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

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

Solve for x

$$x = \frac{\sqrt[3]{\sqrt{3} + 2} + \sqrt[3]{2 - \sqrt{3}}}{2} \approx 1.097911673$$

$$x=0$$

Solution Steps

Square both sides of the equation.

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

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

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

Expand \left(2x^{2}\right)^{2}.

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

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

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

Calculate 2 to the power of 2 and get 4.

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

Subtract 4x^{4} from both sides.

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

Substitute t for x^{2}.

$$-4t^{2}+3t+2=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute -4 for a, 3 for b, and 2 for c in the quadratic formula.

$$t=\frac{-3±\sqrt{3^{2}-4\left(-4\right)\times 2}}{-4\times 2}$$

Do the calculations.

$$t=\frac{-3±\sqrt{41}}{-8}$$

Solve the equation t=\frac{-3±\sqrt{41}}{-8} when ± is plus and when ± is minus.

$$t=\frac{3-\sqrt{41}}{8} t=\frac{\sqrt{41}+3}{8}$$

Since x=t^{2}, the solutions are obtained by evaluating x=±\sqrt{t} for positive t.

$$x=\frac{\sqrt{\frac{\sqrt{41}+3}{2}}}{2} x=-\frac{\sqrt{\frac{\sqrt{41}+3}{2}}}{2}$$

Substitute \frac{\sqrt{\frac{\sqrt{41}+3}{2}}}{2} for x in the equation \sqrt{2x+3x^{2}}=2x^{2}.

$$\sqrt{2\times \frac{\sqrt{\frac{\sqrt{41}+3}{2}}}{2}+3\times \left(\frac{\sqrt{\frac{\sqrt{41}+3}{2}}}{2}\right)^{2}}=2\times \left(\frac{\sqrt{\frac{\sqrt{41}+3}{2}}}{2}\right)^{2}$$

Simplify. The value x=\frac{\sqrt{\frac{\sqrt{41}+3}{2}}}{2} does not satisfy the equation.

$$\frac{1}{2}\left(2\times 2^{\frac{1}{2}}\left(41^{\frac{1}{2}}+3\right)^{\frac{1}{2}}+\frac{3}{2}\times 41^{\frac{1}{2}}+\frac{9}{2}\right)^{\frac{1}{2}}=\frac{1}{4}\times 41^{\frac{1}{2}}+\frac{3}{4}$$

Substitute -\frac{\sqrt{\frac{\sqrt{41}+3}{2}}}{2} for x in the equation \sqrt{2x+3x^{2}}=2x^{2}.

$$\sqrt{2\left(-\frac{\sqrt{\frac{\sqrt{41}+3}{2}}}{2}\right)+3\left(-\frac{\sqrt{\frac{\sqrt{41}+3}{2}}}{2}\right)^{2}}=2\left(-\frac{\sqrt{\frac{\sqrt{41}+3}{2}}}{2}\right)^{2}$$

Simplify. The value x=-\frac{\sqrt{\frac{\sqrt{41}+3}{2}}}{2} does not satisfy the equation.

$$\frac{1}{2}\left(-2\times 2^{\frac{1}{2}}\left(41^{\frac{1}{2}}+3\right)^{\frac{1}{2}}+\frac{3}{2}\times 41^{\frac{1}{2}}+\frac{9}{2}\right)^{\frac{1}{2}}=\frac{1}{4}\times 41^{\frac{1}{2}}+\frac{3}{4}$$

Equation \sqrt{3x^{2}+2x}=2x^{2} has no solutions.

$$x\in \emptyset$$