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

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

Solve for x

$$x=16$$

Solution Steps

Square both sides of the equation.

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

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

$$4x^{2}-32x+64=\left(2\sqrt{x^{2}-7x}\right)^{2}$$

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

$$4x^{2}-32x+64=2^{2}\left(\sqrt{x^{2}-7x}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-32x+64=4\left(\sqrt{x^{2}-7x}\right)^{2}$$

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

$$4x^{2}-32x+64=4\left(x^{2}-7x\right)$$

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

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

Subtract 4x^{2} from both sides.

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

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

$$-32x+64=-28x$$

Add 28x to both sides.

$$-32x+64+28x=0$$

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

$$-4x+64=0$$

Subtract 64 from both sides. Anything subtracted from zero gives its negation.

$$-4x=-64$$

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

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

Simplify. The value x=16 satisfies the equation.

$$24=24$$

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

$$x=16$$