Solve sqrt{2x-7}+x=5

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

Solve for x

$$x=4$$

Solution Steps

Subtract x from both sides of the equation.

$$\sqrt{2x-7}=5-x$$

Square both sides of the equation.

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

Calculate \sqrt{2x-7} to the power of 2 and get 2x-7.

$$2x-7=\left(5-x\right)^{2}$$

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

$$2x-7=25-10x+x^{2}$$

Subtract 25 from both sides.

$$2x-7-25=-10x+x^{2}$$

Subtract 25 from -7 to get -32.

$$2x-32=-10x+x^{2}$$

Add 10x to both sides.

$$2x-32+10x=x^{2}$$

Combine 2x and 10x to get 12x.

$$12x-32=x^{2}$$

Subtract x^{2} from both sides.

$$12x-32-x^{2}=0$$

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

$$-x^{2}+12x-32=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-32. To find a and b, set up a system to be solved.

$$a+b=12 ab=-\left(-32\right)=32$$

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 32.

$$1,32 2,16 4,8$$

Calculate the sum for each pair.

$$1+32=33 2+16=18 4+8=12$$

The solution is the pair that gives sum 12.

$$a=8 b=4$$

Rewrite -x^{2}+12x-32 as \left(-x^{2}+8x\right)+\left(4x-32\right).

$$\left(-x^{2}+8x\right)+\left(4x-32\right)$$

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

$$-x\left(x-8\right)+4\left(x-8\right)$$

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

$$\left(x-8\right)\left(-x+4\right)$$

To find equation solutions, solve x-8=0 and -x+4=0.

$$x=8 x=4$$

Substitute 8 for x in the equation \sqrt{2x-7}+x=5.

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

Simplify. The value x=8 does not satisfy the equation.

$$11=5$$

Substitute 4 for x in the equation \sqrt{2x-7}+x=5.

$$\sqrt{2\times 4-7}+4=5$$

Simplify. The value x=4 satisfies the equation.

$$5=5$$

Equation \sqrt{2x-7}=5-x has a unique solution.

$$x=4$$