Solve 3x+10sqrt{x}+140=500

$$3x+10 \sqrt{ x } +140=500$$

Solve for x

$$x = \frac{1130 - 10 \sqrt{1105}}{9} \approx 88.620510803$$

Solution Steps

Subtract 3x+140 from both sides of the equation.

$$10\sqrt{x}=500-\left(3x+140\right)$$

To find the opposite of 3x+140, find the opposite of each term.

$$10\sqrt{x}=500-3x-140$$

Subtract 140 from 500 to get 360.

$$10\sqrt{x}=360-3x$$

Square both sides of the equation.

$$\left(10\sqrt{x}\right)^{2}=\left(360-3x\right)^{2}$$

Expand \left(10\sqrt{x}\right)^{2}.

$$10^{2}\left(\sqrt{x}\right)^{2}=\left(360-3x\right)^{2}$$

Calculate 10 to the power of 2 and get 100.

$$100\left(\sqrt{x}\right)^{2}=\left(360-3x\right)^{2}$$

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

$$100x=\left(360-3x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(360-3x\right)^{2}.

$$100x=129600-2160x+9x^{2}$$

Subtract 129600 from both sides.

$$100x-129600=-2160x+9x^{2}$$

Add 2160x to both sides.

$$100x-129600+2160x=9x^{2}$$

Combine 100x and 2160x to get 2260x.

$$2260x-129600=9x^{2}$$

Subtract 9x^{2} from both sides.

$$2260x-129600-9x^{2}=0$$

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.

$$-9x^{2}+2260x-129600=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 2260 for b, and -129600 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2260±\sqrt{2260^{2}-4\left(-9\right)\left(-129600\right)}}{2\left(-9\right)}$$

Square 2260.

$$x=\frac{-2260±\sqrt{5107600-4\left(-9\right)\left(-129600\right)}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-2260±\sqrt{5107600+36\left(-129600\right)}}{2\left(-9\right)}$$

Multiply 36 times -129600.

$$x=\frac{-2260±\sqrt{5107600-4665600}}{2\left(-9\right)}$$

Add 5107600 to -4665600.

$$x=\frac{-2260±\sqrt{442000}}{2\left(-9\right)}$$

Take the square root of 442000.

$$x=\frac{-2260±20\sqrt{1105}}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{-2260±20\sqrt{1105}}{-18}$$

Now solve the equation x=\frac{-2260±20\sqrt{1105}}{-18} when ± is plus. Add -2260 to 20\sqrt{1105}.

$$x=\frac{20\sqrt{1105}-2260}{-18}$$

Divide -2260+20\sqrt{1105} by -18.

$$x=\frac{1130-10\sqrt{1105}}{9}$$

Now solve the equation x=\frac{-2260±20\sqrt{1105}}{-18} when ± is minus. Subtract 20\sqrt{1105} from -2260.

$$x=\frac{-20\sqrt{1105}-2260}{-18}$$

Divide -2260-20\sqrt{1105} by -18.

$$x=\frac{10\sqrt{1105}+1130}{9}$$

The equation is now solved.

$$x=\frac{1130-10\sqrt{1105}}{9} x=\frac{10\sqrt{1105}+1130}{9}$$

Substitute \frac{1130-10\sqrt{1105}}{9} for x in the equation 3x+10\sqrt{x}+140=500.

$$3\times \frac{1130-10\sqrt{1105}}{9}+10\sqrt{\frac{1130-10\sqrt{1105}}{9}}+140=500$$

Simplify. The value x=\frac{1130-10\sqrt{1105}}{9} satisfies the equation.

$$500=500$$

Substitute \frac{10\sqrt{1105}+1130}{9} for x in the equation 3x+10\sqrt{x}+140=500.

$$3\times \frac{10\sqrt{1105}+1130}{9}+10\sqrt{\frac{10\sqrt{1105}+1130}{9}}+140=500$$

Simplify. The value x=\frac{10\sqrt{1105}+1130}{9} does not satisfy the equation.

$$\frac{20}{3}\times 1105^{\frac{1}{2}}+\frac{1600}{3}=500$$

Equation 10\sqrt{x}=360-3x has a unique solution.

$$x=\frac{1130-10\sqrt{1105}}{9}$$