Question

Solve the equation by completing the square. Round to the nearest tenth.

$x^2-2x-10=0$

Short answer

$\mathit{x}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.3}\mathbf{,}\mathbf{4}\mathbf{.4}$are the solutions of the given equation.

Step-by-step solution

Step 1. Define the standard form of the quadratic equation.

A quadratic equation, which is written in the form, $a{x}^2+bx+c=0$, where,$a\ne 0$ is called the standard form of the quadratic equation.

Step 2. Define the perfect square a−b2.

${\left(a-b\right)}^2=a^2-2ab+b^2\mathrm{...........}\left(1\right)$

Step 3. Solve the equation x2−2x−10=0 by completing the square.

Use (1) to solve the equation.

$$\begin{gathered} x^2-2x-10=0 \\ {x}^2-2\left(x\right)\left(1\right)-10=0 \\ {x}^2-2\left(x\right)\left(1\right)+1^2-10=1^2 \\ {\left(x-1\right)}^2-10=1 \end{gathered}$$

$$\begin{gathered} {\left(x-1\right)}^2=1+10 \\ {\left(x-1\right)}^2=11 \\ x-1=\pm \sqrt{11} \\ x=1\pm \sqrt{11} \\ x=1-\sqrt{11},1+\sqrt{11} \end{gathered}$$

Use calculator.

$\begin{array}{l}x=-2.31662479,4.3662479\\ x\simeq -2.3,4.4\end{array}$

Hence, $x=-2.3,4.4$are the solutions of the given equation.