Solve 3x+sqrt{5x}=33

$$3 x + \sqrt { 5 x } = 33$$

Solve for x

$$x = \frac{203 - \sqrt{2005}}{18} \approx 8.790154091$$

Solution Steps

Subtract 3x from both sides of the equation.

$$\sqrt{5x}=33-3x$$

Square both sides of the equation.

$$\left(\sqrt{5x}\right)^{2}=\left(33-3x\right)^{2}$$

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

$$5x=\left(33-3x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(33-3x\right)^{2}.

$$5x=1089-198x+9x^{2}$$

Subtract 1089 from both sides.

$$5x-1089=-198x+9x^{2}$$

Add 198x to both sides.

$$5x-1089+198x=9x^{2}$$

Combine 5x and 198x to get 203x.

$$203x-1089=9x^{2}$$

Subtract 9x^{2} from both sides.

$$203x-1089-9x^{2}=0$$

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.

$$-9x^{2}+203x-1089=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 203 for b, and -1089 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-203±\sqrt{203^{2}-4\left(-9\right)\left(-1089\right)}}{2\left(-9\right)}$$

Square 203.

$$x=\frac{-203±\sqrt{41209-4\left(-9\right)\left(-1089\right)}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-203±\sqrt{41209+36\left(-1089\right)}}{2\left(-9\right)}$$

Multiply 36 times -1089.

$$x=\frac{-203±\sqrt{41209-39204}}{2\left(-9\right)}$$

Add 41209 to -39204.

$$x=\frac{-203±\sqrt{2005}}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{-203±\sqrt{2005}}{-18}$$

Now solve the equation x=\frac{-203±\sqrt{2005}}{-18} when ± is plus. Add -203 to \sqrt{2005}.

$$x=\frac{\sqrt{2005}-203}{-18}$$

Divide -203+\sqrt{2005} by -18.

$$x=\frac{203-\sqrt{2005}}{18}$$

Now solve the equation x=\frac{-203±\sqrt{2005}}{-18} when ± is minus. Subtract \sqrt{2005} from -203.

$$x=\frac{-\sqrt{2005}-203}{-18}$$

Divide -203-\sqrt{2005} by -18.

$$x=\frac{\sqrt{2005}+203}{18}$$

The equation is now solved.

$$x=\frac{203-\sqrt{2005}}{18} x=\frac{\sqrt{2005}+203}{18}$$

Substitute \frac{203-\sqrt{2005}}{18} for x in the equation 3x+\sqrt{5x}=33.

$$3\times \frac{203-\sqrt{2005}}{18}+\sqrt{5\times \frac{203-\sqrt{2005}}{18}}=33$$

Simplify. The value x=\frac{203-\sqrt{2005}}{18} satisfies the equation.

$$33=33$$

Substitute \frac{\sqrt{2005}+203}{18} for x in the equation 3x+\sqrt{5x}=33.

$$3\times \frac{\sqrt{2005}+203}{18}+\sqrt{5\times \frac{\sqrt{2005}+203}{18}}=33$$

Simplify. The value x=\frac{\sqrt{2005}+203}{18} does not satisfy the equation.

$$\frac{1}{3}\times 2005^{\frac{1}{2}}+\frac{104}{3}=33$$

Equation \sqrt{5x}=33-3x has a unique solution.

$$x=\frac{203-\sqrt{2005}}{18}$$