Solve 6/x+2+1/x-2=1

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

Solve for x

$$x=1$$

$$x=6$$

Steps Using the Quadratic Formula

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

$$\left(x-2\right)\times 6+x+2=\left(x-2\right)\left(x+2\right)$$

Use the distributive property to multiply x-2 by 6.

$$6x-12+x+2=\left(x-2\right)\left(x+2\right)$$

Combine 6x and x to get 7x.

$$7x-12+2=\left(x-2\right)\left(x+2\right)$$

Add -12 and 2 to get -10.

$$7x-10=\left(x-2\right)\left(x+2\right)$$

Consider \left(x-2\right)\left(x+2\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 2.

$$7x-10=x^{2}-4$$

Subtract x^{2} from both sides.

$$7x-10-x^{2}=-4$$

Add 4 to both sides.

$$7x-10-x^{2}+4=0$$

Add -10 and 4 to get -6.

$$7x-6-x^{2}=0$$

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.

$$-x^{2}+7x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 7 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\left(-1\right)\left(-6\right)}}{2\left(-1\right)}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\left(-1\right)\left(-6\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-7±\sqrt{49+4\left(-6\right)}}{2\left(-1\right)}$$

Multiply 4 times -6.

$$x=\frac{-7±\sqrt{49-24}}{2\left(-1\right)}$$

Add 49 to -24.

$$x=\frac{-7±\sqrt{25}}{2\left(-1\right)}$$

Take the square root of 25.

$$x=\frac{-7±5}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-7±5}{-2}$$

Now solve the equation x=\frac{-7±5}{-2} when ± is plus. Add -7 to 5.

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

Divide -2 by -2.

$$x=1$$

Now solve the equation x=\frac{-7±5}{-2} when ± is minus. Subtract 5 from -7.

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

Divide -12 by -2.

$$x=6$$

The equation is now solved.

$$x=1 x=6$$

Steps for Completing the Square

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

$$\left(x-2\right)\times 6+x+2=\left(x-2\right)\left(x+2\right)$$

Use the distributive property to multiply x-2 by 6.

$$6x-12+x+2=\left(x-2\right)\left(x+2\right)$$

Combine 6x and x to get 7x.

$$7x-12+2=\left(x-2\right)\left(x+2\right)$$

Add -12 and 2 to get -10.

$$7x-10=\left(x-2\right)\left(x+2\right)$$

Consider \left(x-2\right)\left(x+2\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 2.

$$7x-10=x^{2}-4$$

Subtract x^{2} from both sides.

$$7x-10-x^{2}=-4$$

Add 10 to both sides.

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

Add -4 and 10 to get 6.

$$7x-x^{2}=6$$

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.

$$-x^{2}+7x=6$$

Divide both sides by -1.

$$\frac{-x^{2}+7x}{-1}=\frac{6}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{7}{-1}x=\frac{6}{-1}$$

Divide 7 by -1.

$$x^{2}-7x=\frac{6}{-1}$$

Divide 6 by -1.

$$x^{2}-7x=-6$$

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.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=-6+\left(-\frac{7}{2}\right)^{2}$$

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

$$x^{2}-7x+\frac{49}{4}=-6+\frac{49}{4}$$

Add -6 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{25}{4}$$

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}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{5}{2} x-\frac{7}{2}=-\frac{5}{2}$$

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

$$x=6 x=1$$