Solve sqrt{3x^2+7x-4}=-x

$$\sqrt{ 3 { x }^{ 2 } +7x-4 } = -x$$

Solve for x

$$x=-4$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{3x^{2}+7x-4}\right)^{2}=\left(-x\right)^{2}$$

Calculate \sqrt{3x^{2}+7x-4} to the power of 2 and get 3x^{2}+7x-4.

$$3x^{2}+7x-4=\left(-x\right)^{2}$$

Calculate -x to the power of 2 and get x^{2}.

$$3x^{2}+7x-4=x^{2}$$

Subtract x^{2} from both sides.

$$3x^{2}+7x-4-x^{2}=0$$

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

$$2x^{2}+7x-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=7 ab=2\left(-4\right)=-8$$

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 -8.

$$-1,8 -2,4$$

Calculate the sum for each pair.

$$-1+8=7 -2+4=2$$

The solution is the pair that gives sum 7.

$$a=-1 b=8$$

Rewrite 2x^{2}+7x-4 as \left(2x^{2}-x\right)+\left(8x-4\right).

$$\left(2x^{2}-x\right)+\left(8x-4\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(2x-1\right)+4\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(x+4\right)$$

To find equation solutions, solve 2x-1=0 and x+4=0.

$$x=\frac{1}{2} x=-4$$

Substitute \frac{1}{2} for x in the equation \sqrt{3x^{2}+7x-4}=-x.

$$\sqrt{3\times \left(\frac{1}{2}\right)^{2}+7\times \frac{1}{2}-4}=-\frac{1}{2}$$

Simplify. The value x=\frac{1}{2} does not satisfy the equation because the left and the right hand side have opposite signs.

$$\frac{1}{2}=-\frac{1}{2}$$

Substitute -4 for x in the equation \sqrt{3x^{2}+7x-4}=-x.

$$\sqrt{3\left(-4\right)^{2}+7\left(-4\right)-4}=-\left(-4\right)$$

Simplify. The value x=-4 satisfies the equation.

$$4=4$$

Equation \sqrt{3x^{2}+7x-4}=-x has a unique solution.

$$x=-4$$