Solve sqrt{x+1}=x-5

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

Solve for x

$$x=8$$

Solution Steps

Square both sides of the equation.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$x+1=x^{2}-10x+25$$

Subtract x^{2} from both sides.

$$x+1-x^{2}=-10x+25$$

Add 10x to both sides.

$$x+1-x^{2}+10x=25$$

Combine x and 10x to get 11x.

$$11x+1-x^{2}=25$$

Subtract 25 from both sides.

$$11x+1-x^{2}-25=0$$

Subtract 25 from 1 to get -24.

$$11x-24-x^{2}=0$$

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

$$-x^{2}+11x-24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-24. To find a and b, set up a system to be solved.

$$a+b=11 ab=-\left(-24\right)=24$$

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 24.

$$1,24 2,12 3,8 4,6$$

Calculate the sum for each pair.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

The solution is the pair that gives sum 11.

$$a=8 b=3$$

Rewrite -x^{2}+11x-24 as \left(-x^{2}+8x\right)+\left(3x-24\right).

$$\left(-x^{2}+8x\right)+\left(3x-24\right)$$

Factor out -x in the first and 3 in the second group.

$$-x\left(x-8\right)+3\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(-x+3\right)$$

To find equation solutions, solve x-8=0 and -x+3=0.

$$x=8 x=3$$

Substitute 8 for x in the equation \sqrt{x+1}=x-5.

$$\sqrt{8+1}=8-5$$

Simplify. The value x=8 satisfies the equation.

$$3=3$$

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

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

Simplify. The value x=3 does not satisfy the equation because the left and the right hand side have opposite signs.

$$2=-2$$

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

$$x=8$$