Solve sqrt{3x^2-4x-4}=sqrt{2x+5}

$$\sqrt { 3 x ^ { 2 } - 4 x - 4 } = \sqrt { 2 x + 5 }$$

Solve for x

$$x=-1$$

$$x=3$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{3x^{2}-4x-4}\right)^{2}=\left(\sqrt{2x+5}\right)^{2}$$

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

$$3x^{2}-4x-4=\left(\sqrt{2x+5}\right)^{2}$$

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

$$3x^{2}-4x-4=2x+5$$

Subtract 2x from both sides.

$$3x^{2}-4x-4-2x=5$$

Combine -4x and -2x to get -6x.

$$3x^{2}-6x-4=5$$

Subtract 5 from both sides.

$$3x^{2}-6x-4-5=0$$

Subtract 5 from -4 to get -9.

$$3x^{2}-6x-9=0$$

Divide both sides by 3.

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

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

$$a+b=-2 ab=1\left(-3\right)=-3$$

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. The only such pair is the system solution.

$$a=-3 b=1$$

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

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

Factor out x in x^{2}-3x.

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

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(x+1\right)$$

To find equation solutions, solve x-3=0 and x+1=0.

$$x=3 x=-1$$

Substitute 3 for x in the equation \sqrt{3x^{2}-4x-4}=\sqrt{2x+5}.

$$\sqrt{3\times 3^{2}-4\times 3-4}=\sqrt{2\times 3+5}$$

Simplify. The value x=3 satisfies the equation.

$$11^{\frac{1}{2}}=11^{\frac{1}{2}}$$

Substitute -1 for x in the equation \sqrt{3x^{2}-4x-4}=\sqrt{2x+5}.

$$\sqrt{3\left(-1\right)^{2}-4\left(-1\right)-4}=\sqrt{2\left(-1\right)+5}$$

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

$$3^{\frac{1}{2}}=3^{\frac{1}{2}}$$

List all solutions of \sqrt{3x^{2}-4x-4}=\sqrt{2x+5}.

$$x=3 x=-1$$