Solve x^2+14/x-2=9x/x-2

$$\frac { x ^ { 2 } + 14 } { x - 2 } = \frac { 9 x } { x - 2 }$$

Solve for x

$$x=7$$

Steps Using Factoring

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2.

$$x^{2}+14=9x$$

Subtract 9x from both sides.

$$x^{2}+14-9x=0$$

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

$$x^{2}-9x+14=0$$

To solve the equation, factor x^{2}-9x+14 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.

$$a+b=-9 ab=14$$

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 14.

$$-1,-14 -2,-7$$

Calculate the sum for each pair.

$$-1-14=-15 -2-7=-9$$

The solution is the pair that gives sum -9.

$$a=-7 b=-2$$

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

$$\left(x-7\right)\left(x-2\right)$$

To find equation solutions, solve x-7=0 and x-2=0.

$$x=7 x=2$$

Variable x cannot be equal to 2.

$$x=7$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2.

$$x^{2}+14=9x$$

Subtract 9x from both sides.

$$x^{2}+14-9x=0$$

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

$$x^{2}-9x+14=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+14. To find a and b, set up a system to be solved.

$$a+b=-9 ab=1\times 14=14$$

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 14.

$$-1,-14 -2,-7$$

Calculate the sum for each pair.

$$-1-14=-15 -2-7=-9$$

The solution is the pair that gives sum -9.

$$a=-7 b=-2$$

Rewrite x^{2}-9x+14 as \left(x^{2}-7x\right)+\left(-2x+14\right).

$$\left(x^{2}-7x\right)+\left(-2x+14\right)$$

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

$$x\left(x-7\right)-2\left(x-7\right)$$

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

$$\left(x-7\right)\left(x-2\right)$$

To find equation solutions, solve x-7=0 and x-2=0.

$$x=7 x=2$$

Variable x cannot be equal to 2.

$$x=7$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2.

$$x^{2}+14=9x$$

Subtract 9x from both sides.

$$x^{2}+14-9x=0$$

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.

$$x^{2}-9x+14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -9 for b, and 14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 14}}{2}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 14}}{2}$$

Multiply -4 times 14.

$$x=\frac{-\left(-9\right)±\sqrt{81-56}}{2}$$

Add 81 to -56.

$$x=\frac{-\left(-9\right)±\sqrt{25}}{2}$$

Take the square root of 25.

$$x=\frac{-\left(-9\right)±5}{2}$$

The opposite of -9 is 9.

$$x=\frac{9±5}{2}$$

Now solve the equation x=\frac{9±5}{2} when ± is plus. Add 9 to 5.

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

Divide 14 by 2.

$$x=7$$

Now solve the equation x=\frac{9±5}{2} when ± is minus. Subtract 5 from 9.

$$x=\frac{4}{2}$$

Divide 4 by 2.

$$x=2$$

The equation is now solved.

$$x=7 x=2$$

Variable x cannot be equal to 2.

$$x=7$$

Steps for Completing the Square

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2.

$$x^{2}+14=9x$$

Subtract 9x from both sides.

$$x^{2}+14-9x=0$$

Subtract 14 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-9x=-14$$

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.

$$x^{2}-9x+\left(-\frac{9}{2}\right)^{2}=-14+\left(-\frac{9}{2}\right)^{2}$$

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

$$x^{2}-9x+\frac{81}{4}=-14+\frac{81}{4}$$

Add -14 to \frac{81}{4}.

$$x^{2}-9x+\frac{81}{4}=\frac{25}{4}$$

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}.

$$\left(x-\frac{9}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x-\frac{9}{2}=\frac{5}{2} x-\frac{9}{2}=-\frac{5}{2}$$

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

$$x=7 x=2$$

Variable x cannot be equal to 2.

$$x=7$$