Solve sqrt{2x+7}+4=x

$$\sqrt { 2 x + 7 } + 4 = x$$

Solve for x

$$x=9$$

Solution Steps

Subtract 4 from both sides of the equation.

$$\sqrt{2x+7}=x-4$$

Square both sides of the equation.

$$\left(\sqrt{2x+7}\right)^{2}=\left(x-4\right)^{2}$$

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

$$2x+7=\left(x-4\right)^{2}$$

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

$$2x+7=x^{2}-8x+16$$

Subtract x^{2} from both sides.

$$2x+7-x^{2}=-8x+16$$

Add 8x to both sides.

$$2x+7-x^{2}+8x=16$$

Combine 2x and 8x to get 10x.

$$10x+7-x^{2}=16$$

Subtract 16 from both sides.

$$10x+7-x^{2}-16=0$$

Subtract 16 from 7 to get -9.

$$10x-9-x^{2}=0$$

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

$$-x^{2}+10x-9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-9. To find a and b, set up a system to be solved.

$$a+b=10 ab=-\left(-9\right)=9$$

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 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 10.

$$a=9 b=1$$

Rewrite -x^{2}+10x-9 as \left(-x^{2}+9x\right)+\left(x-9\right).

$$\left(-x^{2}+9x\right)+\left(x-9\right)$$

Factor out -x in -x^{2}+9x.

$$-x\left(x-9\right)+x-9$$

Factor out common term x-9 by using distributive property.

$$\left(x-9\right)\left(-x+1\right)$$

To find equation solutions, solve x-9=0 and -x+1=0.

$$x=9 x=1$$

Substitute 9 for x in the equation \sqrt{2x+7}+4=x.

$$\sqrt{2\times 9+7}+4=9$$

Simplify. The value x=9 satisfies the equation.

$$9=9$$

Substitute 1 for x in the equation \sqrt{2x+7}+4=x.

$$\sqrt{2\times 1+7}+4=1$$

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

$$7=1$$

Equation \sqrt{2x+7}=x-4 has a unique solution.

$$x=9$$