Solve sqrt{x^2-4x-5}=sqrt{-3x^2+7x}

$$\sqrt { x ^ { 2 } - 4 x - 5 } = \sqrt { - 3 x ^ { 2 } + 7 x }$$

Solve for x (complex solution)

$$x = \frac{\sqrt{201} + 11}{8} \approx 3.14718086$$

$$x=\frac{11-\sqrt{201}}{8}\approx -0.39718086$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x^{2}-4x-5}\right)^{2}=\left(\sqrt{-3x^{2}+7x}\right)^{2}$$

Calculate \sqrt{x^{2}-4x-5} to the power of 2 and get x^{2}-4x-5.

$$x^{2}-4x-5=\left(\sqrt{-3x^{2}+7x}\right)^{2}$$

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

$$x^{2}-4x-5=-3x^{2}+7x$$

Add 3x^{2} to both sides.

$$x^{2}-4x-5+3x^{2}=7x$$

Combine x^{2} and 3x^{2} to get 4x^{2}.

$$4x^{2}-4x-5=7x$$

Subtract 7x from both sides.

$$4x^{2}-4x-5-7x=0$$

Combine -4x and -7x to get -11x.

$$4x^{2}-11x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -11 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 4\left(-5\right)}}{2\times 4}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 4\left(-5\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-11\right)±\sqrt{121-16\left(-5\right)}}{2\times 4}$$

Multiply -16 times -5.

$$x=\frac{-\left(-11\right)±\sqrt{121+80}}{2\times 4}$$

Add 121 to 80.

$$x=\frac{-\left(-11\right)±\sqrt{201}}{2\times 4}$$

The opposite of -11 is 11.

$$x=\frac{11±\sqrt{201}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{11±\sqrt{201}}{8}$$

Now solve the equation x=\frac{11±\sqrt{201}}{8} when ± is plus. Add 11 to \sqrt{201}.

$$x=\frac{\sqrt{201}+11}{8}$$

Now solve the equation x=\frac{11±\sqrt{201}}{8} when ± is minus. Subtract \sqrt{201} from 11.

$$x=\frac{11-\sqrt{201}}{8}$$

The equation is now solved.

$$x=\frac{\sqrt{201}+11}{8} x=\frac{11-\sqrt{201}}{8}$$

Substitute \frac{\sqrt{201}+11}{8} for x in the equation \sqrt{x^{2}-4x-5}=\sqrt{-3x^{2}+7x}.

$$\sqrt{\left(\frac{\sqrt{201}+11}{8}\right)^{2}-4\times \frac{\sqrt{201}+11}{8}-5}=\sqrt{-3\times \left(\frac{\sqrt{201}+11}{8}\right)^{2}+7\times \frac{\sqrt{201}+11}{8}}$$

Simplify. The value x=\frac{\sqrt{201}+11}{8} satisfies the equation.

$$\frac{1}{8}i\times 335^{\frac{1}{2}}+\frac{1}{8}i\times 15^{\frac{1}{2}}=\frac{1}{8}i\times 335^{\frac{1}{2}}+\frac{1}{8}i\times 15^{\frac{1}{2}}$$

Substitute \frac{11-\sqrt{201}}{8} for x in the equation \sqrt{x^{2}-4x-5}=\sqrt{-3x^{2}+7x}.

$$\sqrt{\left(\frac{11-\sqrt{201}}{8}\right)^{2}-4\times \frac{11-\sqrt{201}}{8}-5}=\sqrt{-3\times \left(\frac{11-\sqrt{201}}{8}\right)^{2}+7\times \frac{11-\sqrt{201}}{8}}$$

Simplify. The value x=\frac{11-\sqrt{201}}{8} satisfies the equation.

$$\frac{1}{8}i\times 335^{\frac{1}{2}}-\frac{1}{8}i\times 15^{\frac{1}{2}}=\frac{1}{8}i\times 335^{\frac{1}{2}}-\frac{1}{8}i\times 15^{\frac{1}{2}}$$

List all solutions of \sqrt{x^{2}-4x-5}=\sqrt{7x-3x^{2}}.

$$x=\frac{\sqrt{201}+11}{8} x=\frac{11-\sqrt{201}}{8}$$