Solve sqrt{x^2-5}=sqrt{x+1}

$$\sqrt { x ^ { 2 } - 5 } = \sqrt { x + 1 }$$

Solve for x (complex solution)

$$x=-2$$

$$x=3$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x^{2}-5}\right)^{2}=\left(\sqrt{x+1}\right)^{2}$$

Calculate \sqrt{x^{2}-5} to the power of 2 and get x^{2}-5.

$$x^{2}-5=\left(\sqrt{x+1}\right)^{2}$$

Calculate \sqrt{x+1} to the power of 2 and get x+1.

$$x^{2}-5=x+1$$

Subtract x from both sides.

$$x^{2}-5-x=1$$

Subtract 1 from both sides.

$$x^{2}-5-x-1=0$$

Subtract 1 from -5 to get -6.

$$x^{2}-6-x=0$$

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

$$x^{2}-x-6=0$$

To solve the equation, factor x^{2}-x-6 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.

$$a+b=-1 ab=-6$$

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 -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=-3 b=2$$

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

$$\left(x-3\right)\left(x+2\right)$$

To find equation solutions, solve x-3=0 and x+2=0.

$$x=3 x=-2$$

Substitute 3 for x in the equation \sqrt{x^{2}-5}=\sqrt{x+1}.

$$\sqrt{3^{2}-5}=\sqrt{3+1}$$

Simplify. The value x=3 satisfies the equation.

$$2=2$$

Substitute -2 for x in the equation \sqrt{x^{2}-5}=\sqrt{x+1}.

$$\sqrt{\left(-2\right)^{2}-5}=\sqrt{-2+1}$$

Simplify. The value x=-2 satisfies the equation.

$$i=i$$

List all solutions of \sqrt{x^{2}-5}=\sqrt{x+1}.

$$x=3 x=-2$$

Solve for x

$$x=3$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x^{2}-5}\right)^{2}=\left(\sqrt{x+1}\right)^{2}$$

Calculate \sqrt{x^{2}-5} to the power of 2 and get x^{2}-5.

$$x^{2}-5=\left(\sqrt{x+1}\right)^{2}$$

Calculate \sqrt{x+1} to the power of 2 and get x+1.

$$x^{2}-5=x+1$$

Subtract x from both sides.

$$x^{2}-5-x=1$$

Subtract 1 from both sides.

$$x^{2}-5-x-1=0$$

Subtract 1 from -5 to get -6.

$$x^{2}-6-x=0$$

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

$$x^{2}-x-6=0$$

To solve the equation, factor x^{2}-x-6 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.

$$a+b=-1 ab=-6$$

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 -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=-3 b=2$$

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

$$\left(x-3\right)\left(x+2\right)$$

To find equation solutions, solve x-3=0 and x+2=0.

$$x=3 x=-2$$

Substitute 3 for x in the equation \sqrt{x^{2}-5}=\sqrt{x+1}.

$$\sqrt{3^{2}-5}=\sqrt{3+1}$$

Simplify. The value x=3 satisfies the equation.

$$2=2$$

Substitute -2 for x in the equation \sqrt{x^{2}-5}=\sqrt{x+1}. The expression \sqrt{\left(-2\right)^{2}-5} is undefined because the radicand cannot be negative.

$$\sqrt{\left(-2\right)^{2}-5}=\sqrt{-2+1}$$

Equation \sqrt{x^{2}-5}=\sqrt{x+1} has a unique solution.

$$x=3$$