Solve x^2/x+80=40

$$\frac { x ^ { 2 } } { x + 80 } = 40$$

Solve for x

$$x=-40$$

$$x=80$$

Steps Using Factoring

Variable x cannot be equal to -80 since division by zero is not defined. Multiply both sides of the equation by x+80.

$$x^{2}=40\left(x+80\right)$$

Use the distributive property to multiply 40 by x+80.

$$x^{2}=40x+3200$$

Subtract 40x from both sides.

$$x^{2}-40x=3200$$

Subtract 3200 from both sides.

$$x^{2}-40x-3200=0$$

To solve the equation, factor x^{2}-40x-3200 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.

$$a+b=-40 ab=-3200$$

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 -3200.

$$1,-3200 2,-1600 4,-800 5,-640 8,-400 10,-320 16,-200 20,-160 25,-128 32,-100 40,-80 50,-64$$

Calculate the sum for each pair.

$$1-3200=-3199 2-1600=-1598 4-800=-796 5-640=-635 8-400=-392 10-320=-310 16-200=-184 20-160=-140 25-128=-103 32-100=-68 40-80=-40 50-64=-14$$

The solution is the pair that gives sum -40.

$$a=-80 b=40$$

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

$$\left(x-80\right)\left(x+40\right)$$

To find equation solutions, solve x-80=0 and x+40=0.

$$x=80 x=-40$$

Steps Using Factoring By Grouping

Variable x cannot be equal to -80 since division by zero is not defined. Multiply both sides of the equation by x+80.

$$x^{2}=40\left(x+80\right)$$

Use the distributive property to multiply 40 by x+80.

$$x^{2}=40x+3200$$

Subtract 40x from both sides.

$$x^{2}-40x=3200$$

Subtract 3200 from both sides.

$$x^{2}-40x-3200=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-3200. To find a and b, set up a system to be solved.

$$a+b=-40 ab=1\left(-3200\right)=-3200$$

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 -3200.

$$1,-3200 2,-1600 4,-800 5,-640 8,-400 10,-320 16,-200 20,-160 25,-128 32,-100 40,-80 50,-64$$

Calculate the sum for each pair.

$$1-3200=-3199 2-1600=-1598 4-800=-796 5-640=-635 8-400=-392 10-320=-310 16-200=-184 20-160=-140 25-128=-103 32-100=-68 40-80=-40 50-64=-14$$

The solution is the pair that gives sum -40.

$$a=-80 b=40$$

Rewrite x^{2}-40x-3200 as \left(x^{2}-80x\right)+\left(40x-3200\right).

$$\left(x^{2}-80x\right)+\left(40x-3200\right)$$

Factor out x in the first and 40 in the second group.

$$x\left(x-80\right)+40\left(x-80\right)$$

Factor out common term x-80 by using distributive property.

$$\left(x-80\right)\left(x+40\right)$$

To find equation solutions, solve x-80=0 and x+40=0.

$$x=80 x=-40$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -80 since division by zero is not defined. Multiply both sides of the equation by x+80.

$$x^{2}=40\left(x+80\right)$$

Use the distributive property to multiply 40 by x+80.

$$x^{2}=40x+3200$$

Subtract 40x from both sides.

$$x^{2}-40x=3200$$

Subtract 3200 from both sides.

$$x^{2}-40x-3200=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -40 for b, and -3200 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\left(-3200\right)}}{2}$$

Square -40.

$$x=\frac{-\left(-40\right)±\sqrt{1600-4\left(-3200\right)}}{2}$$

Multiply -4 times -3200.

$$x=\frac{-\left(-40\right)±\sqrt{1600+12800}}{2}$$

Add 1600 to 12800.

$$x=\frac{-\left(-40\right)±\sqrt{14400}}{2}$$

Take the square root of 14400.

$$x=\frac{-\left(-40\right)±120}{2}$$

The opposite of -40 is 40.

$$x=\frac{40±120}{2}$$

Now solve the equation x=\frac{40±120}{2} when ± is plus. Add 40 to 120.

$$x=\frac{160}{2}$$

Divide 160 by 2.

$$x=80$$

Now solve the equation x=\frac{40±120}{2} when ± is minus. Subtract 120 from 40.

$$x=-\frac{80}{2}$$

Divide -80 by 2.

$$x=-40$$

The equation is now solved.

$$x=80 x=-40$$

Steps for Completing the Square

Variable x cannot be equal to -80 since division by zero is not defined. Multiply both sides of the equation by x+80.

$$x^{2}=40\left(x+80\right)$$

Use the distributive property to multiply 40 by x+80.

$$x^{2}=40x+3200$$

Subtract 40x from both sides.

$$x^{2}-40x=3200$$

Divide -40, the coefficient of the x term, by 2 to get -20. Then add the square of -20 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-40x+\left(-20\right)^{2}=3200+\left(-20\right)^{2}$$

Square -20.

$$x^{2}-40x+400=3200+400$$

Add 3200 to 400.

$$x^{2}-40x+400=3600$$

Factor x^{2}-40x+400. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-20\right)^{2}=3600$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-20\right)^{2}}=\sqrt{3600}$$

Simplify.

$$x-20=60 x-20=-60$$

Add 20 to both sides of the equation.

$$x=80 x=-40$$