# Given that sqrt(x^2+4)=-3x determine the value or values of x?

Jun 27, 2016

$x = \pm \frac{\sqrt{2}}{2}$

#### Explanation:

The x-intercept is at $y = 0$

Write as:$\text{ } 0 = 3 x - \sqrt{{x}^{2} + 4}$

Add $\sqrt{{x}^{2} + 4} \text{ }$ to both sides

$\sqrt{{x}^{2} + 4} = 3 x$

Square both sides

${x}^{2} + 4 = 9 {x}^{2}$

Subtract$\text{ } {x}^{2}$ from both sides

$8 {x}^{2} = 4$

Divide both sides by 8

${x}^{2} = \frac{1}{2}$

Square root both sides

$x = \pm \frac{1}{\sqrt{2}}$
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Mathematicians consider it bad practice to have roots in the denominator if at all avoidable.

Multiply b1 but in the form of $1 = \frac{\sqrt{2}}{\sqrt{2}}$

$x = \pm \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$