Solve 3x=16+sqrt{x}

$$3 x = 16 + \sqrt { x }$$

Solve for x

$$x = \frac{\sqrt{193} + 97}{18} \approx 6.160691333$$

Solution Steps

Subtract 16 from both sides of the equation.

$$3x-16=\sqrt{x}$$

Square both sides of the equation.

$$\left(3x-16\right)^{2}=\left(\sqrt{x}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-16\right)^{2}.

$$9x^{2}-96x+256=\left(\sqrt{x}\right)^{2}$$

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

$$9x^{2}-96x+256=x$$

Subtract x from both sides.

$$9x^{2}-96x+256-x=0$$

Combine -96x and -x to get -97x.

$$9x^{2}-97x+256=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -97 for b, and 256 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-97\right)±\sqrt{\left(-97\right)^{2}-4\times 9\times 256}}{2\times 9}$$

Square -97.

$$x=\frac{-\left(-97\right)±\sqrt{9409-4\times 9\times 256}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-97\right)±\sqrt{9409-36\times 256}}{2\times 9}$$

Multiply -36 times 256.

$$x=\frac{-\left(-97\right)±\sqrt{9409-9216}}{2\times 9}$$

Add 9409 to -9216.

$$x=\frac{-\left(-97\right)±\sqrt{193}}{2\times 9}$$

The opposite of -97 is 97.

$$x=\frac{97±\sqrt{193}}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{97±\sqrt{193}}{18}$$

Now solve the equation x=\frac{97±\sqrt{193}}{18} when ± is plus. Add 97 to \sqrt{193}.

$$x=\frac{\sqrt{193}+97}{18}$$

Now solve the equation x=\frac{97±\sqrt{193}}{18} when ± is minus. Subtract \sqrt{193} from 97.

$$x=\frac{97-\sqrt{193}}{18}$$

The equation is now solved.

$$x=\frac{\sqrt{193}+97}{18} x=\frac{97-\sqrt{193}}{18}$$

Substitute \frac{\sqrt{193}+97}{18} for x in the equation 3x=16+\sqrt{x}.

$$3\times \frac{\sqrt{193}+97}{18}=16+\sqrt{\frac{\sqrt{193}+97}{18}}$$

Simplify. The value x=\frac{\sqrt{193}+97}{18} satisfies the equation.

$$\frac{1}{6}\times 193^{\frac{1}{2}}+\frac{97}{6}=\frac{97}{6}+\frac{1}{6}\times 193^{\frac{1}{2}}$$

Substitute \frac{97-\sqrt{193}}{18} for x in the equation 3x=16+\sqrt{x}.

$$3\times \frac{97-\sqrt{193}}{18}=16+\sqrt{\frac{97-\sqrt{193}}{18}}$$

Simplify. The value x=\frac{97-\sqrt{193}}{18} does not satisfy the equation.

$$\frac{97}{6}-\frac{1}{6}\times 193^{\frac{1}{2}}=\frac{95}{6}+\frac{1}{6}\times 193^{\frac{1}{2}}$$

Equation 3x-16=\sqrt{x} has a unique solution.

$$x=\frac{\sqrt{193}+97}{18}$$