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

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

Solve for x

$$x=\frac{1}{2}=0.5$$

Solution Steps

Square both sides of the equation.

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

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

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

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

$$4-3x=x+2$$

Subtract x from both sides.

$$4-3x-x=2$$

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

$$4-4x=2$$

Subtract 4 from both sides.

$$-4x=2-4$$

Subtract 4 from 2 to get -2.

$$-4x=-2$$

Divide both sides by -4.

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

Reduce the fraction \frac{-2}{-4} to lowest terms by extracting and canceling out -2.

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

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

$$\sqrt{4-3\times \frac{1}{2}}=\sqrt{\frac{1}{2}+2}$$

Simplify. The value x=\frac{1}{2} satisfies the equation.

$$\frac{1}{2}\times 10^{\frac{1}{2}}=\frac{1}{2}\times 10^{\frac{1}{2}}$$

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

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