Solve x^2-9/x^2-6x+9-1/x-3=2

$$\frac { x ^ { 2 } - 9 } { x ^ { 2 } - 6 x + 9 } - \frac { 1 } { x - 3 } = 2$$

Solve for x

$$x=8$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)^{2}, the least common multiple of x^{2}-6x+9,x-3.

$$x^{2}-9-\left(x-3\right)=2\left(x-3\right)^{2}$$

To find the opposite of x-3, find the opposite of each term.

$$x^{2}-9-x+3=2\left(x-3\right)^{2}$$

Add -9 and 3 to get -6.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6-x=2\left(x^{2}-6x+9\right)$$

Use the distributive property to multiply 2 by x^{2}-6x+9.

$$x^{2}-6-x=2x^{2}-12x+18$$

Subtract 2x^{2} from both sides.

$$x^{2}-6-x-2x^{2}=-12x+18$$

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

$$-x^{2}-6-x=-12x+18$$

Add 12x to both sides.

$$-x^{2}-6-x+12x=18$$

Combine -x and 12x to get 11x.

$$-x^{2}-6+11x=18$$

Subtract 18 from both sides.

$$-x^{2}-6+11x-18=0$$

Subtract 18 from -6 to get -24.

$$-x^{2}-24+11x=0$$

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

$$-x^{2}+11x-24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-24. To find a and b, set up a system to be solved.

$$a+b=11 ab=-\left(-24\right)=24$$

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 24.

$$1,24 2,12 3,8 4,6$$

Calculate the sum for each pair.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

The solution is the pair that gives sum 11.

$$a=8 b=3$$

Rewrite -x^{2}+11x-24 as \left(-x^{2}+8x\right)+\left(3x-24\right).

$$\left(-x^{2}+8x\right)+\left(3x-24\right)$$

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

$$-x\left(x-8\right)+3\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(-x+3\right)$$

To find equation solutions, solve x-8=0 and -x+3=0.

$$x=8 x=3$$

Variable x cannot be equal to 3.

$$x=8$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)^{2}, the least common multiple of x^{2}-6x+9,x-3.

$$x^{2}-9-\left(x-3\right)=2\left(x-3\right)^{2}$$

To find the opposite of x-3, find the opposite of each term.

$$x^{2}-9-x+3=2\left(x-3\right)^{2}$$

Add -9 and 3 to get -6.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6-x=2\left(x^{2}-6x+9\right)$$

Use the distributive property to multiply 2 by x^{2}-6x+9.

$$x^{2}-6-x=2x^{2}-12x+18$$

Subtract 2x^{2} from both sides.

$$x^{2}-6-x-2x^{2}=-12x+18$$

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

$$-x^{2}-6-x=-12x+18$$

Add 12x to both sides.

$$-x^{2}-6-x+12x=18$$

Combine -x and 12x to get 11x.

$$-x^{2}-6+11x=18$$

Subtract 18 from both sides.

$$-x^{2}-6+11x-18=0$$

Subtract 18 from -6 to get -24.

$$-x^{2}-24+11x=0$$

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.

$$-x^{2}+11x-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 11 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\left(-1\right)\left(-24\right)}}{2\left(-1\right)}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\left(-1\right)\left(-24\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-11±\sqrt{121+4\left(-24\right)}}{2\left(-1\right)}$$

Multiply 4 times -24.

$$x=\frac{-11±\sqrt{121-96}}{2\left(-1\right)}$$

Add 121 to -96.

$$x=\frac{-11±\sqrt{25}}{2\left(-1\right)}$$

Take the square root of 25.

$$x=\frac{-11±5}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-11±5}{-2}$$

Now solve the equation x=\frac{-11±5}{-2} when ± is plus. Add -11 to 5.

$$x=-\frac{6}{-2}$$

Divide -6 by -2.

$$x=3$$

Now solve the equation x=\frac{-11±5}{-2} when ± is minus. Subtract 5 from -11.

$$x=-\frac{16}{-2}$$

Divide -16 by -2.

$$x=8$$

The equation is now solved.

$$x=3 x=8$$

Variable x cannot be equal to 3.

$$x=8$$

Steps for Completing the Square

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)^{2}, the least common multiple of x^{2}-6x+9,x-3.

$$x^{2}-9-\left(x-3\right)=2\left(x-3\right)^{2}$$

To find the opposite of x-3, find the opposite of each term.

$$x^{2}-9-x+3=2\left(x-3\right)^{2}$$

Add -9 and 3 to get -6.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6-x=2\left(x^{2}-6x+9\right)$$

Use the distributive property to multiply 2 by x^{2}-6x+9.

$$x^{2}-6-x=2x^{2}-12x+18$$

Subtract 2x^{2} from both sides.

$$x^{2}-6-x-2x^{2}=-12x+18$$

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

$$-x^{2}-6-x=-12x+18$$

Add 12x to both sides.

$$-x^{2}-6-x+12x=18$$

Combine -x and 12x to get 11x.

$$-x^{2}-6+11x=18$$

Add 6 to both sides.

$$-x^{2}+11x=18+6$$

Add 18 and 6 to get 24.

$$-x^{2}+11x=24$$

Divide both sides by -1.

$$\frac{-x^{2}+11x}{-1}=\frac{24}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{11}{-1}x=\frac{24}{-1}$$

Divide 11 by -1.

$$x^{2}-11x=\frac{24}{-1}$$

Divide 24 by -1.

$$x^{2}-11x=-24$$

Divide -11, the coefficient of the x term, by 2 to get -\frac{11}{2}. Then add the square of -\frac{11}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-11x+\left(-\frac{11}{2}\right)^{2}=-24+\left(-\frac{11}{2}\right)^{2}$$

Square -\frac{11}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-11x+\frac{121}{4}=-24+\frac{121}{4}$$

Add -24 to \frac{121}{4}.

$$x^{2}-11x+\frac{121}{4}=\frac{25}{4}$$

Factor x^{2}-11x+\frac{121}{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}.

$$\left(x-\frac{11}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x-\frac{11}{2}=\frac{5}{2} x-\frac{11}{2}=-\frac{5}{2}$$

Add \frac{11}{2} to both sides of the equation.

$$x=8 x=3$$

Variable x cannot be equal to 3.

$$x=8$$