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

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

Solve for x

$$x=-4$$

Steps Using Factoring By Grouping

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

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

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

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

Subtract 3 from 6 to get 3.

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

Consider \left(x-3\right)\left(x+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$3-x=x^{2}-9$$

Subtract x^{2} from both sides.

$$3-x-x^{2}=-9$$

Add 9 to both sides.

$$3-x-x^{2}+9=0$$

Add 3 and 9 to get 12.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-12=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -1.

$$a=3 b=-4$$

Rewrite -x^{2}-x+12 as \left(-x^{2}+3x\right)+\left(-4x+12\right).

$$\left(-x^{2}+3x\right)+\left(-4x+12\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(-x+3\right)+4\left(-x+3\right)$$

Factor out common term -x+3 by using distributive property.

$$\left(-x+3\right)\left(x+4\right)$$

To find equation solutions, solve -x+3=0 and x+4=0.

$$x=3 x=-4$$

Variable x cannot be equal to 3.

$$x=-4$$

Steps Using the Quadratic Formula

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

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

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

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

Subtract 3 from 6 to get 3.

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

Consider \left(x-3\right)\left(x+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$3-x=x^{2}-9$$

Subtract x^{2} from both sides.

$$3-x-x^{2}=-9$$

Add 9 to both sides.

$$3-x-x^{2}+9=0$$

Add 3 and 9 to get 12.

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

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.

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

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

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-1\right)\times 12}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-1\right)±\sqrt{1+4\times 12}}{2\left(-1\right)}$$

Multiply 4 times 12.

$$x=\frac{-\left(-1\right)±\sqrt{1+48}}{2\left(-1\right)}$$

Add 1 to 48.

$$x=\frac{-\left(-1\right)±\sqrt{49}}{2\left(-1\right)}$$

Take the square root of 49.

$$x=\frac{-\left(-1\right)±7}{2\left(-1\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±7}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{1±7}{-2}$$

Now solve the equation x=\frac{1±7}{-2} when ± is plus. Add 1 to 7.

$$x=\frac{8}{-2}$$

Divide 8 by -2.

$$x=-4$$

Now solve the equation x=\frac{1±7}{-2} when ± is minus. Subtract 7 from 1.

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

Divide -6 by -2.

$$x=3$$

The equation is now solved.

$$x=-4 x=3$$

Variable x cannot be equal to 3.

$$x=-4$$

Steps for Completing the Square

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

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

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

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

Subtract 3 from 6 to get 3.

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

Consider \left(x-3\right)\left(x+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$3-x=x^{2}-9$$

Subtract x^{2} from both sides.

$$3-x-x^{2}=-9$$

Subtract 3 from both sides.

$$-x-x^{2}=-9-3$$

Subtract 3 from -9 to get -12.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{1}{-1}\right)x=-\frac{12}{-1}$$

Divide -1 by -1.

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

Divide -12 by -1.

$$x^{2}+x=12$$

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=12+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=12+\frac{1}{4}$$

Add 12 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{49}{4}$$

Factor x^{2}+x+\frac{1}{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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{7}{2} x+\frac{1}{2}=-\frac{7}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=3 x=-4$$

Variable x cannot be equal to 3.

$$x=-4$$