Solve x^2+12x+20/x^2+6x+8=13/7

$$\frac { x ^ { 2 } + 12 x + 20 } { x ^ { 2 } + 6 x + 8 } = \frac { 13 } { 7 }$$

Solve for x

$$x=3$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -4,-2 since division by zero is not defined. Multiply both sides of the equation by 7\left(x+2\right)\left(x+4\right), the least common multiple of x^{2}+6x+8,7.

$$7\left(x^{2}+12x+20\right)=13\left(x+2\right)\left(x+4\right)$$

Use the distributive property to multiply 7 by x^{2}+12x+20.

$$7x^{2}+84x+140=13\left(x+2\right)\left(x+4\right)$$

Use the distributive property to multiply 13 by x+2.

$$7x^{2}+84x+140=\left(13x+26\right)\left(x+4\right)$$

Use the distributive property to multiply 13x+26 by x+4 and combine like terms.

$$7x^{2}+84x+140=13x^{2}+78x+104$$

Subtract 13x^{2} from both sides.

$$7x^{2}+84x+140-13x^{2}=78x+104$$

Combine 7x^{2} and -13x^{2} to get -6x^{2}.

$$-6x^{2}+84x+140=78x+104$$

Subtract 78x from both sides.

$$-6x^{2}+84x+140-78x=104$$

Combine 84x and -78x to get 6x.

$$-6x^{2}+6x+140=104$$

Subtract 104 from both sides.

$$-6x^{2}+6x+140-104=0$$

Subtract 104 from 140 to get 36.

$$-6x^{2}+6x+36=0$$

Divide both sides by 6.

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

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

$$a+b=1 ab=-6=-6$$

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 -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 1.

$$a=3 b=-2$$

Rewrite -x^{2}+x+6 as \left(-x^{2}+3x\right)+\left(-2x+6\right).

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

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

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

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

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

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

$$x=3 x=-2$$

Variable x cannot be equal to -2.

$$x=3$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -4,-2 since division by zero is not defined. Multiply both sides of the equation by 7\left(x+2\right)\left(x+4\right), the least common multiple of x^{2}+6x+8,7.

$$7\left(x^{2}+12x+20\right)=13\left(x+2\right)\left(x+4\right)$$

Use the distributive property to multiply 7 by x^{2}+12x+20.

$$7x^{2}+84x+140=13\left(x+2\right)\left(x+4\right)$$

Use the distributive property to multiply 13 by x+2.

$$7x^{2}+84x+140=\left(13x+26\right)\left(x+4\right)$$

Use the distributive property to multiply 13x+26 by x+4 and combine like terms.

$$7x^{2}+84x+140=13x^{2}+78x+104$$

Subtract 13x^{2} from both sides.

$$7x^{2}+84x+140-13x^{2}=78x+104$$

Combine 7x^{2} and -13x^{2} to get -6x^{2}.

$$-6x^{2}+84x+140=78x+104$$

Subtract 78x from both sides.

$$-6x^{2}+84x+140-78x=104$$

Combine 84x and -78x to get 6x.

$$-6x^{2}+6x+140=104$$

Subtract 104 from both sides.

$$-6x^{2}+6x+140-104=0$$

Subtract 104 from 140 to get 36.

$$-6x^{2}+6x+36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, 6 for b, and 36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-6\right)\times 36}}{2\left(-6\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-6\right)\times 36}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-6±\sqrt{36+24\times 36}}{2\left(-6\right)}$$

Multiply 24 times 36.

$$x=\frac{-6±\sqrt{36+864}}{2\left(-6\right)}$$

Add 36 to 864.

$$x=\frac{-6±\sqrt{900}}{2\left(-6\right)}$$

Take the square root of 900.

$$x=\frac{-6±30}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{-6±30}{-12}$$

Now solve the equation x=\frac{-6±30}{-12} when ± is plus. Add -6 to 30.

$$x=\frac{24}{-12}$$

Divide 24 by -12.

$$x=-2$$

Now solve the equation x=\frac{-6±30}{-12} when ± is minus. Subtract 30 from -6.

$$x=-\frac{36}{-12}$$

Divide -36 by -12.

$$x=3$$

The equation is now solved.

$$x=-2 x=3$$

Variable x cannot be equal to -2.

$$x=3$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -4,-2 since division by zero is not defined. Multiply both sides of the equation by 7\left(x+2\right)\left(x+4\right), the least common multiple of x^{2}+6x+8,7.

$$7\left(x^{2}+12x+20\right)=13\left(x+2\right)\left(x+4\right)$$

Use the distributive property to multiply 7 by x^{2}+12x+20.

$$7x^{2}+84x+140=13\left(x+2\right)\left(x+4\right)$$

Use the distributive property to multiply 13 by x+2.

$$7x^{2}+84x+140=\left(13x+26\right)\left(x+4\right)$$

Use the distributive property to multiply 13x+26 by x+4 and combine like terms.

$$7x^{2}+84x+140=13x^{2}+78x+104$$

Subtract 13x^{2} from both sides.

$$7x^{2}+84x+140-13x^{2}=78x+104$$

Combine 7x^{2} and -13x^{2} to get -6x^{2}.

$$-6x^{2}+84x+140=78x+104$$

Subtract 78x from both sides.

$$-6x^{2}+84x+140-78x=104$$

Combine 84x and -78x to get 6x.

$$-6x^{2}+6x+140=104$$

Subtract 140 from both sides.

$$-6x^{2}+6x=104-140$$

Subtract 140 from 104 to get -36.

$$-6x^{2}+6x=-36$$

Divide both sides by -6.

$$\frac{-6x^{2}+6x}{-6}=-\frac{36}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$x^{2}+\frac{6}{-6}x=-\frac{36}{-6}$$

Divide 6 by -6.

$$x^{2}-x=-\frac{36}{-6}$$

Divide -36 by -6.

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

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.

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

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

$$x^{2}-x+\frac{1}{4}=6+\frac{1}{4}$$

Add 6 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{25}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{5}{2} x-\frac{1}{2}=-\frac{5}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=3 x=-2$$

Variable x cannot be equal to -2.

$$x=3$$