Q24. - Algebra 1

Question

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

$2x^2+5x-3=0$

Step-by-Step Solution

Answer

$\mathit{x}\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{,}\frac{\mathbf{1}}{\mathbf{2}}$ 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 quadratic formula.

For the quadratic equation $a{x}^2+bx+c=0,$ where, $a\ne 0$ is given by 

 $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Step 3. Solve the equation 2 x 2 + 5 x − 3 = 0 by completing the square.

Compare the quadratic equation $2x^2+5x-3=0$ with the quadratic equation $a{x}^2+bx+c=0$.

$a=2,b=5,c=-3$

Substitute, $a=2,b=5$ and $c=-3$ in $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

$$\begin{gathered} x=\frac{-5\pm \sqrt{5^2-4\left(2\right)\left(-3\right)}}{2\left(2\right)} \\ x=\frac{-5\pm \sqrt{25+24}}{4} \\ x=\frac{-5\pm \sqrt{49}}{4} \\ x=\frac{-5\pm 7}{4} \\ x=\frac{-5-7}{4},\frac{-5+7}{4} \\ x=\frac{-12}{4},\frac{2}{4} \\ x=-3,\frac{1}{2} \end{gathered}$$

Hence, $x=-3,\frac{1}{2}$ are the solutions of the given equation.