Solve 15/x=72-6x/2x^2+2

$$\frac { 15 } { x } = \frac { 72 - 6 x } { 2 x ^ { 2 } } + 2$$

Solve for x

$$x=3$$

$$x=6$$

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 2x^{2}, the least common multiple of x,2x^{2}.

$$2x\times 15=72-6x+2x^{2}\times 2$$

Multiply 2 and 15 to get 30.

$$30x=72-6x+2x^{2}\times 2$$

Multiply 2 and 2 to get 4.

$$30x=72-6x+4x^{2}$$

Subtract 72 from both sides.

$$30x-72=-6x+4x^{2}$$

Add 6x to both sides.

$$30x-72+6x=4x^{2}$$

Combine 30x and 6x to get 36x.

$$36x-72=4x^{2}$$

Subtract 4x^{2} from both sides.

$$36x-72-4x^{2}=0$$

Divide both sides by 4.

$$9x-18-x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+9x-18=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-18. To find a and b, set up a system to be solved.

$$a+b=9 ab=-\left(-18\right)=18$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 18.

$$1,18 2,9 3,6$$

Calculate the sum for each pair.

$$1+18=19 2+9=11 3+6=9$$

The solution is the pair that gives sum 9.

$$a=6 b=3$$

Rewrite -x^{2}+9x-18 as \left(-x^{2}+6x\right)+\left(3x-18\right).

$$\left(-x^{2}+6x\right)+\left(3x-18\right)$$

Factor out -x in the first and 3 in the second group.

$$-x\left(x-6\right)+3\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(-x+3\right)$$

To find equation solutions, solve x-6=0 and -x+3=0.

$$x=6 x=3$$

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 2x^{2}, the least common multiple of x,2x^{2}.

$$2x\times 15=72-6x+2x^{2}\times 2$$

Multiply 2 and 15 to get 30.

$$30x=72-6x+2x^{2}\times 2$$

Multiply 2 and 2 to get 4.

$$30x=72-6x+4x^{2}$$

Subtract 72 from both sides.

$$30x-72=-6x+4x^{2}$$

Add 6x to both sides.

$$30x-72+6x=4x^{2}$$

Combine 30x and 6x to get 36x.

$$36x-72=4x^{2}$$

Subtract 4x^{2} from both sides.

$$36x-72-4x^{2}=0$$

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.

$$-4x^{2}+36x-72=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 36 for b, and -72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-36±\sqrt{36^{2}-4\left(-4\right)\left(-72\right)}}{2\left(-4\right)}$$

Square 36.

$$x=\frac{-36±\sqrt{1296-4\left(-4\right)\left(-72\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-36±\sqrt{1296+16\left(-72\right)}}{2\left(-4\right)}$$

Multiply 16 times -72.

$$x=\frac{-36±\sqrt{1296-1152}}{2\left(-4\right)}$$

Add 1296 to -1152.

$$x=\frac{-36±\sqrt{144}}{2\left(-4\right)}$$

Take the square root of 144.

$$x=\frac{-36±12}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-36±12}{-8}$$

Now solve the equation x=\frac{-36±12}{-8} when ± is plus. Add -36 to 12.

$$x=-\frac{24}{-8}$$

Divide -24 by -8.

$$x=3$$

Now solve the equation x=\frac{-36±12}{-8} when ± is minus. Subtract 12 from -36.

$$x=-\frac{48}{-8}$$

Divide -48 by -8.

$$x=6$$

The equation is now solved.

$$x=3 x=6$$

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 2x^{2}, the least common multiple of x,2x^{2}.

$$2x\times 15=72-6x+2x^{2}\times 2$$

Multiply 2 and 15 to get 30.

$$30x=72-6x+2x^{2}\times 2$$

Multiply 2 and 2 to get 4.

$$30x=72-6x+4x^{2}$$

Add 6x to both sides.

$$30x+6x=72+4x^{2}$$

Combine 30x and 6x to get 36x.

$$36x=72+4x^{2}$$

Subtract 4x^{2} from both sides.

$$36x-4x^{2}=72$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-4x^{2}+36x=72$$

Divide both sides by -4.

$$\frac{-4x^{2}+36x}{-4}=\frac{72}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{36}{-4}x=\frac{72}{-4}$$

Divide 36 by -4.

$$x^{2}-9x=\frac{72}{-4}$$

Divide 72 by -4.

$$x^{2}-9x=-18$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-9x+\left(-\frac{9}{2}\right)^{2}=-18+\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-9x+\frac{81}{4}=-18+\frac{81}{4}$$

Add -18 to \frac{81}{4}.

$$x^{2}-9x+\frac{81}{4}=\frac{9}{4}$$

Factor x^{2}-9x+\frac{81}{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}.

$$\left(x-\frac{9}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$x-\frac{9}{2}=\frac{3}{2} x-\frac{9}{2}=-\frac{3}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$x=6 x=3$$