by completing the square method.

Answer to Problem 20MCQ

The solutions are: $-2.3\text{and}4.3$

Explanation of Solution

Given:

The equation: $x^2-2x-10=0$

Concept Used:

The perfect square term: ${(x\pm \frac{b}{2})}^2=x^2\pm 2·x·\frac{b}{2}+{(\frac{b}{2})}^2$

Calculation:

The perfect square term: ${(x\pm \frac{b}{2})}^2=x^2\pm 2·x·\frac{b}{2}+{(\frac{b}{2})}^2$

The equation: $x^2-2x-10=0$

  $\begin{array}{l}{x}^2-2x-10=0[\text{original}\text{equation}]\\ x^2-2x-10+1=0+1[\text{Since}{(}^{\frac{2}{2}}=1\text{add}\text{to}\text{each}\text{side}]\\ x^2-2x+1-10+10=0+1+10[\text{add}\text{10}\text{to}\text{each}\text{side}]\\ $ {(x-1)}^2=11$\sqrt{{(x-1)}^2}=\pm \sqrt{11}[\text{Take}\text{the}\text{square}\text{root}\text{of}\text{each}\text{side}]\\ x-1=\pm \sqrt{11}\\ x-1+1=1\pm \sqrt{11}[\text{add}\text{11}\text{to}\text{both}\text{sides}]\\ x=1-\sqrt{11}\text{or}x=1+\sqrt{11}[\text{Seperate}\text{the}\text{solution}]\\ x=-2.31662\text{or}x=4.31662\\ x\approx -2.3\text{o}rx=4.3\end{array}$

The solutions are: $-2.3\text{and}4.3$

Thus, the solutions are: $-2.3\text{and}4.3$