Solve 5/x-2+1/1-14/x^2-4x+4=0

$$\frac{ 5 }{ x-2 } + \frac{ 1 }{ 1 } - \frac{ 14 }{ { x }^{ 2 } -4x+4 } = 0$$

Solve for x

$$x=-5$$

$$x=4$$

Steps Using Factoring

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 x-2,x^{2}-4x+4.

$$\left(x-2\right)\times 5+\left(x-2\right)^{2}-14=0$$

Use the distributive property to multiply x-2 by 5.

$$5x-10+\left(x-2\right)^{2}-14=0$$

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

$$5x-10+x^{2}-4x+4-14=0$$

Combine 5x and -4x to get x.

$$x-10+x^{2}+4-14=0$$

Add -10 and 4 to get -6.

$$x-6+x^{2}-14=0$$

Subtract 14 from -6 to get -20.

$$x-20+x^{2}=0$$

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

$$x^{2}+x-20=0$$

To solve the equation, factor x^{2}+x-20 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.

$$a+b=1 ab=-20$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -20.

$$-1,20 -2,10 -4,5$$

Calculate the sum for each pair.

$$-1+20=19 -2+10=8 -4+5=1$$

The solution is the pair that gives sum 1.

$$a=-4 b=5$$

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

$$\left(x-4\right)\left(x+5\right)$$

To find equation solutions, solve x-4=0 and x+5=0.

$$x=4 x=-5$$

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 x-2,x^{2}-4x+4.

$$\left(x-2\right)\times 5+\left(x-2\right)^{2}-14=0$$

Use the distributive property to multiply x-2 by 5.

$$5x-10+\left(x-2\right)^{2}-14=0$$

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

$$5x-10+x^{2}-4x+4-14=0$$

Combine 5x and -4x to get x.

$$x-10+x^{2}+4-14=0$$

Add -10 and 4 to get -6.

$$x-6+x^{2}-14=0$$

Subtract 14 from -6 to get -20.

$$x-20+x^{2}=0$$

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

$$x^{2}+x-20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-20. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-20\right)=-20$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -20.

$$-1,20 -2,10 -4,5$$

Calculate the sum for each pair.

$$-1+20=19 -2+10=8 -4+5=1$$

The solution is the pair that gives sum 1.

$$a=-4 b=5$$

Rewrite x^{2}+x-20 as \left(x^{2}-4x\right)+\left(5x-20\right).

$$\left(x^{2}-4x\right)+\left(5x-20\right)$$

Factor out x in the first and 5 in the second group.

$$x\left(x-4\right)+5\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(x+5\right)$$

To find equation solutions, solve x-4=0 and x+5=0.

$$x=4 x=-5$$

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 x-2,x^{2}-4x+4.

$$\left(x-2\right)\times 5+\left(x-2\right)^{2}-14=0$$

Use the distributive property to multiply x-2 by 5.

$$5x-10+\left(x-2\right)^{2}-14=0$$

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

$$5x-10+x^{2}-4x+4-14=0$$

Combine 5x and -4x to get x.

$$x-10+x^{2}+4-14=0$$

Add -10 and 4 to get -6.

$$x-6+x^{2}-14=0$$

Subtract 14 from -6 to get -20.

$$x-20+x^{2}=0$$

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.

$$x^{2}+x-20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-20\right)}}{2}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-20\right)}}{2}$$

Multiply -4 times -20.

$$x=\frac{-1±\sqrt{1+80}}{2}$$

Add 1 to 80.

$$x=\frac{-1±\sqrt{81}}{2}$$

Take the square root of 81.

$$x=\frac{-1±9}{2}$$

Now solve the equation x=\frac{-1±9}{2} when ± is plus. Add -1 to 9.

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

Divide 8 by 2.

$$x=4$$

Now solve the equation x=\frac{-1±9}{2} when ± is minus. Subtract 9 from -1.

$$x=-\frac{10}{2}$$

Divide -10 by 2.

$$x=-5$$

The equation is now solved.

$$x=4 x=-5$$

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 x-2,x^{2}-4x+4.

$$\left(x-2\right)\times 5+\left(x-2\right)^{2}-14=0$$

Use the distributive property to multiply x-2 by 5.

$$5x-10+\left(x-2\right)^{2}-14=0$$

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

$$5x-10+x^{2}-4x+4-14=0$$

Combine 5x and -4x to get x.

$$x-10+x^{2}+4-14=0$$

Add -10 and 4 to get -6.

$$x-6+x^{2}-14=0$$

Subtract 14 from -6 to get -20.

$$x-20+x^{2}=0$$

Add 20 to both sides. Anything plus zero gives itself.

$$x+x^{2}=20$$

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.

$$x^{2}+x=20$$

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=20+\left(\frac{1}{2}\right)^{2}$$

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

$$x^{2}+x+\frac{1}{4}=20+\frac{1}{4}$$

Add 20 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{81}{4}$$

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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

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

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

$$x=4 x=-5$$