Solve 1=-90x/x+90+x

$$1 = \frac { - 90 x } { x + 90 } + x$$

Solve for x

$$x=-9$$

$$x=10$$

Steps Using Factoring By Grouping

Variable x cannot be equal to -90 since division by zero is not defined. Multiply both sides of the equation by x+90.

$$x+90=-90x+\left(x+90\right)x$$

Use the distributive property to multiply x+90 by x.

$$x+90=-90x+x^{2}+90x$$

Combine -90x and 90x to get 0.

$$x+90=x^{2}$$

Subtract x^{2} from both sides.

$$x+90-x^{2}=0$$

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

$$-x^{2}+x+90=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+90. To find a and b, set up a system to be solved.

$$a+b=1 ab=-90=-90$$

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 -90.

$$-1,90 -2,45 -3,30 -5,18 -6,15 -9,10$$

Calculate the sum for each pair.

$$-1+90=89 -2+45=43 -3+30=27 -5+18=13 -6+15=9 -9+10=1$$

The solution is the pair that gives sum 1.

$$a=10 b=-9$$

Rewrite -x^{2}+x+90 as \left(-x^{2}+10x\right)+\left(-9x+90\right).

$$\left(-x^{2}+10x\right)+\left(-9x+90\right)$$

Factor out -x in the first and -9 in the second group.

$$-x\left(x-10\right)-9\left(x-10\right)$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(-x-9\right)$$

To find equation solutions, solve x-10=0 and -x-9=0.

$$x=10 x=-9$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -90 since division by zero is not defined. Multiply both sides of the equation by x+90.

$$x+90=-90x+\left(x+90\right)x$$

Use the distributive property to multiply x+90 by x.

$$x+90=-90x+x^{2}+90x$$

Combine -90x and 90x to get 0.

$$x+90=x^{2}$$

Subtract x^{2} from both sides.

$$x+90-x^{2}=0$$

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.

$$-x^{2}+x+90=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 1 for b, and 90 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-1\right)\times 90}}{2\left(-1\right)}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-1\right)\times 90}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-1±\sqrt{1+4\times 90}}{2\left(-1\right)}$$

Multiply 4 times 90.

$$x=\frac{-1±\sqrt{1+360}}{2\left(-1\right)}$$

Add 1 to 360.

$$x=\frac{-1±\sqrt{361}}{2\left(-1\right)}$$

Take the square root of 361.

$$x=\frac{-1±19}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-1±19}{-2}$$

Now solve the equation x=\frac{-1±19}{-2} when ± is plus. Add -1 to 19.

$$x=\frac{18}{-2}$$

Divide 18 by -2.

$$x=-9$$

Now solve the equation x=\frac{-1±19}{-2} when ± is minus. Subtract 19 from -1.

$$x=-\frac{20}{-2}$$

Divide -20 by -2.

$$x=10$$

The equation is now solved.

$$x=-9 x=10$$

Steps for Completing the Square

Variable x cannot be equal to -90 since division by zero is not defined. Multiply both sides of the equation by x+90.

$$x+90=-90x+\left(x+90\right)x$$

Use the distributive property to multiply x+90 by x.

$$x+90=-90x+x^{2}+90x$$

Combine -90x and 90x to get 0.

$$x+90=x^{2}$$

Subtract x^{2} from both sides.

$$x+90-x^{2}=0$$

Subtract 90 from both sides. Anything subtracted from zero gives its negation.

$$x-x^{2}=-90$$

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.

$$-x^{2}+x=-90$$

Divide both sides by -1.

$$\frac{-x^{2}+x}{-1}=-\frac{90}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{1}{-1}x=-\frac{90}{-1}$$

Divide 1 by -1.

$$x^{2}-x=-\frac{90}{-1}$$

Divide -90 by -1.

$$x^{2}-x=90$$

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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=90+\left(-\frac{1}{2}\right)^{2}$$

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

$$x^{2}-x+\frac{1}{4}=90+\frac{1}{4}$$

Add 90 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{361}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{361}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{361}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{19}{2} x-\frac{1}{2}=-\frac{19}{2}$$

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

$$x=10 x=-9$$