Solve 10+sqrt{25x}=10x*15

$$10+ \sqrt{ 25x } =10x \times 15$$

Solve for x

$$x=\frac{\sqrt{241}+121}{1800}\approx 0.075846764$$

Solution Steps

Subtract 10 from both sides of the equation.

$$\sqrt{25x}=10x\times 15-10$$

Multiply 10 and 15 to get 150.

$$\sqrt{25x}=150x-10$$

Square both sides of the equation.

$$\left(\sqrt{25x}\right)^{2}=\left(150x-10\right)^{2}$$

Calculate \sqrt{25x} to the power of 2 and get 25x.

$$25x=\left(150x-10\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(150x-10\right)^{2}.

$$25x=22500x^{2}-3000x+100$$

Subtract 22500x^{2} from both sides.

$$25x-22500x^{2}=-3000x+100$$

Add 3000x to both sides.

$$25x-22500x^{2}+3000x=100$$

Combine 25x and 3000x to get 3025x.

$$3025x-22500x^{2}=100$$

Subtract 100 from both sides.

$$3025x-22500x^{2}-100=0$$

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.

$$-22500x^{2}+3025x-100=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -22500 for a, 3025 for b, and -100 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3025±\sqrt{3025^{2}-4\left(-22500\right)\left(-100\right)}}{2\left(-22500\right)}$$

Square 3025.

$$x=\frac{-3025±\sqrt{9150625-4\left(-22500\right)\left(-100\right)}}{2\left(-22500\right)}$$

Multiply -4 times -22500.

$$x=\frac{-3025±\sqrt{9150625+90000\left(-100\right)}}{2\left(-22500\right)}$$

Multiply 90000 times -100.

$$x=\frac{-3025±\sqrt{9150625-9000000}}{2\left(-22500\right)}$$

Add 9150625 to -9000000.

$$x=\frac{-3025±\sqrt{150625}}{2\left(-22500\right)}$$

Take the square root of 150625.

$$x=\frac{-3025±25\sqrt{241}}{2\left(-22500\right)}$$

Multiply 2 times -22500.

$$x=\frac{-3025±25\sqrt{241}}{-45000}$$

Now solve the equation x=\frac{-3025±25\sqrt{241}}{-45000} when ± is plus. Add -3025 to 25\sqrt{241}.

$$x=\frac{25\sqrt{241}-3025}{-45000}$$

Divide -3025+25\sqrt{241} by -45000.

$$x=\frac{121-\sqrt{241}}{1800}$$

Now solve the equation x=\frac{-3025±25\sqrt{241}}{-45000} when ± is minus. Subtract 25\sqrt{241} from -3025.

$$x=\frac{-25\sqrt{241}-3025}{-45000}$$

Divide -3025-25\sqrt{241} by -45000.

$$x=\frac{\sqrt{241}+121}{1800}$$

The equation is now solved.

$$x=\frac{121-\sqrt{241}}{1800} x=\frac{\sqrt{241}+121}{1800}$$

Substitute \frac{121-\sqrt{241}}{1800} for x in the equation 10+\sqrt{25x}=10x\times 15.

$$10+\sqrt{25\times \frac{121-\sqrt{241}}{1800}}=10\times \frac{121-\sqrt{241}}{1800}\times 15$$

Simplify. The value x=\frac{121-\sqrt{241}}{1800} does not satisfy the equation.

$$\frac{119}{12}+\frac{1}{12}\times 241^{\frac{1}{2}}=\frac{121}{12}-\frac{1}{12}\times 241^{\frac{1}{2}}$$

Substitute \frac{\sqrt{241}+121}{1800} for x in the equation 10+\sqrt{25x}=10x\times 15.

$$10+\sqrt{25\times \frac{\sqrt{241}+121}{1800}}=10\times \frac{\sqrt{241}+121}{1800}\times 15$$

Simplify. The value x=\frac{\sqrt{241}+121}{1800} satisfies the equation.

$$\frac{121}{12}+\frac{1}{12}\times 241^{\frac{1}{2}}=\frac{1}{12}\times 241^{\frac{1}{2}}+\frac{121}{12}$$

Equation \sqrt{25x}=150x-10 has a unique solution.

$$x=\frac{\sqrt{241}+121}{1800}$$