Solve 1-15/x+50/x^2=0

$$1 - \frac { 15 } { x } + \frac { 50 } { x ^ { 2 } } = 0$$

Solve for x

$$x=5$$

$$x=10$$

Steps Using Factoring

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}, the least common multiple of x,x^{2}.

$$x^{2}-x\times 15+50=0$$

Multiply -1 and 15 to get -15.

$$x^{2}-15x+50=0$$

To solve the equation, factor x^{2}-15x+50 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-15 ab=50$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 50.

$$-1,-50 -2,-25 -5,-10$$

Calculate the sum for each pair.

$$-1-50=-51 -2-25=-27 -5-10=-15$$

The solution is the pair that gives sum -15.

$$a=-10 b=-5$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-10\right)\left(x-5\right)$$

To find equation solutions, solve x-10=0 and x-5=0.

$$x=10 x=5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}, the least common multiple of x,x^{2}.

$$x^{2}-x\times 15+50=0$$

Multiply -1 and 15 to get -15.

$$x^{2}-15x+50=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+50. To find a and b, set up a system to be solved.

$$a+b=-15 ab=1\times 50=50$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 50.

$$-1,-50 -2,-25 -5,-10$$

Calculate the sum for each pair.

$$-1-50=-51 -2-25=-27 -5-10=-15$$

The solution is the pair that gives sum -15.

$$a=-10 b=-5$$

Rewrite x^{2}-15x+50 as \left(x^{2}-10x\right)+\left(-5x+50\right).

$$\left(x^{2}-10x\right)+\left(-5x+50\right)$$

Factor out x in the first and -5 in the second group.

$$x\left(x-10\right)-5\left(x-10\right)$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(x-5\right)$$

To find equation solutions, solve x-10=0 and x-5=0.

$$x=10 x=5$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}, the least common multiple of x,x^{2}.

$$x^{2}-x\times 15+50=0$$

Multiply -1 and 15 to get -15.

$$x^{2}-15x+50=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -15 for b, and 50 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 50}}{2}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 50}}{2}$$

Multiply -4 times 50.

$$x=\frac{-\left(-15\right)±\sqrt{225-200}}{2}$$

Add 225 to -200.

$$x=\frac{-\left(-15\right)±\sqrt{25}}{2}$$

Take the square root of 25.

$$x=\frac{-\left(-15\right)±5}{2}$$

The opposite of -15 is 15.

$$x=\frac{15±5}{2}$$

Now solve the equation x=\frac{15±5}{2} when ± is plus. Add 15 to 5.

$$x=\frac{20}{2}$$

Divide 20 by 2.

$$x=10$$

Now solve the equation x=\frac{15±5}{2} when ± is minus. Subtract 5 from 15.

$$x=\frac{10}{2}$$

Divide 10 by 2.

$$x=5$$

The equation is now solved.

$$x=10 x=5$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}, the least common multiple of x,x^{2}.

$$x^{2}-x\times 15+50=0$$

Subtract 50 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-x\times 15=-50$$

Multiply -1 and 15 to get -15.

$$x^{2}-15x=-50$$

Divide -15, the coefficient of the x term, by 2 to get -\frac{15}{2}. Then add the square of -\frac{15}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-15x+\left(-\frac{15}{2}\right)^{2}=-50+\left(-\frac{15}{2}\right)^{2}$$

Square -\frac{15}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-15x+\frac{225}{4}=-50+\frac{225}{4}$$

Add -50 to \frac{225}{4}.

$$x^{2}-15x+\frac{225}{4}=\frac{25}{4}$$

Factor x^{2}-15x+\frac{225}{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}.

$$\left(x-\frac{15}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{15}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x-\frac{15}{2}=\frac{5}{2} x-\frac{15}{2}=-\frac{5}{2}$$

Add \frac{15}{2} to both sides of the equation.

$$x=10 x=5$$