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

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

Solve for x

$$x=1$$

Solution Steps

Subtract -2 from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

$$2x^{2}+7x=x^{2}+4x+4$$

Subtract x^{2} from both sides.

$$2x^{2}+7x-x^{2}=4x+4$$

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

$$x^{2}+7x=4x+4$$

Subtract 4x from both sides.

$$x^{2}+7x-4x=4$$

Combine 7x and -4x to get 3x.

$$x^{2}+3x=4$$

Subtract 4 from both sides.

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

To solve the equation, factor x^{2}+3x-4 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.

$$a+b=3 ab=-4$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -4.

$$-1,4 -2,2$$

Calculate the sum for each pair.

$$-1+4=3 -2+2=0$$

The solution is the pair that gives sum 3.

$$a=-1 b=4$$

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

$$\left(x-1\right)\left(x+4\right)$$

To find equation solutions, solve x-1=0 and x+4=0.

$$x=1 x=-4$$

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

$$\sqrt{2\times 1^{2}+7\times 1}-2=1$$

Simplify. The value x=1 satisfies the equation.

$$1=1$$

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

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

Simplify. The value x=-4 does not satisfy the equation.

$$0=-4$$

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

$$x=1$$