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

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

Solve for x

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

Solution Steps

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

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

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

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

Subtract 140 from 500 to get 360.

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

Square both sides of the equation.

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

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

$$129600-2160x+9x^{2}=\left(10\sqrt{x}\right)^{2}$$

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

$$129600-2160x+9x^{2}=10^{2}\left(\sqrt{x}\right)^{2}$$

Calculate 10 to the power of 2 and get 100.

$$129600-2160x+9x^{2}=100\left(\sqrt{x}\right)^{2}$$

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

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

Subtract 100x from both sides.

$$129600-2160x+9x^{2}-100x=0$$

Combine -2160x and -100x to get -2260x.

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

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}.

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

Square -2260.

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

Multiply -4 times 9.

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

Multiply -36 times 129600.

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

Add 5107600 to -4665600.

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

Take the square root of 442000.

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

The opposite of -2260 is 2260.

$$x=\frac{2260±20\sqrt{1105}}{2\times 9}$$

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.

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

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

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

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

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

The equation is now solved.

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

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

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

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

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

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

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

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

$$500=500$$

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

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