Solve sqrt{x+5}=sqrt{8-x}+1

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

Solve for x

$$x=4$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x+5}\right)^{2}=\left(\sqrt{8-x}+1\right)^{2}$$

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

$$x+5=\left(\sqrt{8-x}+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{8-x}+1\right)^{2}.

$$x+5=\left(\sqrt{8-x}\right)^{2}+2\sqrt{8-x}+1$$

Calculate \sqrt{8-x} to the power of 2 and get 8-x.

$$x+5=8-x+2\sqrt{8-x}+1$$

Add 8 and 1 to get 9.

$$x+5=9-x+2\sqrt{8-x}$$

Subtract 9-x from both sides of the equation.

$$x+5-\left(9-x\right)=2\sqrt{8-x}$$

To find the opposite of 9-x, find the opposite of each term.

$$x+5-9+x=2\sqrt{8-x}$$

Subtract 9 from 5 to get -4.

$$x-4+x=2\sqrt{8-x}$$

Combine x and x to get 2x.

$$2x-4=2\sqrt{8-x}$$

Square both sides of the equation.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-4\right)^{2}.

$$4x^{2}-16x+16=\left(2\sqrt{8-x}\right)^{2}$$

Expand \left(2\sqrt{8-x}\right)^{2}.

$$4x^{2}-16x+16=2^{2}\left(\sqrt{8-x}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-16x+16=4\left(\sqrt{8-x}\right)^{2}$$

Calculate \sqrt{8-x} to the power of 2 and get 8-x.

$$4x^{2}-16x+16=4\left(8-x\right)$$

Use the distributive property to multiply 4 by 8-x.

$$4x^{2}-16x+16=32-4x$$

Subtract 32 from both sides.

$$4x^{2}-16x+16-32=-4x$$

Subtract 32 from 16 to get -16.

$$4x^{2}-16x-16=-4x$$

Add 4x to both sides.

$$4x^{2}-16x-16+4x=0$$

Combine -16x and 4x to get -12x.

$$4x^{2}-12x-16=0$$

Divide both sides by 4.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=-3 ab=1\left(-4\right)=-4$$

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 x^{2}-3x-4 as \left(x^{2}-4x\right)+\left(x-4\right).

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

Factor out x in x^{2}-4x.

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

Factor out common term x-4 by using distributive property.

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

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+5}=\sqrt{8-x}+1.

$$\sqrt{4+5}=\sqrt{8-4}+1$$

Simplify. The value x=4 satisfies the equation.

$$3=3$$

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

$$\sqrt{-1+5}=\sqrt{8-\left(-1\right)}+1$$

Simplify. The value x=-1 does not satisfy the equation.

$$2=4$$

Substitute 4 for x in the equation \sqrt{x+5}=\sqrt{8-x}+1.

$$\sqrt{4+5}=\sqrt{8-4}+1$$

Simplify. The value x=4 satisfies the equation.

$$3=3$$

Equation \sqrt{x+5}=\sqrt{8-x}+1 has a unique solution.

$$x=4$$