Solve sqrt{2x^2+x-6}=x+2

$$\sqrt{ 2 { x }^{ 2 } +x-6 } =x+2$$

Solve for x

$$x=5$$

$$x=-2$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{2x^{2}+x-6}\right)^{2}=\left(x+2\right)^{2}$$

Calculate \sqrt{2x^{2}+x-6} to the power of 2 and get 2x^{2}+x-6.

$$2x^{2}+x-6=\left(x+2\right)^{2}$$

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

$$2x^{2}+x-6=x^{2}+4x+4$$

Subtract x^{2} from both sides.

$$2x^{2}+x-6-x^{2}=4x+4$$

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

$$x^{2}+x-6=4x+4$$

Subtract 4x from both sides.

$$x^{2}+x-6-4x=4$$

Combine x and -4x to get -3x.

$$x^{2}-3x-6=4$$

Subtract 4 from both sides.

$$x^{2}-3x-6-4=0$$

Subtract 4 from -6 to get -10.

$$x^{2}-3x-10=0$$

To solve the equation, factor x^{2}-3x-10 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-3 ab=-10$$

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 -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -3.

$$a=-5 b=2$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-5\right)\left(x+2\right)$$

To find equation solutions, solve x-5=0 and x+2=0.

$$x=5 x=-2$$

Substitute 5 for x in the equation \sqrt{2x^{2}+x-6}=x+2.

$$\sqrt{2\times 5^{2}+5-6}=5+2$$

Simplify. The value x=5 satisfies the equation.

$$7=7$$

Substitute -2 for x in the equation \sqrt{2x^{2}+x-6}=x+2.

$$\sqrt{2\left(-2\right)^{2}-2-6}=-2+2$$

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

$$0=0$$

List all solutions of \sqrt{2x^{2}+x-6}=x+2.

$$x=5 x=-2$$