Solve sqrt{x}-sqrt{2x-8}-2=0

$$\sqrt { x } - \sqrt { 2 x - 8 } - 2 = 0$$

Solve for x

$$x=4$$

Solution Steps

Subtract -\sqrt{2x-8}-2 from both sides of the equation.

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

To find the opposite of -\sqrt{2x-8}-2, find the opposite of each term.

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

The opposite of -\sqrt{2x-8} is \sqrt{2x-8}.

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

The opposite of -2 is 2.

$$\sqrt{x}=\sqrt{2x-8}+2$$

Square both sides of the equation.

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

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

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{2x-8}+2\right)^{2}.

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

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

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

Add -8 and 4 to get -4.

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

Subtract 2x-4 from both sides of the equation.

$$x-\left(2x-4\right)=4\sqrt{2x-8}$$

To find the opposite of 2x-4, find the opposite of each term.

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

Combine x and -2x to get -x.

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

Square both sides of the equation.

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

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

$$x^{2}-8x+16=\left(4\sqrt{2x-8}\right)^{2}$$

Expand \left(4\sqrt{2x-8}\right)^{2}.

$$x^{2}-8x+16=4^{2}\left(\sqrt{2x-8}\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$x^{2}-8x+16=16\left(\sqrt{2x-8}\right)^{2}$$

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

$$x^{2}-8x+16=16\left(2x-8\right)$$

Use the distributive property to multiply 16 by 2x-8.

$$x^{2}-8x+16=32x-128$$

Subtract 32x from both sides.

$$x^{2}-8x+16-32x=-128$$

Combine -8x and -32x to get -40x.

$$x^{2}-40x+16=-128$$

Add 128 to both sides.

$$x^{2}-40x+16+128=0$$

Add 16 and 128 to get 144.

$$x^{2}-40x+144=0$$

To solve the equation, factor x^{2}-40x+144 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-40 ab=144$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 144.

$$-1,-144 -2,-72 -3,-48 -4,-36 -6,-24 -8,-18 -9,-16 -12,-12$$

Calculate the sum for each pair.

$$-1-144=-145 -2-72=-74 -3-48=-51 -4-36=-40 -6-24=-30 -8-18=-26 -9-16=-25 -12-12=-24$$

The solution is the pair that gives sum -40.

$$a=-36 b=-4$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-36\right)\left(x-4\right)$$

To find equation solutions, solve x-36=0 and x-4=0.

$$x=36 x=4$$

Substitute 36 for x in the equation \sqrt{x}-\sqrt{2x-8}-2=0.

$$\sqrt{36}-\sqrt{2\times 36-8}-2=0$$

Simplify. The value x=36 does not satisfy the equation.

$$-4=0$$

Substitute 4 for x in the equation \sqrt{x}-\sqrt{2x-8}-2=0.

$$\sqrt{4}-\sqrt{2\times 4-8}-2=0$$

Simplify. The value x=4 satisfies the equation.

$$0=0$$

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

$$x=4$$