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

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

Solve for x

$$x = \frac{\sqrt{17} + 7}{8} \approx 1.390388203$$

Solution Steps

Subtract 2 from both sides of the equation.

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

Square both sides of the equation.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-2\right)^{2}.

$$9x^{2}-12x+4=\left(\sqrt{x^{2}+2x}\right)^{2}$$

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

$$9x^{2}-12x+4=x^{2}+2x$$

Subtract x^{2} from both sides.

$$9x^{2}-12x+4-x^{2}=2x$$

Combine 9x^{2} and -x^{2} to get 8x^{2}.

$$8x^{2}-12x+4=2x$$

Subtract 2x from both sides.

$$8x^{2}-12x+4-2x=0$$

Combine -12x and -2x to get -14x.

$$8x^{2}-14x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, -14 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 8\times 4}}{2\times 8}$$

Square -14.

$$x=\frac{-\left(-14\right)±\sqrt{196-4\times 8\times 4}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-14\right)±\sqrt{196-32\times 4}}{2\times 8}$$

Multiply -32 times 4.

$$x=\frac{-\left(-14\right)±\sqrt{196-128}}{2\times 8}$$

Add 196 to -128.

$$x=\frac{-\left(-14\right)±\sqrt{68}}{2\times 8}$$

Take the square root of 68.

$$x=\frac{-\left(-14\right)±2\sqrt{17}}{2\times 8}$$

The opposite of -14 is 14.

$$x=\frac{14±2\sqrt{17}}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{14±2\sqrt{17}}{16}$$

Now solve the equation x=\frac{14±2\sqrt{17}}{16} when ± is plus. Add 14 to 2\sqrt{17}.

$$x=\frac{2\sqrt{17}+14}{16}$$

Divide 14+2\sqrt{17} by 16.

$$x=\frac{\sqrt{17}+7}{8}$$

Now solve the equation x=\frac{14±2\sqrt{17}}{16} when ± is minus. Subtract 2\sqrt{17} from 14.

$$x=\frac{14-2\sqrt{17}}{16}$$

Divide 14-2\sqrt{17} by 16.

$$x=\frac{7-\sqrt{17}}{8}$$

The equation is now solved.

$$x=\frac{\sqrt{17}+7}{8} x=\frac{7-\sqrt{17}}{8}$$

Substitute \frac{\sqrt{17}+7}{8} for x in the equation 3x=2+\sqrt{x^{2}+2x}.

$$3\times \frac{\sqrt{17}+7}{8}=2+\sqrt{\left(\frac{\sqrt{17}+7}{8}\right)^{2}+2\times \frac{\sqrt{17}+7}{8}}$$

Simplify. The value x=\frac{\sqrt{17}+7}{8} satisfies the equation.

$$\frac{3}{8}\times 17^{\frac{1}{2}}+\frac{21}{8}=\frac{21}{8}+\frac{3}{8}\times 17^{\frac{1}{2}}$$

Substitute \frac{7-\sqrt{17}}{8} for x in the equation 3x=2+\sqrt{x^{2}+2x}.

$$3\times \frac{7-\sqrt{17}}{8}=2+\sqrt{\left(\frac{7-\sqrt{17}}{8}\right)^{2}+2\times \frac{7-\sqrt{17}}{8}}$$

Simplify. The value x=\frac{7-\sqrt{17}}{8} does not satisfy the equation.

$$\frac{21}{8}-\frac{3}{8}\times 17^{\frac{1}{2}}=\frac{11}{8}+\frac{3}{8}\times 17^{\frac{1}{2}}$$

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

$$x=\frac{\sqrt{17}+7}{8}$$