Solve x+4/x+6=x+14/x^2+2x-24

$$\frac { x + 4 } { x + 6 } = \frac { x + 14 } { x ^ { 2 } + 2 x - 24 }$$

Solve for x

$$x=-5$$

$$x=6$$

Steps Using Factoring

Variable x cannot be equal to any of the values -6,4 since division by zero is not defined. Multiply both sides of the equation by \left(x-4\right)\left(x+6\right), the least common multiple of x+6,x^{2}+2x-24.

$$\left(x-4\right)\left(x+4\right)=x+14$$

Consider \left(x-4\right)\left(x+4\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 4.

$$x^{2}-16=x+14$$

Subtract x from both sides.

$$x^{2}-16-x=14$$

Subtract 14 from both sides.

$$x^{2}-16-x-14=0$$

Subtract 14 from -16 to get -30.

$$x^{2}-30-x=0$$

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

$$x^{2}-x-30=0$$

To solve the equation, factor x^{2}-x-30 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.

$$a+b=-1 ab=-30$$

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 -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -1.

$$a=-6 b=5$$

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

$$\left(x-6\right)\left(x+5\right)$$

To find equation solutions, solve x-6=0 and x+5=0.

$$x=6 x=-5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -6,4 since division by zero is not defined. Multiply both sides of the equation by \left(x-4\right)\left(x+6\right), the least common multiple of x+6,x^{2}+2x-24.

$$\left(x-4\right)\left(x+4\right)=x+14$$

Consider \left(x-4\right)\left(x+4\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 4.

$$x^{2}-16=x+14$$

Subtract x from both sides.

$$x^{2}-16-x=14$$

Subtract 14 from both sides.

$$x^{2}-16-x-14=0$$

Subtract 14 from -16 to get -30.

$$x^{2}-30-x=0$$

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

$$x^{2}-x-30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-30. To find a and b, set up a system to be solved.

$$a+b=-1 ab=1\left(-30\right)=-30$$

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 -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -1.

$$a=-6 b=5$$

Rewrite x^{2}-x-30 as \left(x^{2}-6x\right)+\left(5x-30\right).

$$\left(x^{2}-6x\right)+\left(5x-30\right)$$

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

$$x\left(x-6\right)+5\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(x+5\right)$$

To find equation solutions, solve x-6=0 and x+5=0.

$$x=6 x=-5$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -6,4 since division by zero is not defined. Multiply both sides of the equation by \left(x-4\right)\left(x+6\right), the least common multiple of x+6,x^{2}+2x-24.

$$\left(x-4\right)\left(x+4\right)=x+14$$

Consider \left(x-4\right)\left(x+4\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 4.

$$x^{2}-16=x+14$$

Subtract x from both sides.

$$x^{2}-16-x=14$$

Subtract 14 from both sides.

$$x^{2}-16-x-14=0$$

Subtract 14 from -16 to get -30.

$$x^{2}-30-x=0$$

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.

$$x^{2}-x-30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -1 for b, and -30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-30\right)}}{2}$$

Multiply -4 times -30.

$$x=\frac{-\left(-1\right)±\sqrt{1+120}}{2}$$

Add 1 to 120.

$$x=\frac{-\left(-1\right)±\sqrt{121}}{2}$$

Take the square root of 121.

$$x=\frac{-\left(-1\right)±11}{2}$$

The opposite of -1 is 1.

$$x=\frac{1±11}{2}$$

Now solve the equation x=\frac{1±11}{2} when ± is plus. Add 1 to 11.

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

Divide 12 by 2.

$$x=6$$

Now solve the equation x=\frac{1±11}{2} when ± is minus. Subtract 11 from 1.

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

Divide -10 by 2.

$$x=-5$$

The equation is now solved.

$$x=6 x=-5$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -6,4 since division by zero is not defined. Multiply both sides of the equation by \left(x-4\right)\left(x+6\right), the least common multiple of x+6,x^{2}+2x-24.

$$\left(x-4\right)\left(x+4\right)=x+14$$

Consider \left(x-4\right)\left(x+4\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 4.

$$x^{2}-16=x+14$$

Subtract x from both sides.

$$x^{2}-16-x=14$$

Add 16 to both sides.

$$x^{2}-x=14+16$$

Add 14 and 16 to get 30.

$$x^{2}-x=30$$

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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=30+\left(-\frac{1}{2}\right)^{2}$$

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

$$x^{2}-x+\frac{1}{4}=30+\frac{1}{4}$$

Add 30 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{121}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{121}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{121}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{11}{2} x-\frac{1}{2}=-\frac{11}{2}$$

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

$$x=6 x=-5$$