Solve sqrt{2x}=x-4

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

Solve for x

$$x=8$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Subtract x^{2} from both sides.

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

Add 8x to both sides.

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

Combine 2x and 8x to get 10x.

$$10x-x^{2}=16$$

Subtract 16 from both sides.

$$10x-x^{2}-16=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+10x-16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-16. To find a and b, set up a system to be solved.

$$a+b=10 ab=-\left(-16\right)=16$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 16.

$$1,16 2,8 4,4$$

Calculate the sum for each pair.

$$1+16=17 2+8=10 4+4=8$$

The solution is the pair that gives sum 10.

$$a=8 b=2$$

Rewrite -x^{2}+10x-16 as \left(-x^{2}+8x\right)+\left(2x-16\right).

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

Factor out -x in the first and 2 in the second group.

$$-x\left(x-8\right)+2\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(-x+2\right)$$

To find equation solutions, solve x-8=0 and -x+2=0.

$$x=8 x=2$$

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

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

Simplify. The value x=8 satisfies the equation.

$$4=4$$

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

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

Simplify. The value x=2 does not satisfy the equation because the left and the right hand side have opposite signs.

$$2=-2$$

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

$$x=8$$