Q21. - Algebra 1

Question

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

$2x^2+4x-5=7$

Step-by-Step Solution

Answer

$x=-3.6,1.6$ are the solutions of the given equation.

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 + b 2 .

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

Step 3. Solve the equation 2 x 2 + 4 x − 5 = 7 by completing the square.

Use (1) to solve the equation.

     $$\begin{gathered} 2x^2+4x-5=7 \\ 2x^2+4x=7+5 \\ 2\left(x^2+2x\right)=12 \\ x^2+2x=\frac{12}{2} \\ x^2+2x=6 \end{gathered}$$

$$\begin{gathered} x^2+2\left(x\right)\left(1\right)=6 \\ {x}^2+2\left(x\right)\left(1\right)+1^2=6+1^2 \\ {\left(x+1\right)}^2=6+1 \\ {\left(x+1\right)}^2=7 \\ x+1=\pm \sqrt{7} \\ x=-1\pm \sqrt{7} \end{gathered}$$

Use calculator.

$\begin{array}{l}x=-3.645751311,1.645751311\\ x\simeq -3.6,1.6\end{array}$

Hence, $x=-3.6,1.6$ are the solutions of the given equation.