Solve x-2sqrt{2x}+1=0

$$x - 2 \sqrt { 2 x } + 1 = 0$$

Solve for x

$$x=2\sqrt{2}+3\approx 5.828427125$$

$$x=3-2\sqrt{2}\approx 0.171572875$$

Solution Steps

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

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

Subtract x from both sides of the equation.

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

Square both sides of the equation.

$$\left(-2\sqrt{2x}\right)^{2}=\left(-1-x\right)^{2}$$

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

$$\left(-2\right)^{2}\left(\sqrt{2x}\right)^{2}=\left(-1-x\right)^{2}$$

Calculate -2 to the power of 2 and get 4.

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

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

$$4\times 2x=\left(-1-x\right)^{2}$$

Multiply 4 and 2 to get 8.

$$8x=\left(-1-x\right)^{2}$$

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

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

Subtract 2x from both sides.

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

Combine 8x and -2x to get 6x.

$$6x=1+x^{2}$$

Subtract x^{2} from both sides.

$$6x-x^{2}=1$$

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.

$$-x^{2}+6x=1$$

Subtract 1 from both sides of the equation.

$$-x^{2}+6x-1=1-1$$

Subtracting 1 from itself leaves 0.

$$-x^{2}+6x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 6 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-1\right)\left(-1\right)}}{2\left(-1\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-1\right)\left(-1\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-6±\sqrt{36+4\left(-1\right)}}{2\left(-1\right)}$$

Multiply 4 times -1.

$$x=\frac{-6±\sqrt{36-4}}{2\left(-1\right)}$$

Add 36 to -4.

$$x=\frac{-6±\sqrt{32}}{2\left(-1\right)}$$

Take the square root of 32.

$$x=\frac{-6±4\sqrt{2}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-6±4\sqrt{2}}{-2}$$

Now solve the equation x=\frac{-6±4\sqrt{2}}{-2} when ± is plus. Add -6 to 4\sqrt{2}.

$$x=\frac{4\sqrt{2}-6}{-2}$$

Divide -6+4\sqrt{2} by -2.

$$x=3-2\sqrt{2}$$

Now solve the equation x=\frac{-6±4\sqrt{2}}{-2} when ± is minus. Subtract 4\sqrt{2} from -6.

$$x=\frac{-4\sqrt{2}-6}{-2}$$

Divide -6-4\sqrt{2} by -2.

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

The equation is now solved.

$$x=3-2\sqrt{2} x=2\sqrt{2}+3$$

Substitute 3-2\sqrt{2} for x in the equation x-2\sqrt{2x}+1=0.

$$3-2\sqrt{2}-2\sqrt{2\left(3-2\sqrt{2}\right)}+1=0$$

Simplify. The value x=3-2\sqrt{2} satisfies the equation.

$$0=0$$

Substitute 2\sqrt{2}+3 for x in the equation x-2\sqrt{2x}+1=0.

$$2\sqrt{2}+3-2\sqrt{2\left(2\sqrt{2}+3\right)}+1=0$$

Simplify. The value x=2\sqrt{2}+3 satisfies the equation.

$$0=0$$

List all solutions of -2\sqrt{2x}=-x-1.

$$x=3-2\sqrt{2} x=2\sqrt{2}+3$$