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

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

Solve for x

$$x=-\frac{1}{5}=-0.2$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Subtract 2x from both sides.

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

Combine -3x and -2x to get -5x.

$$2-5x=3$$

Subtract 2 from both sides.

$$-5x=3-2$$

Subtract 2 from 3 to get 1.

$$-5x=1$$

Divide both sides by -5.

$$x=\frac{1}{-5}$$

Fraction \frac{1}{-5} can be rewritten as -\frac{1}{5} by extracting the negative sign.

$$x=-\frac{1}{5}$$

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

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

Simplify. The value x=-\frac{1}{5} satisfies the equation.

$$\frac{1}{5}\times 65^{\frac{1}{2}}=\frac{1}{5}\times 65^{\frac{1}{2}}$$

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

$$x=-\frac{1}{5}$$