Solve sqrt{7-2x-sqrt{5}+x}=sqrt{4+3x}

$$\sqrt{ 7-2x- \sqrt{ 5 } +x } = \sqrt{ 4+3x }$$

Solve for x

$$x=\frac{3-\sqrt{5}}{4}\approx 0.190983006$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{7-2x-\sqrt{5}+x}\right)^{2}=\left(\sqrt{4+3x}\right)^{2}$$

Combine -2x and x to get -x.

$$\left(\sqrt{7-x-\sqrt{5}}\right)^{2}=\left(\sqrt{4+3x}\right)^{2}$$

Calculate \sqrt{7-x-\sqrt{5}} to the power of 2 and get 7-x-\sqrt{5}.

$$7-x-\sqrt{5}=\left(\sqrt{4+3x}\right)^{2}$$

Calculate \sqrt{4+3x} to the power of 2 and get 4+3x.

$$7-x-\sqrt{5}=4+3x$$

Subtract 3x from both sides.

$$7-x-\sqrt{5}-3x=4$$

Combine -x and -3x to get -4x.

$$7-4x-\sqrt{5}=4$$

Subtract 7 from both sides.

$$-4x-\sqrt{5}=4-7$$

Subtract 7 from 4 to get -3.

$$-4x-\sqrt{5}=-3$$

Add \sqrt{5} to both sides.

$$-4x=-3+\sqrt{5}$$

The equation is in standard form.

$$-4x=\sqrt{5}-3$$

Divide both sides by -4.

$$\frac{-4x}{-4}=\frac{\sqrt{5}-3}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x=\frac{\sqrt{5}-3}{-4}$$

Divide -3+\sqrt{5} by -4.

$$x=\frac{3-\sqrt{5}}{4}$$

Substitute \frac{3-\sqrt{5}}{4} for x in the equation \sqrt{7-2x-\sqrt{5}+x}=\sqrt{4+3x}.

$$\sqrt{7-2\times \frac{3-\sqrt{5}}{4}-\sqrt{5}+\frac{3-\sqrt{5}}{4}}=\sqrt{4+3\times \frac{3-\sqrt{5}}{4}}$$

Simplify. The value x=\frac{3-\sqrt{5}}{4} satisfies the equation.

$$\frac{1}{2}\left(25-3\times 5^{\frac{1}{2}}\right)^{\frac{1}{2}}=\frac{1}{2}\left(25-3\times 5^{\frac{1}{2}}\right)^{\frac{1}{2}}$$

Equation \sqrt{x-2x+7-\sqrt{5}}=\sqrt{3x+4} has a unique solution.

$$x=\frac{3-\sqrt{5}}{4}$$