Solve sqrt{x^2-2x}=sqrt{x+4}

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

Solve for x

$$x=4$$

$$x=-1$$

Solution Steps

Square both sides of the equation.

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

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

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

Calculate \sqrt{x+4} to the power of 2 and get x+4.

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

Subtract x from both sides.

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

Combine -2x and -x to get -3x.

$$x^{2}-3x=4$$

Subtract 4 from both sides.

$$x^{2}-3x-4=0$$

To solve the equation, factor x^{2}-3x-4 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.

$$a+b=-3 ab=-4$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -4.

$$1,-4 2,-2$$

Calculate the sum for each pair.

$$1-4=-3 2-2=0$$

The solution is the pair that gives sum -3.

$$a=-4 b=1$$

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

$$\left(x-4\right)\left(x+1\right)$$

To find equation solutions, solve x-4=0 and x+1=0.

$$x=4 x=-1$$

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

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

Simplify. The value x=4 satisfies the equation.

$$2\times 2^{\frac{1}{2}}=2\times 2^{\frac{1}{2}}$$

Substitute -1 for x in the equation \sqrt{x^{2}-2x}=\sqrt{x+4}.

$$\sqrt{\left(-1\right)^{2}-2\left(-1\right)}=\sqrt{-1+4}$$

Simplify. The value x=-1 satisfies the equation.

$$3^{\frac{1}{2}}=3^{\frac{1}{2}}$$

List all solutions of \sqrt{x^{2}-2x}=\sqrt{x+4}.

$$x=4 x=-1$$