Solve 5x=sqrt{4x^2+8}

$$5 x = \sqrt { 4 x ^ { 2 } + 8 }$$

Solve for x

$$x=\frac{2\sqrt{42}}{21}\approx 0.6172134$$

Solution Steps

Square both sides of the equation.

$$\left(5x\right)^{2}=\left(\sqrt{4x^{2}+8}\right)^{2}$$

Expand \left(5x\right)^{2}.

$$5^{2}x^{2}=\left(\sqrt{4x^{2}+8}\right)^{2}$$

Calculate 5 to the power of 2 and get 25.

$$25x^{2}=\left(\sqrt{4x^{2}+8}\right)^{2}$$

Calculate \sqrt{4x^{2}+8} to the power of 2 and get 4x^{2}+8.

$$25x^{2}=4x^{2}+8$$

Subtract 4x^{2} from both sides.

$$25x^{2}-4x^{2}=8$$

Combine 25x^{2} and -4x^{2} to get 21x^{2}.

$$21x^{2}=8$$

Divide both sides by 21.

$$x^{2}=\frac{8}{21}$$

Take the square root of both sides of the equation.

$$x=\frac{2\sqrt{42}}{21} x=-\frac{2\sqrt{42}}{21}$$

Substitute \frac{2\sqrt{42}}{21} for x in the equation 5x=\sqrt{4x^{2}+8}.

$$5\times \frac{2\sqrt{42}}{21}=\sqrt{4\times \left(\frac{2\sqrt{42}}{21}\right)^{2}+8}$$

Simplify. The value x=\frac{2\sqrt{42}}{21} satisfies the equation.

$$\frac{10}{21}\times 42^{\frac{1}{2}}=\frac{10}{21}\times 42^{\frac{1}{2}}$$

Substitute -\frac{2\sqrt{42}}{21} for x in the equation 5x=\sqrt{4x^{2}+8}.

$$5\left(-\frac{2\sqrt{42}}{21}\right)=\sqrt{4\left(-\frac{2\sqrt{42}}{21}\right)^{2}+8}$$

Simplify. The value x=-\frac{2\sqrt{42}}{21} does not satisfy the equation because the left and the right hand side have opposite signs.

$$-\frac{10}{21}\times 42^{\frac{1}{2}}=\frac{10}{21}\times 42^{\frac{1}{2}}$$

Equation 5x=\sqrt{4x^{2}+8} has a unique solution.

$$x=\frac{2\sqrt{42}}{21}$$