Solve x:x^2-9/x+3=4/7

$$x : \frac { x ^ { 2 } - 9 } { x + 3 } = \frac { 4 } { 7 }$$

Solve for x

$$x=-4$$

Steps Using Factoring By Grouping

Variable x cannot be equal to -3 since division by zero is not defined. Divide x by \frac{x^{2}-9}{x+3} by multiplying x by the reciprocal of \frac{x^{2}-9}{x+3}.

$$\frac{x\left(x+3\right)}{x^{2}-9}=\frac{4}{7}$$

Use the distributive property to multiply x by x+3.

$$\frac{x^{2}+3x}{x^{2}-9}=\frac{4}{7}$$

Subtract \frac{4}{7} from both sides.

$$\frac{x^{2}+3x}{x^{2}-9}-\frac{4}{7}=0$$

Factor x^{2}-9.

$$\frac{x^{2}+3x}{\left(x-3\right)\left(x+3\right)}-\frac{4}{7}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-3\right)\left(x+3\right) and 7 is 7\left(x-3\right)\left(x+3\right). Multiply \frac{x^{2}+3x}{\left(x-3\right)\left(x+3\right)} times \frac{7}{7}. Multiply \frac{4}{7} times \frac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}.

$$\frac{7\left(x^{2}+3x\right)}{7\left(x-3\right)\left(x+3\right)}-\frac{4\left(x-3\right)\left(x+3\right)}{7\left(x-3\right)\left(x+3\right)}=0$$

Since \frac{7\left(x^{2}+3x\right)}{7\left(x-3\right)\left(x+3\right)} and \frac{4\left(x-3\right)\left(x+3\right)}{7\left(x-3\right)\left(x+3\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{7\left(x^{2}+3x\right)-4\left(x-3\right)\left(x+3\right)}{7\left(x-3\right)\left(x+3\right)}=0$$

Do the multiplications in 7\left(x^{2}+3x\right)-4\left(x-3\right)\left(x+3\right).

$$\frac{7x^{2}+21x-4x^{2}-12x+12x+36}{7\left(x-3\right)\left(x+3\right)}=0$$

Combine like terms in 7x^{2}+21x-4x^{2}-12x+12x+36.

$$\frac{3x^{2}+21x+36}{7\left(x-3\right)\left(x+3\right)}=0$$

Variable x cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by 7\left(x-3\right)\left(x+3\right).

$$3x^{2}+21x+36=0$$

Divide both sides by 3.

$$x^{2}+7x+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=7 ab=1\times 12=12$$

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 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 7.

$$a=3 b=4$$

Rewrite x^{2}+7x+12 as \left(x^{2}+3x\right)+\left(4x+12\right).

$$\left(x^{2}+3x\right)+\left(4x+12\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(x+3\right)+4\left(x+3\right)$$

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

$$\left(x+3\right)\left(x+4\right)$$

To find equation solutions, solve x+3=0 and x+4=0.

$$x=-3 x=-4$$

Variable x cannot be equal to -3.

$$x=-4$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -3 since division by zero is not defined. Divide x by \frac{x^{2}-9}{x+3} by multiplying x by the reciprocal of \frac{x^{2}-9}{x+3}.

$$\frac{x\left(x+3\right)}{x^{2}-9}=\frac{4}{7}$$

Use the distributive property to multiply x by x+3.

$$\frac{x^{2}+3x}{x^{2}-9}=\frac{4}{7}$$

Subtract \frac{4}{7} from both sides.

$$\frac{x^{2}+3x}{x^{2}-9}-\frac{4}{7}=0$$

Factor x^{2}-9.

$$\frac{x^{2}+3x}{\left(x-3\right)\left(x+3\right)}-\frac{4}{7}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-3\right)\left(x+3\right) and 7 is 7\left(x-3\right)\left(x+3\right). Multiply \frac{x^{2}+3x}{\left(x-3\right)\left(x+3\right)} times \frac{7}{7}. Multiply \frac{4}{7} times \frac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}.

$$\frac{7\left(x^{2}+3x\right)}{7\left(x-3\right)\left(x+3\right)}-\frac{4\left(x-3\right)\left(x+3\right)}{7\left(x-3\right)\left(x+3\right)}=0$$

Since \frac{7\left(x^{2}+3x\right)}{7\left(x-3\right)\left(x+3\right)} and \frac{4\left(x-3\right)\left(x+3\right)}{7\left(x-3\right)\left(x+3\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{7\left(x^{2}+3x\right)-4\left(x-3\right)\left(x+3\right)}{7\left(x-3\right)\left(x+3\right)}=0$$

Do the multiplications in 7\left(x^{2}+3x\right)-4\left(x-3\right)\left(x+3\right).

$$\frac{7x^{2}+21x-4x^{2}-12x+12x+36}{7\left(x-3\right)\left(x+3\right)}=0$$

Combine like terms in 7x^{2}+21x-4x^{2}-12x+12x+36.

$$\frac{3x^{2}+21x+36}{7\left(x-3\right)\left(x+3\right)}=0$$

Variable x cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by 7\left(x-3\right)\left(x+3\right).

$$3x^{2}+21x+36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 21 for b, and 36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-21±\sqrt{21^{2}-4\times 3\times 36}}{2\times 3}$$

Square 21.

$$x=\frac{-21±\sqrt{441-4\times 3\times 36}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-21±\sqrt{441-12\times 36}}{2\times 3}$$

Multiply -12 times 36.

$$x=\frac{-21±\sqrt{441-432}}{2\times 3}$$

Add 441 to -432.

$$x=\frac{-21±\sqrt{9}}{2\times 3}$$

Take the square root of 9.

$$x=\frac{-21±3}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-21±3}{6}$$

Now solve the equation x=\frac{-21±3}{6} when ± is plus. Add -21 to 3.

$$x=-\frac{18}{6}$$

Divide -18 by 6.

$$x=-3$$

Now solve the equation x=\frac{-21±3}{6} when ± is minus. Subtract 3 from -21.

$$x=-\frac{24}{6}$$

Divide -24 by 6.

$$x=-4$$

The equation is now solved.

$$x=-3 x=-4$$

Variable x cannot be equal to -3.

$$x=-4$$

Steps for Completing the Square

Variable x cannot be equal to -3 since division by zero is not defined. Divide x by \frac{x^{2}-9}{x+3} by multiplying x by the reciprocal of \frac{x^{2}-9}{x+3}.

$$\frac{x\left(x+3\right)}{x^{2}-9}=\frac{4}{7}$$

Use the distributive property to multiply x by x+3.

$$\frac{x^{2}+3x}{x^{2}-9}=\frac{4}{7}$$

Variable x cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by 7\left(x-3\right)\left(x+3\right), the least common multiple of x^{2}-9,7.

$$7\left(x^{2}+3x\right)=4\left(x-3\right)\left(x+3\right)$$

Use the distributive property to multiply 7 by x^{2}+3x.

$$7x^{2}+21x=4\left(x-3\right)\left(x+3\right)$$

Use the distributive property to multiply 4 by x-3.

$$7x^{2}+21x=\left(4x-12\right)\left(x+3\right)$$

Use the distributive property to multiply 4x-12 by x+3 and combine like terms.

$$7x^{2}+21x=4x^{2}-36$$

Subtract 4x^{2} from both sides.

$$7x^{2}+21x-4x^{2}=-36$$

Combine 7x^{2} and -4x^{2} to get 3x^{2}.

$$3x^{2}+21x=-36$$

Divide both sides by 3.

$$\frac{3x^{2}+21x}{3}=-\frac{36}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{21}{3}x=-\frac{36}{3}$$

Divide 21 by 3.

$$x^{2}+7x=-\frac{36}{3}$$

Divide -36 by 3.

$$x^{2}+7x=-12$$

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.

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

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

$$x^{2}+7x+\frac{49}{4}=-12+\frac{49}{4}$$

Add -12 to \frac{49}{4}.

$$x^{2}+7x+\frac{49}{4}=\frac{1}{4}$$

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}.

$$\left(x+\frac{7}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{7}{2} from both sides of the equation.

$$x=-3 x=-4$$

Variable x cannot be equal to -3.

$$x=-4$$