Solve 100sqrt{x}=2x+100

$$100 \sqrt{ x } = 2x+100$$

Solve for x

$$x=1200-250\sqrt{23}\approx 1.042119172$$

$$x=250\sqrt{23}+1200\approx 2398.957880828$$

Solution Steps

Square both sides of the equation.

$$\left(100\sqrt{x}\right)^{2}=\left(2x+100\right)^{2}$$

Expand \left(100\sqrt{x}\right)^{2}.

$$100^{2}\left(\sqrt{x}\right)^{2}=\left(2x+100\right)^{2}$$

Calculate 100 to the power of 2 and get 10000.

$$10000\left(\sqrt{x}\right)^{2}=\left(2x+100\right)^{2}$$

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

$$10000x=\left(2x+100\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+100\right)^{2}.

$$10000x=4x^{2}+400x+10000$$

Subtract 4x^{2} from both sides.

$$10000x-4x^{2}=400x+10000$$

Subtract 400x from both sides.

$$10000x-4x^{2}-400x=10000$$

Combine 10000x and -400x to get 9600x.

$$9600x-4x^{2}=10000$$

Subtract 10000 from both sides.

$$9600x-4x^{2}-10000=0$$

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.

$$-4x^{2}+9600x-10000=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 9600 for b, and -10000 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9600±\sqrt{9600^{2}-4\left(-4\right)\left(-10000\right)}}{2\left(-4\right)}$$

Square 9600.

$$x=\frac{-9600±\sqrt{92160000-4\left(-4\right)\left(-10000\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-9600±\sqrt{92160000+16\left(-10000\right)}}{2\left(-4\right)}$$

Multiply 16 times -10000.

$$x=\frac{-9600±\sqrt{92160000-160000}}{2\left(-4\right)}$$

Add 92160000 to -160000.

$$x=\frac{-9600±\sqrt{92000000}}{2\left(-4\right)}$$

Take the square root of 92000000.

$$x=\frac{-9600±2000\sqrt{23}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-9600±2000\sqrt{23}}{-8}$$

Now solve the equation x=\frac{-9600±2000\sqrt{23}}{-8} when ± is plus. Add -9600 to 2000\sqrt{23}.

$$x=\frac{2000\sqrt{23}-9600}{-8}$$

Divide -9600+2000\sqrt{23} by -8.

$$x=1200-250\sqrt{23}$$

Now solve the equation x=\frac{-9600±2000\sqrt{23}}{-8} when ± is minus. Subtract 2000\sqrt{23} from -9600.

$$x=\frac{-2000\sqrt{23}-9600}{-8}$$

Divide -9600-2000\sqrt{23} by -8.

$$x=250\sqrt{23}+1200$$

The equation is now solved.

$$x=1200-250\sqrt{23} x=250\sqrt{23}+1200$$

Substitute 1200-250\sqrt{23} for x in the equation 100\sqrt{x}=2x+100.

$$100\sqrt{1200-250\sqrt{23}}=2\left(1200-250\sqrt{23}\right)+100$$

Simplify. The value x=1200-250\sqrt{23} satisfies the equation.

$$2500-500\times 23^{\frac{1}{2}}=2500-500\times 23^{\frac{1}{2}}$$

Substitute 250\sqrt{23}+1200 for x in the equation 100\sqrt{x}=2x+100.

$$100\sqrt{250\sqrt{23}+1200}=2\left(250\sqrt{23}+1200\right)+100$$

Simplify. The value x=250\sqrt{23}+1200 satisfies the equation.

$$2500+500\times 23^{\frac{1}{2}}=500\times 23^{\frac{1}{2}}+2500$$

List all solutions of 100\sqrt{x}=2x+100.

$$x=1200-250\sqrt{23} x=250\sqrt{23}+1200$$