Solve x-2/x-7-x/x+7=18-5x^2/x^2-49

$$\frac { x - 2 } { x - 7 } - \frac { x } { x + 7 } = \frac { 18 - 5 x ^ { 2 } } { x ^ { 2 } - 49 }$$

Solve for x

$$x=-4$$

$$x = \frac{8}{5} = 1\frac{3}{5} = 1.6$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -7,7 since division by zero is not defined. Multiply both sides of the equation by \left(x-7\right)\left(x+7\right), the least common multiple of x-7,x+7,x^{2}-49.

$$\left(x+7\right)\left(x-2\right)-\left(x-7\right)x=18-5x^{2}$$

Use the distributive property to multiply x+7 by x-2 and combine like terms.

$$x^{2}+5x-14-\left(x-7\right)x=18-5x^{2}$$

Use the distributive property to multiply x-7 by x.

$$x^{2}+5x-14-\left(x^{2}-7x\right)=18-5x^{2}$$

To find the opposite of x^{2}-7x, find the opposite of each term.

$$x^{2}+5x-14-x^{2}+7x=18-5x^{2}$$

Combine x^{2} and -x^{2} to get 0.

$$5x-14+7x=18-5x^{2}$$

Combine 5x and 7x to get 12x.

$$12x-14=18-5x^{2}$$

Subtract 18 from both sides.

$$12x-14-18=-5x^{2}$$

Subtract 18 from -14 to get -32.

$$12x-32=-5x^{2}$$

Add 5x^{2} to both sides.

$$12x-32+5x^{2}=0$$

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

$$5x^{2}+12x-32=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-32. To find a and b, set up a system to be solved.

$$a+b=12 ab=5\left(-32\right)=-160$$

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 -160.

$$-1,160 -2,80 -4,40 -5,32 -8,20 -10,16$$

Calculate the sum for each pair.

$$-1+160=159 -2+80=78 -4+40=36 -5+32=27 -8+20=12 -10+16=6$$

The solution is the pair that gives sum 12.

$$a=-8 b=20$$

Rewrite 5x^{2}+12x-32 as \left(5x^{2}-8x\right)+\left(20x-32\right).

$$\left(5x^{2}-8x\right)+\left(20x-32\right)$$

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

$$x\left(5x-8\right)+4\left(5x-8\right)$$

Factor out common term 5x-8 by using distributive property.

$$\left(5x-8\right)\left(x+4\right)$$

To find equation solutions, solve 5x-8=0 and x+4=0.

$$x=\frac{8}{5} x=-4$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -7,7 since division by zero is not defined. Multiply both sides of the equation by \left(x-7\right)\left(x+7\right), the least common multiple of x-7,x+7,x^{2}-49.

$$\left(x+7\right)\left(x-2\right)-\left(x-7\right)x=18-5x^{2}$$

Use the distributive property to multiply x+7 by x-2 and combine like terms.

$$x^{2}+5x-14-\left(x-7\right)x=18-5x^{2}$$

Use the distributive property to multiply x-7 by x.

$$x^{2}+5x-14-\left(x^{2}-7x\right)=18-5x^{2}$$

To find the opposite of x^{2}-7x, find the opposite of each term.

$$x^{2}+5x-14-x^{2}+7x=18-5x^{2}$$

Combine x^{2} and -x^{2} to get 0.

$$5x-14+7x=18-5x^{2}$$

Combine 5x and 7x to get 12x.

$$12x-14=18-5x^{2}$$

Subtract 18 from both sides.

$$12x-14-18=-5x^{2}$$

Subtract 18 from -14 to get -32.

$$12x-32=-5x^{2}$$

Add 5x^{2} to both sides.

$$12x-32+5x^{2}=0$$

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.

$$5x^{2}+12x-32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 12 for b, and -32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 5\left(-32\right)}}{2\times 5}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 5\left(-32\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-12±\sqrt{144-20\left(-32\right)}}{2\times 5}$$

Multiply -20 times -32.

$$x=\frac{-12±\sqrt{144+640}}{2\times 5}$$

Add 144 to 640.

$$x=\frac{-12±\sqrt{784}}{2\times 5}$$

Take the square root of 784.

$$x=\frac{-12±28}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-12±28}{10}$$

Now solve the equation x=\frac{-12±28}{10} when ± is plus. Add -12 to 28.

$$x=\frac{16}{10}$$

Reduce the fraction \frac{16}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{8}{5}$$

Now solve the equation x=\frac{-12±28}{10} when ± is minus. Subtract 28 from -12.

$$x=-\frac{40}{10}$$

Divide -40 by 10.

$$x=-4$$

The equation is now solved.

$$x=\frac{8}{5} x=-4$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -7,7 since division by zero is not defined. Multiply both sides of the equation by \left(x-7\right)\left(x+7\right), the least common multiple of x-7,x+7,x^{2}-49.

$$\left(x+7\right)\left(x-2\right)-\left(x-7\right)x=18-5x^{2}$$

Use the distributive property to multiply x+7 by x-2 and combine like terms.

$$x^{2}+5x-14-\left(x-7\right)x=18-5x^{2}$$

Use the distributive property to multiply x-7 by x.

$$x^{2}+5x-14-\left(x^{2}-7x\right)=18-5x^{2}$$

To find the opposite of x^{2}-7x, find the opposite of each term.

$$x^{2}+5x-14-x^{2}+7x=18-5x^{2}$$

Combine x^{2} and -x^{2} to get 0.

$$5x-14+7x=18-5x^{2}$$

Combine 5x and 7x to get 12x.

$$12x-14=18-5x^{2}$$

Add 5x^{2} to both sides.

$$12x-14+5x^{2}=18$$

Add 14 to both sides.

$$12x+5x^{2}=18+14$$

Add 18 and 14 to get 32.

$$12x+5x^{2}=32$$

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.

$$5x^{2}+12x=32$$

Divide both sides by 5.

$$\frac{5x^{2}+12x}{5}=\frac{32}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\frac{12}{5}x=\frac{32}{5}$$

Divide \frac{12}{5}, the coefficient of the x term, by 2 to get \frac{6}{5}. Then add the square of \frac{6}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{12}{5}x+\left(\frac{6}{5}\right)^{2}=\frac{32}{5}+\left(\frac{6}{5}\right)^{2}$$

Square \frac{6}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{12}{5}x+\frac{36}{25}=\frac{32}{5}+\frac{36}{25}$$

Add \frac{32}{5} to \frac{36}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{12}{5}x+\frac{36}{25}=\frac{196}{25}$$

Factor x^{2}+\frac{12}{5}x+\frac{36}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{6}{5}\right)^{2}=\frac{196}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{6}{5}\right)^{2}}=\sqrt{\frac{196}{25}}$$

Simplify.

$$x+\frac{6}{5}=\frac{14}{5} x+\frac{6}{5}=-\frac{14}{5}$$

Subtract \frac{6}{5} from both sides of the equation.

$$x=\frac{8}{5} x=-4$$