Solve sqrt{4x^2-sqrt{8x+5}}=2x+1

$$\sqrt { 4 x ^ { 2 } - \sqrt { 8 x + 5 } } = 2 x + 1$$

Solve for x

$$x=-\frac{1}{2}=-0.5$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{4x^{2}-\sqrt{8x+5}}\right)^{2}=\left(2x+1\right)^{2}$$

Calculate \sqrt{4x^{2}-\sqrt{8x+5}} to the power of 2 and get 4x^{2}-\sqrt{8x+5}.

$$4x^{2}-\sqrt{8x+5}=\left(2x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

$$4x^{2}-\sqrt{8x+5}=4x^{2}+4x+1$$

Subtract 4x^{2} from both sides of the equation.

$$-\sqrt{8x+5}=4x^{2}+4x+1-4x^{2}$$

Combine 4x^{2} and -4x^{2} to get 0.

$$-\sqrt{8x+5}=4x+1$$

Square both sides of the equation.

$$\left(-\sqrt{8x+5}\right)^{2}=\left(4x+1\right)^{2}$$

Expand \left(-\sqrt{8x+5}\right)^{2}.

$$\left(-1\right)^{2}\left(\sqrt{8x+5}\right)^{2}=\left(4x+1\right)^{2}$$

Calculate -1 to the power of 2 and get 1.

$$1\left(\sqrt{8x+5}\right)^{2}=\left(4x+1\right)^{2}$$

Calculate \sqrt{8x+5} to the power of 2 and get 8x+5.

$$1\left(8x+5\right)=\left(4x+1\right)^{2}$$

Use the distributive property to multiply 1 by 8x+5.

$$8x+5=\left(4x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+1\right)^{2}.

$$8x+5=16x^{2}+8x+1$$

Subtract 16x^{2} from both sides.

$$8x+5-16x^{2}=8x+1$$

Subtract 8x from both sides.

$$8x+5-16x^{2}-8x=1$$

Combine 8x and -8x to get 0.

$$5-16x^{2}=1$$

Subtract 5 from both sides.

$$-16x^{2}=1-5$$

Subtract 5 from 1 to get -4.

$$-16x^{2}=-4$$

Divide both sides by -16.

$$x^{2}=\frac{-4}{-16}$$

Reduce the fraction \frac{-4}{-16} to lowest terms by extracting and canceling out -4.

$$x^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

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

Substitute \frac{1}{2} for x in the equation \sqrt{4x^{2}-\sqrt{8x+5}}=2x+1. The expression \sqrt{4\times \left(\frac{1}{2}\right)^{2}-\sqrt{8\times \frac{1}{2}+5}} is undefined because the radicand cannot be negative.

$$\sqrt{4\times \left(\frac{1}{2}\right)^{2}-\sqrt{8\times \frac{1}{2}+5}}=2\times \frac{1}{2}+1$$

Substitute -\frac{1}{2} for x in the equation \sqrt{4x^{2}-\sqrt{8x+5}}=2x+1.

$$\sqrt{4\left(-\frac{1}{2}\right)^{2}-\sqrt{8\left(-\frac{1}{2}\right)+5}}=2\left(-\frac{1}{2}\right)+1$$

Simplify. The value x=-\frac{1}{2} satisfies the equation.

$$0=0$$

Equation \sqrt{4x^{2}-\sqrt{8x+5}}=2x+1 has a unique solution.

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