Solve 8+sqrt{36x}=2x

$$8+ \sqrt{ 36x } =2x$$

Solve for x

$$x=16$$

Solution Steps

Subtract 8 from both sides of the equation.

$$\sqrt{36x}=2x-8$$

Square both sides of the equation.

$$\left(\sqrt{36x}\right)^{2}=\left(2x-8\right)^{2}$$

Calculate \sqrt{36x} to the power of 2 and get 36x.

$$36x=\left(2x-8\right)^{2}$$

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

$$36x=4x^{2}-32x+64$$

Subtract 4x^{2} from both sides.

$$36x-4x^{2}=-32x+64$$

Add 32x to both sides.

$$36x-4x^{2}+32x=64$$

Combine 36x and 32x to get 68x.

$$68x-4x^{2}=64$$

Subtract 64 from both sides.

$$68x-4x^{2}-64=0$$

Divide both sides by 4.

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

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

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

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.

$$a+b=17 ab=-\left(-16\right)=16$$

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 17.

$$a=16 b=1$$

Rewrite -x^{2}+17x-16 as \left(-x^{2}+16x\right)+\left(x-16\right).

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

Factor out -x in -x^{2}+16x.

$$-x\left(x-16\right)+x-16$$

Factor out common term x-16 by using distributive property.

$$\left(x-16\right)\left(-x+1\right)$$

To find equation solutions, solve x-16=0 and -x+1=0.

$$x=16 x=1$$

Substitute 16 for x in the equation 8+\sqrt{36x}=2x.

$$8+\sqrt{36\times 16}=2\times 16$$

Simplify. The value x=16 satisfies the equation.

$$32=32$$

Substitute 1 for x in the equation 8+\sqrt{36x}=2x.

$$8+\sqrt{36\times 1}=2\times 1$$

Simplify. The value x=1 does not satisfy the equation.

$$14=2$$

Equation \sqrt{36x}=2x-8 has a unique solution.

$$x=16$$