Solve 1-12/x+35/x^2=0

$$1 - \frac { 12 } { x } + \frac { 35 } { x ^ { 2 } } = 0$$

Solve for x

$$x=5$$

$$x=7$$

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}.

$$x^{2}-x\times 12+35=0$$

Multiply -1 and 12 to get -12.

$$x^{2}-12x+35=0$$

To solve the equation, factor x^{2}-12x+35 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.

$$a+b=-12 ab=35$$

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 35.

$$-1,-35 -5,-7$$

Calculate the sum for each pair.

$$-1-35=-36 -5-7=-12$$

The solution is the pair that gives sum -12.

$$a=-7 b=-5$$

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

$$\left(x-7\right)\left(x-5\right)$$

To find equation solutions, solve x-7=0 and x-5=0.

$$x=7 x=5$$

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}.

$$x^{2}-x\times 12+35=0$$

Multiply -1 and 12 to get -12.

$$x^{2}-12x+35=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+35. To find a and b, set up a system to be solved.

$$a+b=-12 ab=1\times 35=35$$

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 35.

$$-1,-35 -5,-7$$

Calculate the sum for each pair.

$$-1-35=-36 -5-7=-12$$

The solution is the pair that gives sum -12.

$$a=-7 b=-5$$

Rewrite x^{2}-12x+35 as \left(x^{2}-7x\right)+\left(-5x+35\right).

$$\left(x^{2}-7x\right)+\left(-5x+35\right)$$

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

$$x\left(x-7\right)-5\left(x-7\right)$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(x-5\right)$$

To find equation solutions, solve x-7=0 and x-5=0.

$$x=7 x=5$$

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}.

$$x^{2}-x\times 12+35=0$$

Multiply -1 and 12 to get -12.

$$x^{2}-12x+35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -12 for b, and 35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 35}}{2}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 35}}{2}$$

Multiply -4 times 35.

$$x=\frac{-\left(-12\right)±\sqrt{144-140}}{2}$$

Add 144 to -140.

$$x=\frac{-\left(-12\right)±\sqrt{4}}{2}$$

Take the square root of 4.

$$x=\frac{-\left(-12\right)±2}{2}$$

The opposite of -12 is 12.

$$x=\frac{12±2}{2}$$

Now solve the equation x=\frac{12±2}{2} when ± is plus. Add 12 to 2.

$$x=\frac{14}{2}$$

Divide 14 by 2.

$$x=7$$

Now solve the equation x=\frac{12±2}{2} when ± is minus. Subtract 2 from 12.

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

Divide 10 by 2.

$$x=5$$

The equation is now solved.

$$x=7 x=5$$

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}.

$$x^{2}-x\times 12+35=0$$

Subtract 35 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-x\times 12=-35$$

Multiply -1 and 12 to get -12.

$$x^{2}-12x=-35$$

Divide -12, the coefficient of the x term, by 2 to get -6. Then add the square of -6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-12x+\left(-6\right)^{2}=-35+\left(-6\right)^{2}$$

Square -6.

$$x^{2}-12x+36=-35+36$$

Add -35 to 36.

$$x^{2}-12x+36=1$$

Factor x^{2}-12x+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-6\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-6\right)^{2}}=\sqrt{1}$$

Simplify.

$$x-6=1 x-6=-1$$

Add 6 to both sides of the equation.

$$x=7 x=5$$