Solve 3x^2/x-10=15x/x-10

$$\frac{ 3 { x }^{ 2 } }{ x-10 } = \frac{ 15x }{ x-10 }$$

Solve for x

$$x=5$$

$$x=0$$

Steps Using Factoring

Variable x cannot be equal to 10 since division by zero is not defined. Multiply both sides of the equation by x-10.

$$3x^{2}=15x$$

Subtract 15x from both sides.

$$3x^{2}-15x=0$$

Factor out x.

$$x\left(3x-15\right)=0$$

To find equation solutions, solve x=0 and 3x-15=0.

$$x=0 x=5$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 10 since division by zero is not defined. Multiply both sides of the equation by x-10.

$$3x^{2}=15x$$

Subtract 15x from both sides.

$$3x^{2}-15x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -15 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}}}{2\times 3}$$

Take the square root of \left(-15\right)^{2}.

$$x=\frac{-\left(-15\right)±15}{2\times 3}$$

The opposite of -15 is 15.

$$x=\frac{15±15}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{15±15}{6}$$

Now solve the equation x=\frac{15±15}{6} when ± is plus. Add 15 to 15.

$$x=\frac{30}{6}$$

Divide 30 by 6.

$$x=5$$

Now solve the equation x=\frac{15±15}{6} when ± is minus. Subtract 15 from 15.

$$x=\frac{0}{6}$$

Divide 0 by 6.

$$x=0$$

The equation is now solved.

$$x=5 x=0$$

Steps for Completing the Square

Variable x cannot be equal to 10 since division by zero is not defined. Multiply both sides of the equation by x-10.

$$3x^{2}=15x$$

Subtract 15x from both sides.

$$3x^{2}-15x=0$$

Divide both sides by 3.

$$\frac{3x^{2}-15x}{3}=\frac{0}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\left(-\frac{15}{3}\right)x=\frac{0}{3}$$

Divide -15 by 3.

$$x^{2}-5x=\frac{0}{3}$$

Divide 0 by 3.

$$x^{2}-5x=0$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=\frac{25}{4}$$

Factor x^{2}-5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{5}{2} x-\frac{5}{2}=-\frac{5}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=5 x=0$$