Solve sqrt{3x-8}+1=sqrt{x+5}

$$\sqrt{ 3x-8 } +1 = \sqrt{ x+5 }$$

Solve for x

$$x=4$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{3x-8}+1\right)^{2}=\left(\sqrt{x+5}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{3x-8}+1\right)^{2}.

$$\left(\sqrt{3x-8}\right)^{2}+2\sqrt{3x-8}+1=\left(\sqrt{x+5}\right)^{2}$$

Calculate \sqrt{3x-8} to the power of 2 and get 3x-8.

$$3x-8+2\sqrt{3x-8}+1=\left(\sqrt{x+5}\right)^{2}$$

Add -8 and 1 to get -7.

$$3x-7+2\sqrt{3x-8}=\left(\sqrt{x+5}\right)^{2}$$

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

$$3x-7+2\sqrt{3x-8}=x+5$$

Subtract 3x-7 from both sides of the equation.

$$2\sqrt{3x-8}=x+5-\left(3x-7\right)$$

To find the opposite of 3x-7, find the opposite of each term.

$$2\sqrt{3x-8}=x+5-3x+7$$

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

$$2\sqrt{3x-8}=-2x+5+7$$

Add 5 and 7 to get 12.

$$2\sqrt{3x-8}=-2x+12$$

Square both sides of the equation.

$$\left(2\sqrt{3x-8}\right)^{2}=\left(-2x+12\right)^{2}$$

Expand \left(2\sqrt{3x-8}\right)^{2}.

$$2^{2}\left(\sqrt{3x-8}\right)^{2}=\left(-2x+12\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{3x-8}\right)^{2}=\left(-2x+12\right)^{2}$$

Calculate \sqrt{3x-8} to the power of 2 and get 3x-8.

$$4\left(3x-8\right)=\left(-2x+12\right)^{2}$$

Use the distributive property to multiply 4 by 3x-8.

$$12x-32=\left(-2x+12\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-2x+12\right)^{2}.

$$12x-32=4x^{2}-48x+144$$

Subtract 4x^{2} from both sides.

$$12x-32-4x^{2}=-48x+144$$

Add 48x to both sides.

$$12x-32-4x^{2}+48x=144$$

Combine 12x and 48x to get 60x.

$$60x-32-4x^{2}=144$$

Subtract 144 from both sides.

$$60x-32-4x^{2}-144=0$$

Subtract 144 from -32 to get -176.

$$60x-176-4x^{2}=0$$

Divide both sides by 4.

$$15x-44-x^{2}=0$$

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

$$-x^{2}+15x-44=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-44. To find a and b, set up a system to be solved.

$$a+b=15 ab=-\left(-44\right)=44$$

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 44.

$$1,44 2,22 4,11$$

Calculate the sum for each pair.

$$1+44=45 2+22=24 4+11=15$$

The solution is the pair that gives sum 15.

$$a=11 b=4$$

Rewrite -x^{2}+15x-44 as \left(-x^{2}+11x\right)+\left(4x-44\right).

$$\left(-x^{2}+11x\right)+\left(4x-44\right)$$

Factor out -x in the first and 4 in the second group.

$$-x\left(x-11\right)+4\left(x-11\right)$$

Factor out common term x-11 by using distributive property.

$$\left(x-11\right)\left(-x+4\right)$$

To find equation solutions, solve x-11=0 and -x+4=0.

$$x=11 x=4$$

Substitute 11 for x in the equation \sqrt{3x-8}+1=\sqrt{x+5}.

$$\sqrt{3\times 11-8}+1=\sqrt{11+5}$$

Simplify. The value x=11 does not satisfy the equation.

$$6=4$$

Substitute 4 for x in the equation \sqrt{3x-8}+1=\sqrt{x+5}.

$$\sqrt{3\times 4-8}+1=\sqrt{4+5}$$

Simplify. The value x=4 satisfies the equation.

$$3=3$$

Equation \sqrt{3x-8}+1=\sqrt{x+5} has a unique solution.

$$x=4$$