Solve 1.5*10^-5=(x)(x)/1-x

$$1.5 \times 10 ^ { - 5 } = \frac { ( x ) ( x ) } { 1 - x }$$

Solve for x

$$x=\frac{\sqrt{2400009}-3}{400000}\approx 0.003865491$$

$$x=\frac{-\sqrt{2400009}-3}{400000}\approx -0.003880491$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by -x+1.

$$1.5\times 10^{-5}\left(-x+1\right)=xx$$

Multiply x and x to get x^{2}.

$$1.5\times 10^{-5}\left(-x+1\right)=x^{2}$$

Calculate 10 to the power of -5 and get \frac{1}{100000}.

$$1.5\times \frac{1}{100000}\left(-x+1\right)=x^{2}$$

Multiply 1.5 and \frac{1}{100000} to get \frac{3}{200000}.

$$\frac{3}{200000}\left(-x+1\right)=x^{2}$$

Use the distributive property to multiply \frac{3}{200000} by -x+1.

$$-\frac{3}{200000}x+\frac{3}{200000}=x^{2}$$

Subtract x^{2} from both sides.

$$-\frac{3}{200000}x+\frac{3}{200000}-x^{2}=0$$

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.

$$-x^{2}-\frac{3}{200000}x+\frac{3}{200000}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -\frac{3}{200000} for b, and \frac{3}{200000} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{3}{200000}\right)±\sqrt{\left(-\frac{3}{200000}\right)^{2}-4\left(-1\right)\times \frac{3}{200000}}}{2\left(-1\right)}$$

Square -\frac{3}{200000} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{3}{200000}\right)±\sqrt{\frac{9}{40000000000}-4\left(-1\right)\times \frac{3}{200000}}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-\frac{3}{200000}\right)±\sqrt{\frac{9}{40000000000}+4\times \frac{3}{200000}}}{2\left(-1\right)}$$

Multiply 4 times \frac{3}{200000}.

$$x=\frac{-\left(-\frac{3}{200000}\right)±\sqrt{\frac{9}{40000000000}+\frac{3}{50000}}}{2\left(-1\right)}$$

Add \frac{9}{40000000000} to \frac{3}{50000} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{3}{200000}\right)±\sqrt{\frac{2400009}{40000000000}}}{2\left(-1\right)}$$

Take the square root of \frac{2400009}{40000000000}.

$$x=\frac{-\left(-\frac{3}{200000}\right)±\frac{\sqrt{2400009}}{200000}}{2\left(-1\right)}$$

The opposite of -\frac{3}{200000} is \frac{3}{200000}.

$$x=\frac{\frac{3}{200000}±\frac{\sqrt{2400009}}{200000}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{\frac{3}{200000}±\frac{\sqrt{2400009}}{200000}}{-2}$$

Now solve the equation x=\frac{\frac{3}{200000}±\frac{\sqrt{2400009}}{200000}}{-2} when ± is plus. Add \frac{3}{200000} to \frac{\sqrt{2400009}}{200000}.

$$x=\frac{\sqrt{2400009}+3}{-2\times 200000}$$

Divide \frac{3+\sqrt{2400009}}{200000} by -2.

$$x=\frac{-\sqrt{2400009}-3}{400000}$$

Now solve the equation x=\frac{\frac{3}{200000}±\frac{\sqrt{2400009}}{200000}}{-2} when ± is minus. Subtract \frac{\sqrt{2400009}}{200000} from \frac{3}{200000}.

$$x=\frac{3-\sqrt{2400009}}{-2\times 200000}$$

Divide \frac{3-\sqrt{2400009}}{200000} by -2.

$$x=\frac{\sqrt{2400009}-3}{400000}$$

The equation is now solved.

$$x=\frac{-\sqrt{2400009}-3}{400000} x=\frac{\sqrt{2400009}-3}{400000}$$

Steps for Completing the Square

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by -x+1.

$$1.5\times 10^{-5}\left(-x+1\right)=xx$$

Multiply x and x to get x^{2}.

$$1.5\times 10^{-5}\left(-x+1\right)=x^{2}$$

Calculate 10 to the power of -5 and get \frac{1}{100000}.

$$1.5\times \frac{1}{100000}\left(-x+1\right)=x^{2}$$

Multiply 1.5 and \frac{1}{100000} to get \frac{3}{200000}.

$$\frac{3}{200000}\left(-x+1\right)=x^{2}$$

Use the distributive property to multiply \frac{3}{200000} by -x+1.

$$-\frac{3}{200000}x+\frac{3}{200000}=x^{2}$$

Subtract x^{2} from both sides.

$$-\frac{3}{200000}x+\frac{3}{200000}-x^{2}=0$$

Subtract \frac{3}{200000} from both sides. Anything subtracted from zero gives its negation.

$$-\frac{3}{200000}x-x^{2}=-\frac{3}{200000}$$

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.

$$-x^{2}-\frac{3}{200000}x=-\frac{3}{200000}$$

Divide both sides by -1.

$$\frac{-x^{2}-\frac{3}{200000}x}{-1}=-\frac{\frac{3}{200000}}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{\frac{3}{200000}}{-1}\right)x=-\frac{\frac{3}{200000}}{-1}$$

Divide -\frac{3}{200000} by -1.

$$x^{2}+\frac{3}{200000}x=-\frac{\frac{3}{200000}}{-1}$$

Divide -\frac{3}{200000} by -1.

$$x^{2}+\frac{3}{200000}x=\frac{3}{200000}$$

Divide \frac{3}{200000}, the coefficient of the x term, by 2 to get \frac{3}{400000}. Then add the square of \frac{3}{400000} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{200000}x+\left(\frac{3}{400000}\right)^{2}=\frac{3}{200000}+\left(\frac{3}{400000}\right)^{2}$$

Square \frac{3}{400000} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{200000}x+\frac{9}{160000000000}=\frac{3}{200000}+\frac{9}{160000000000}$$

Add \frac{3}{200000} to \frac{9}{160000000000} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{3}{200000}x+\frac{9}{160000000000}=\frac{2400009}{160000000000}$$

Factor x^{2}+\frac{3}{200000}x+\frac{9}{160000000000}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{400000}\right)^{2}=\frac{2400009}{160000000000}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{400000}\right)^{2}}=\sqrt{\frac{2400009}{160000000000}}$$

Simplify.

$$x+\frac{3}{400000}=\frac{\sqrt{2400009}}{400000} x+\frac{3}{400000}=-\frac{\sqrt{2400009}}{400000}$$

Subtract \frac{3}{400000} from both sides of the equation.

$$x=\frac{\sqrt{2400009}-3}{400000} x=\frac{-\sqrt{2400009}-3}{400000}$$