How do you solve 2x^2-5=0?

Aug 21, 2015

Answer:

${x}_{1 , 2} = \pm \frac{\sqrt{10}}{2}$

Explanation:

You need to take three steps in order to solve this equation

• add $5$ to both sides of the equation

$2 {x}^{2} - \textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} = 0 + 5$

$2 {x}^{2} = 5$

• divide both sides of the equation by $2$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} {x}^{2}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} = \frac{5}{2}$

${x}^{2} = \frac{5}{2}$

• take the square root of both sides to solve for $x$

$\sqrt{{x}^{2}} = \sqrt{\frac{5}{2}}$

${x}_{1 , 2} = \pm \frac{\sqrt{5}}{\sqrt{2}}$

You can simplify this further by rationalizing the denominator if the fraction. To do that, multiply the fraction by $1 = \frac{\sqrt{2}}{\sqrt{2}}$

${x}_{1 , 2} = \pm \frac{\sqrt{5} \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}}$

${x}_{1 , 2} = \pm \frac{\sqrt{10}}{2}$

This means that your equation has two solutions,

${x}_{1} = \textcolor{g r e e n}{\frac{\sqrt{10}}{2}} \text{ }$ or $\text{ } {x}_{2} = \textcolor{g r e e n}{- \frac{\sqrt{10}}{2}}$