Solve 1/left(x-2right)^2-1/x-2-6=0

$$\frac{ 1 }{ { \left(x-2 \right) }^{ 2 } } - \frac{ 1 }{ x-2 } -6=0$$

Solve for x

$$x = \frac{7}{3} = 2\frac{1}{3} \approx 2.333333333$$

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

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 \left(x-2\right)^{2}, the least common multiple of \left(x-2\right)^{2},x-2.

$$1-\left(x-2\right)+\left(x-2\right)^{2}\left(-6\right)=0$$

To find the opposite of x-2, find the opposite of each term.

$$1-x+2+\left(x-2\right)^{2}\left(-6\right)=0$$

Add 1 and 2 to get 3.

$$3-x+\left(x-2\right)^{2}\left(-6\right)=0$$

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

$$3-x+\left(x^{2}-4x+4\right)\left(-6\right)=0$$

Use the distributive property to multiply x^{2}-4x+4 by -6.

$$3-x-6x^{2}+24x-24=0$$

Combine -x and 24x to get 23x.

$$3+23x-6x^{2}-24=0$$

Subtract 24 from 3 to get -21.

$$-21+23x-6x^{2}=0$$

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

$$-6x^{2}+23x-21=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -6x^{2}+ax+bx-21. To find a and b, set up a system to be solved.

$$a+b=23 ab=-6\left(-21\right)=126$$

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 126.

$$1,126 2,63 3,42 6,21 7,18 9,14$$

Calculate the sum for each pair.

$$1+126=127 2+63=65 3+42=45 6+21=27 7+18=25 9+14=23$$

The solution is the pair that gives sum 23.

$$a=14 b=9$$

Rewrite -6x^{2}+23x-21 as \left(-6x^{2}+14x\right)+\left(9x-21\right).

$$\left(-6x^{2}+14x\right)+\left(9x-21\right)$$

Factor out 2x in the first and -3 in the second group.

$$2x\left(-3x+7\right)-3\left(-3x+7\right)$$

Factor out common term -3x+7 by using distributive property.

$$\left(-3x+7\right)\left(2x-3\right)$$

To find equation solutions, solve -3x+7=0 and 2x-3=0.

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

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 \left(x-2\right)^{2}, the least common multiple of \left(x-2\right)^{2},x-2.

$$1-\left(x-2\right)+\left(x-2\right)^{2}\left(-6\right)=0$$

To find the opposite of x-2, find the opposite of each term.

$$1-x+2+\left(x-2\right)^{2}\left(-6\right)=0$$

Add 1 and 2 to get 3.

$$3-x+\left(x-2\right)^{2}\left(-6\right)=0$$

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

$$3-x+\left(x^{2}-4x+4\right)\left(-6\right)=0$$

Use the distributive property to multiply x^{2}-4x+4 by -6.

$$3-x-6x^{2}+24x-24=0$$

Combine -x and 24x to get 23x.

$$3+23x-6x^{2}-24=0$$

Subtract 24 from 3 to get -21.

$$-21+23x-6x^{2}=0$$

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.

$$-6x^{2}+23x-21=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, 23 for b, and -21 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-23±\sqrt{23^{2}-4\left(-6\right)\left(-21\right)}}{2\left(-6\right)}$$

Square 23.

$$x=\frac{-23±\sqrt{529-4\left(-6\right)\left(-21\right)}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-23±\sqrt{529+24\left(-21\right)}}{2\left(-6\right)}$$

Multiply 24 times -21.

$$x=\frac{-23±\sqrt{529-504}}{2\left(-6\right)}$$

Add 529 to -504.

$$x=\frac{-23±\sqrt{25}}{2\left(-6\right)}$$

Take the square root of 25.

$$x=\frac{-23±5}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{-23±5}{-12}$$

Now solve the equation x=\frac{-23±5}{-12} when ± is plus. Add -23 to 5.

$$x=-\frac{18}{-12}$$

Reduce the fraction \frac{-18}{-12} to lowest terms by extracting and canceling out 6.

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

Now solve the equation x=\frac{-23±5}{-12} when ± is minus. Subtract 5 from -23.

$$x=-\frac{28}{-12}$$

Reduce the fraction \frac{-28}{-12} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

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

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 \left(x-2\right)^{2}, the least common multiple of \left(x-2\right)^{2},x-2.

$$1-\left(x-2\right)+\left(x-2\right)^{2}\left(-6\right)=0$$

To find the opposite of x-2, find the opposite of each term.

$$1-x+2+\left(x-2\right)^{2}\left(-6\right)=0$$

Add 1 and 2 to get 3.

$$3-x+\left(x-2\right)^{2}\left(-6\right)=0$$

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

$$3-x+\left(x^{2}-4x+4\right)\left(-6\right)=0$$

Use the distributive property to multiply x^{2}-4x+4 by -6.

$$3-x-6x^{2}+24x-24=0$$

Combine -x and 24x to get 23x.

$$3+23x-6x^{2}-24=0$$

Subtract 24 from 3 to get -21.

$$-21+23x-6x^{2}=0$$

Add 21 to both sides. Anything plus zero gives itself.

$$23x-6x^{2}=21$$

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.

$$-6x^{2}+23x=21$$

Divide both sides by -6.

$$\frac{-6x^{2}+23x}{-6}=\frac{21}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$x^{2}+\frac{23}{-6}x=\frac{21}{-6}$$

Divide 23 by -6.

$$x^{2}-\frac{23}{6}x=\frac{21}{-6}$$

Reduce the fraction \frac{21}{-6} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{23}{6}x=-\frac{7}{2}$$

Divide -\frac{23}{6}, the coefficient of the x term, by 2 to get -\frac{23}{12}. Then add the square of -\frac{23}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{23}{6}x+\left(-\frac{23}{12}\right)^{2}=-\frac{7}{2}+\left(-\frac{23}{12}\right)^{2}$$

Square -\frac{23}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{23}{6}x+\frac{529}{144}=-\frac{7}{2}+\frac{529}{144}$$

Add -\frac{7}{2} to \frac{529}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{23}{6}x+\frac{529}{144}=\frac{25}{144}$$

Factor x^{2}-\frac{23}{6}x+\frac{529}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{23}{12}\right)^{2}=\frac{25}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{23}{12}\right)^{2}}=\sqrt{\frac{25}{144}}$$

Simplify.

$$x-\frac{23}{12}=\frac{5}{12} x-\frac{23}{12}=-\frac{5}{12}$$

Add \frac{23}{12} to both sides of the equation.

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