Solve 19-3x+sqrt{x}=3

$$19 - 3 x + \sqrt { x } = 3$$

Solve for x

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

Solution Steps

Subtract 19-3x from both sides of the equation.

$$\sqrt{x}=3-\left(19-3x\right)$$

To find the opposite of 19-3x, find the opposite of each term.

$$\sqrt{x}=3-19-\left(-3x\right)$$

The opposite of -3x is 3x.

$$\sqrt{x}=3-19+3x$$

Subtract 19 from 3 to get -16.

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

Square both sides of the equation.

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

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

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

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

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

Subtract 256 from both sides.

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

Add 96x to both sides.

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

Combine x and 96x to get 97x.

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

Subtract 9x^{2} from both sides.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

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}.

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

Square 97.

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

Multiply -4 times -9.

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

Multiply 36 times -256.

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

Add 9409 to -9216.

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

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}.

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

Divide -97+\sqrt{193} by -18.

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

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

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

Divide -97-\sqrt{193} by -18.

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

The equation is now solved.

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

Substitute \frac{97-\sqrt{193}}{18} for x in the equation 19-3x+\sqrt{x}=3.

$$19-3\times \frac{97-\sqrt{193}}{18}+\sqrt{\frac{97-\sqrt{193}}{18}}=3$$

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

$$\frac{8}{3}+\frac{1}{3}\times 193^{\frac{1}{2}}=3$$

Substitute \frac{\sqrt{193}+97}{18} for x in the equation 19-3x+\sqrt{x}=3.

$$19-3\times \frac{\sqrt{193}+97}{18}+\sqrt{\frac{\sqrt{193}+97}{18}}=3$$

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

$$3=3$$

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

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