Solve x^2+20/x(x-2)=14-x/x-2

$$\frac { x ^ { 2 } + 20 } { x ( x - 2 ) } = \frac { 14 - x } { x - 2 }$$

Solve for x

$$x=5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values 0,2 since division by zero is not defined. Multiply both sides of the equation by x\left(x-2\right), the least common multiple of x\left(x-2\right),x-2.

$$x^{2}+20=x\left(14-x\right)$$

Use the distributive property to multiply x by 14-x.

$$x^{2}+20=14x-x^{2}$$

Subtract 14x from both sides.

$$x^{2}+20-14x=-x^{2}$$

Add x^{2} to both sides.

$$x^{2}+20-14x+x^{2}=0$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+20-14x=0$$

Divide both sides by 2.

$$x^{2}+10-7x=0$$

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

$$x^{2}-7x+10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+10. To find a and b, set up a system to be solved.

$$a+b=-7 ab=1\times 10=10$$

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 10.

$$-1,-10 -2,-5$$

Calculate the sum for each pair.

$$-1-10=-11 -2-5=-7$$

The solution is the pair that gives sum -7.

$$a=-5 b=-2$$

Rewrite x^{2}-7x+10 as \left(x^{2}-5x\right)+\left(-2x+10\right).

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

Factor out x in the first and -2 in the second group.

$$x\left(x-5\right)-2\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x-2\right)$$

To find equation solutions, solve x-5=0 and x-2=0.

$$x=5 x=2$$

Variable x cannot be equal to 2.

$$x=5$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,2 since division by zero is not defined. Multiply both sides of the equation by x\left(x-2\right), the least common multiple of x\left(x-2\right),x-2.

$$x^{2}+20=x\left(14-x\right)$$

Use the distributive property to multiply x by 14-x.

$$x^{2}+20=14x-x^{2}$$

Subtract 14x from both sides.

$$x^{2}+20-14x=-x^{2}$$

Add x^{2} to both sides.

$$x^{2}+20-14x+x^{2}=0$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+20-14x=0$$

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.

$$2x^{2}-14x+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -14 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 2\times 20}}{2\times 2}$$

Square -14.

$$x=\frac{-\left(-14\right)±\sqrt{196-4\times 2\times 20}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-14\right)±\sqrt{196-8\times 20}}{2\times 2}$$

Multiply -8 times 20.

$$x=\frac{-\left(-14\right)±\sqrt{196-160}}{2\times 2}$$

Add 196 to -160.

$$x=\frac{-\left(-14\right)±\sqrt{36}}{2\times 2}$$

Take the square root of 36.

$$x=\frac{-\left(-14\right)±6}{2\times 2}$$

The opposite of -14 is 14.

$$x=\frac{14±6}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{14±6}{4}$$

Now solve the equation x=\frac{14±6}{4} when ± is plus. Add 14 to 6.

$$x=\frac{20}{4}$$

Divide 20 by 4.

$$x=5$$

Now solve the equation x=\frac{14±6}{4} when ± is minus. Subtract 6 from 14.

$$x=\frac{8}{4}$$

Divide 8 by 4.

$$x=2$$

The equation is now solved.

$$x=5 x=2$$

Variable x cannot be equal to 2.

$$x=5$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,2 since division by zero is not defined. Multiply both sides of the equation by x\left(x-2\right), the least common multiple of x\left(x-2\right),x-2.

$$x^{2}+20=x\left(14-x\right)$$

Use the distributive property to multiply x by 14-x.

$$x^{2}+20=14x-x^{2}$$

Subtract 14x from both sides.

$$x^{2}+20-14x=-x^{2}$$

Add x^{2} to both sides.

$$x^{2}+20-14x+x^{2}=0$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+20-14x=0$$

Subtract 20 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}-14x=-20$$

Divide both sides by 2.

$$\frac{2x^{2}-14x}{2}=-\frac{20}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{14}{2}\right)x=-\frac{20}{2}$$

Divide -14 by 2.

$$x^{2}-7x=-\frac{20}{2}$$

Divide -20 by 2.

$$x^{2}-7x=-10$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=-10+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=-10+\frac{49}{4}$$

Add -10 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{9}{4}$$

Factor x^{2}-7x+\frac{49}{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}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{3}{2} x-\frac{7}{2}=-\frac{3}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$x=5 x=2$$

Variable x cannot be equal to 2.

$$x=5$$