Solve 2x-2sqrt{x^2-7x}=8

$$2x-2 \sqrt{ { x }^{ 2 } -7x } = 8$$

Solve for x

$$x=16$$

Solution Steps

Subtract 2x from both sides of the equation.

$$-2\sqrt{x^{2}-7x}=8-2x$$

Square both sides of the equation.

$$\left(-2\sqrt{x^{2}-7x}\right)^{2}=\left(8-2x\right)^{2}$$

Expand \left(-2\sqrt{x^{2}-7x}\right)^{2}.

$$\left(-2\right)^{2}\left(\sqrt{x^{2}-7x}\right)^{2}=\left(8-2x\right)^{2}$$

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

$$4\left(\sqrt{x^{2}-7x}\right)^{2}=\left(8-2x\right)^{2}$$

Calculate \sqrt{x^{2}-7x} to the power of 2 and get x^{2}-7x.

$$4\left(x^{2}-7x\right)=\left(8-2x\right)^{2}$$

Use the distributive property to multiply 4 by x^{2}-7x.

$$4x^{2}-28x=\left(8-2x\right)^{2}$$

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

$$4x^{2}-28x=64-32x+4x^{2}$$

Add 32x to both sides.

$$4x^{2}-28x+32x=64+4x^{2}$$

Combine -28x and 32x to get 4x.

$$4x^{2}+4x=64+4x^{2}$$

Subtract 4x^{2} from both sides.

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

Combine 4x^{2} and -4x^{2} to get 0.

$$4x=64$$

Divide both sides by 4.

$$x=\frac{64}{4}$$

Divide 64 by 4 to get 16.

$$x=16$$

Substitute 16 for x in the equation 2x-2\sqrt{x^{2}-7x}=8.

$$2\times 16-2\sqrt{16^{2}-7\times 16}=8$$

Simplify. The value x=16 satisfies the equation.

$$8=8$$

Equation -2\sqrt{x^{2}-7x}=8-2x has a unique solution.

$$x=16$$