Solve sqrt{2x^2+7x+5}=x+1

$$\sqrt { 2 x ^ { 2 } + 7 x + 5 } = x + 1$$

Solve for x

$$x=-1$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{2x^{2}+7x+5}\right)^{2}=\left(x+1\right)^{2}$$

Calculate \sqrt{2x^{2}+7x+5} to the power of 2 and get 2x^{2}+7x+5.

$$2x^{2}+7x+5=\left(x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$2x^{2}+7x+5=x^{2}+2x+1$$

Subtract x^{2} from both sides.

$$2x^{2}+7x+5-x^{2}=2x+1$$

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

$$x^{2}+7x+5=2x+1$$

Subtract 2x from both sides.

$$x^{2}+7x+5-2x=1$$

Combine 7x and -2x to get 5x.

$$x^{2}+5x+5=1$$

Subtract 1 from both sides.

$$x^{2}+5x+5-1=0$$

Subtract 1 from 5 to get 4.

$$x^{2}+5x+4=0$$

To solve the equation, factor x^{2}+5x+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.

$$a+b=5 ab=4$$

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 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 5.

$$a=1 b=4$$

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+5}=x+1.

$$\sqrt{2\left(-1\right)^{2}+7\left(-1\right)+5}=-1+1$$

Simplify. The value x=-1 satisfies the equation.

$$0=0$$

Substitute -4 for x in the equation \sqrt{2x^{2}+7x+5}=x+1.

$$\sqrt{2\left(-4\right)^{2}+7\left(-4\right)+5}=-4+1$$

Simplify. The value x=-4 does not satisfy the equation because the left and the right hand side have opposite signs.

$$3=-3$$

Equation \sqrt{2x^{2}+7x+5}=x+1 has a unique solution.

$$x=-1$$