Solve 1200/x-1200/x+8=5

$$\frac { 1200 } { x } - \frac { 1200 } { x + 8 } = 5$$

Solve for x

$$x=-48$$

$$x=40$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -8,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+8\right), the least common multiple of x,x+8.

$$\left(x+8\right)\times 1200-x\times 1200=5x\left(x+8\right)$$

Use the distributive property to multiply x+8 by 1200.

$$1200x+9600-x\times 1200=5x\left(x+8\right)$$

Use the distributive property to multiply 5x by x+8.

$$1200x+9600-x\times 1200=5x^{2}+40x$$

Subtract 5x^{2} from both sides.

$$1200x+9600-x\times 1200-5x^{2}=40x$$

Subtract 40x from both sides.

$$1200x+9600-x\times 1200-5x^{2}-40x=0$$

Combine 1200x and -40x to get 1160x.

$$1160x+9600-x\times 1200-5x^{2}=0$$

Multiply -1 and 1200 to get -1200.

$$1160x+9600-1200x-5x^{2}=0$$

Combine 1160x and -1200x to get -40x.

$$-40x+9600-5x^{2}=0$$

Divide both sides by 5.

$$-8x+1920-x^{2}=0$$

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

$$-x^{2}-8x+1920=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+1920. To find a and b, set up a system to be solved.

$$a+b=-8 ab=-1920=-1920$$

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 -1920.

$$1,-1920 2,-960 3,-640 4,-480 5,-384 6,-320 8,-240 10,-192 12,-160 15,-128 16,-120 20,-96 24,-80 30,-64 32,-60 40,-48$$

Calculate the sum for each pair.

$$1-1920=-1919 2-960=-958 3-640=-637 4-480=-476 5-384=-379 6-320=-314 8-240=-232 10-192=-182 12-160=-148 15-128=-113 16-120=-104 20-96=-76 24-80=-56 30-64=-34 32-60=-28 40-48=-8$$

The solution is the pair that gives sum -8.

$$a=40 b=-48$$

Rewrite -x^{2}-8x+1920 as \left(-x^{2}+40x\right)+\left(-48x+1920\right).

$$\left(-x^{2}+40x\right)+\left(-48x+1920\right)$$

Factor out x in the first and 48 in the second group.

$$x\left(-x+40\right)+48\left(-x+40\right)$$

Factor out common term -x+40 by using distributive property.

$$\left(-x+40\right)\left(x+48\right)$$

To find equation solutions, solve -x+40=0 and x+48=0.

$$x=40 x=-48$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -8,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+8\right), the least common multiple of x,x+8.

$$\left(x+8\right)\times 1200-x\times 1200=5x\left(x+8\right)$$

Use the distributive property to multiply x+8 by 1200.

$$1200x+9600-x\times 1200=5x\left(x+8\right)$$

Use the distributive property to multiply 5x by x+8.

$$1200x+9600-x\times 1200=5x^{2}+40x$$

Subtract 5x^{2} from both sides.

$$1200x+9600-x\times 1200-5x^{2}=40x$$

Subtract 40x from both sides.

$$1200x+9600-x\times 1200-5x^{2}-40x=0$$

Combine 1200x and -40x to get 1160x.

$$1160x+9600-x\times 1200-5x^{2}=0$$

Multiply -1 and 1200 to get -1200.

$$1160x+9600-1200x-5x^{2}=0$$

Combine 1160x and -1200x to get -40x.

$$-40x+9600-5x^{2}=0$$

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.

$$-5x^{2}-40x+9600=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, -40 for b, and 9600 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\left(-5\right)\times 9600}}{2\left(-5\right)}$$

Square -40.

$$x=\frac{-\left(-40\right)±\sqrt{1600-4\left(-5\right)\times 9600}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-\left(-40\right)±\sqrt{1600+20\times 9600}}{2\left(-5\right)}$$

Multiply 20 times 9600.

$$x=\frac{-\left(-40\right)±\sqrt{1600+192000}}{2\left(-5\right)}$$

Add 1600 to 192000.

$$x=\frac{-\left(-40\right)±\sqrt{193600}}{2\left(-5\right)}$$

Take the square root of 193600.

$$x=\frac{-\left(-40\right)±440}{2\left(-5\right)}$$

The opposite of -40 is 40.

$$x=\frac{40±440}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{40±440}{-10}$$

Now solve the equation x=\frac{40±440}{-10} when ± is plus. Add 40 to 440.

$$x=\frac{480}{-10}$$

Divide 480 by -10.

$$x=-48$$

Now solve the equation x=\frac{40±440}{-10} when ± is minus. Subtract 440 from 40.

$$x=-\frac{400}{-10}$$

Divide -400 by -10.

$$x=40$$

The equation is now solved.

$$x=-48 x=40$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -8,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+8\right), the least common multiple of x,x+8.

$$\left(x+8\right)\times 1200-x\times 1200=5x\left(x+8\right)$$

Use the distributive property to multiply x+8 by 1200.

$$1200x+9600-x\times 1200=5x\left(x+8\right)$$

Use the distributive property to multiply 5x by x+8.

$$1200x+9600-x\times 1200=5x^{2}+40x$$

Subtract 5x^{2} from both sides.

$$1200x+9600-x\times 1200-5x^{2}=40x$$

Subtract 40x from both sides.

$$1200x+9600-x\times 1200-5x^{2}-40x=0$$

Combine 1200x and -40x to get 1160x.

$$1160x+9600-x\times 1200-5x^{2}=0$$

Subtract 9600 from both sides. Anything subtracted from zero gives its negation.

$$1160x-x\times 1200-5x^{2}=-9600$$

Multiply -1 and 1200 to get -1200.

$$1160x-1200x-5x^{2}=-9600$$

Combine 1160x and -1200x to get -40x.

$$-40x-5x^{2}=-9600$$

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.

$$-5x^{2}-40x=-9600$$

Divide both sides by -5.

$$\frac{-5x^{2}-40x}{-5}=-\frac{9600}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\left(-\frac{40}{-5}\right)x=-\frac{9600}{-5}$$

Divide -40 by -5.

$$x^{2}+8x=-\frac{9600}{-5}$$

Divide -9600 by -5.

$$x^{2}+8x=1920$$

Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+8x+4^{2}=1920+4^{2}$$

Square 4.

$$x^{2}+8x+16=1920+16$$

Add 1920 to 16.

$$x^{2}+8x+16=1936$$

Factor x^{2}+8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+4\right)^{2}=1936$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{1936}$$

Simplify.

$$x+4=44 x+4=-44$$

Subtract 4 from both sides of the equation.

$$x=40 x=-48$$