Solve sqrt{x^2+2x+9}-7=2x

$$\sqrt { x ^ { 2 } + 2 x + 9 } - 7 = 2 x$$

Solve for x

$$x=-2$$

Solution Steps

Subtract -7 from both sides of the equation.

$$\sqrt{x^{2}+2x+9}=2x+7$$

Square both sides of the equation.

$$\left(\sqrt{x^{2}+2x+9}\right)^{2}=\left(2x+7\right)^{2}$$

Calculate \sqrt{x^{2}+2x+9} to the power of 2 and get x^{2}+2x+9.

$$x^{2}+2x+9=\left(2x+7\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+7\right)^{2}.

$$x^{2}+2x+9=4x^{2}+28x+49$$

Subtract 4x^{2} from both sides.

$$x^{2}+2x+9-4x^{2}=28x+49$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+2x+9=28x+49$$

Subtract 28x from both sides.

$$-3x^{2}+2x+9-28x=49$$

Combine 2x and -28x to get -26x.

$$-3x^{2}-26x+9=49$$

Subtract 49 from both sides.

$$-3x^{2}-26x+9-49=0$$

Subtract 49 from 9 to get -40.

$$-3x^{2}-26x-40=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx-40. To find a and b, set up a system to be solved.

$$a+b=-26 ab=-3\left(-40\right)=120$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 120.

$$-1,-120 -2,-60 -3,-40 -4,-30 -5,-24 -6,-20 -8,-15 -10,-12$$

Calculate the sum for each pair.

$$-1-120=-121 -2-60=-62 -3-40=-43 -4-30=-34 -5-24=-29 -6-20=-26 -8-15=-23 -10-12=-22$$

The solution is the pair that gives sum -26.

$$a=-6 b=-20$$

Rewrite -3x^{2}-26x-40 as \left(-3x^{2}-6x\right)+\left(-20x-40\right).

$$\left(-3x^{2}-6x\right)+\left(-20x-40\right)$$

Factor out 3x in the first and 20 in the second group.

$$3x\left(-x-2\right)+20\left(-x-2\right)$$

Factor out common term -x-2 by using distributive property.

$$\left(-x-2\right)\left(3x+20\right)$$

To find equation solutions, solve -x-2=0 and 3x+20=0.

$$x=-2 x=-\frac{20}{3}$$

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

$$\sqrt{\left(-2\right)^{2}+2\left(-2\right)+9}-7=2\left(-2\right)$$

Simplify. The value x=-2 satisfies the equation.

$$-4=-4$$

Substitute -\frac{20}{3} for x in the equation \sqrt{x^{2}+2x+9}-7=2x.

$$\sqrt{\left(-\frac{20}{3}\right)^{2}+2\left(-\frac{20}{3}\right)+9}-7=2\left(-\frac{20}{3}\right)$$

Simplify. The value x=-\frac{20}{3} does not satisfy the equation.

$$-\frac{2}{3}=-\frac{40}{3}$$

Equation \sqrt{x^{2}+2x+9}=2x+7 has a unique solution.

$$x=-2$$