Solve sqrt{sqrt{x+2}}=sqrt{2x-4}

$$\sqrt{ \sqrt{ x+2 } } = \sqrt{ 2x-4 }$$

Solve for x

$$x = \frac{\sqrt{65} + 17}{8} \approx 3.132782219$$

Solution Steps

Square both sides of the equation.

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

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

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

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

$$\sqrt{x+2}=2x-4$$

Square both sides of the equation.

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

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

$$x+2=\left(2x-4\right)^{2}$$

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

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

Subtract 4x^{2} from both sides.

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

Add 16x to both sides.

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

Combine x and 16x to get 17x.

$$17x+2-4x^{2}=16$$

Subtract 16 from both sides.

$$17x+2-4x^{2}-16=0$$

Subtract 16 from 2 to get -14.

$$17x-14-4x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-4x^{2}+17x-14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 17 for b, and -14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-17±\sqrt{17^{2}-4\left(-4\right)\left(-14\right)}}{2\left(-4\right)}$$

Square 17.

$$x=\frac{-17±\sqrt{289-4\left(-4\right)\left(-14\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-17±\sqrt{289+16\left(-14\right)}}{2\left(-4\right)}$$

Multiply 16 times -14.

$$x=\frac{-17±\sqrt{289-224}}{2\left(-4\right)}$$

Add 289 to -224.

$$x=\frac{-17±\sqrt{65}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-17±\sqrt{65}}{-8}$$

Now solve the equation x=\frac{-17±\sqrt{65}}{-8} when ± is plus. Add -17 to \sqrt{65}.

$$x=\frac{\sqrt{65}-17}{-8}$$

Divide -17+\sqrt{65} by -8.

$$x=\frac{17-\sqrt{65}}{8}$$

Now solve the equation x=\frac{-17±\sqrt{65}}{-8} when ± is minus. Subtract \sqrt{65} from -17.

$$x=\frac{-\sqrt{65}-17}{-8}$$

Divide -17-\sqrt{65} by -8.

$$x=\frac{\sqrt{65}+17}{8}$$

The equation is now solved.

$$x=\frac{17-\sqrt{65}}{8} x=\frac{\sqrt{65}+17}{8}$$

Substitute \frac{17-\sqrt{65}}{8} for x in the equation \sqrt{\sqrt{x+2}}=\sqrt{2x-4}. The expression \sqrt{2\times \frac{17-\sqrt{65}}{8}-4} is undefined because the radicand cannot be negative.

$$\sqrt{\sqrt{\frac{17-\sqrt{65}}{8}+2}}=\sqrt{2\times \frac{17-\sqrt{65}}{8}-4}$$

Substitute \frac{\sqrt{65}+17}{8} for x in the equation \sqrt{\sqrt{x+2}}=\sqrt{2x-4}.

$$\sqrt{\sqrt{\frac{\sqrt{65}+17}{8}+2}}=\sqrt{2\times \frac{\sqrt{65}+17}{8}-4}$$

Simplify. The value x=\frac{\sqrt{65}+17}{8} satisfies the equation.

$$\frac{1}{2}\left(1+65^{\frac{1}{2}}\right)^{\frac{1}{2}}=\frac{1}{2}\left(65^{\frac{1}{2}}+1\right)^{\frac{1}{2}}$$

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

$$x=\frac{\sqrt{65}+17}{8}$$