Solve x-x^2=1-6x^2

$$x - x ^ { 2 } = 1 - 6 x ^ { 2 }$$

Solve for x

$$x=\frac{\sqrt{21}-1}{10}\approx 0.358257569$$

$$x=\frac{-\sqrt{21}-1}{10}\approx -0.558257569$$

Steps Using the Quadratic Formula

Subtract 1 from both sides.

$$x-x^{2}-1=-6x^{2}$$

Add 6x^{2} to both sides.

$$x-x^{2}-1+6x^{2}=0$$

Combine -x^{2} and 6x^{2} to get 5x^{2}.

$$x+5x^{2}-1=0$$

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.

$$5x^{2}+x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 1 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 5\left(-1\right)}}{2\times 5}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 5\left(-1\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-1±\sqrt{1-20\left(-1\right)}}{2\times 5}$$

Multiply -20 times -1.

$$x=\frac{-1±\sqrt{1+20}}{2\times 5}$$

Add 1 to 20.

$$x=\frac{-1±\sqrt{21}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-1±\sqrt{21}}{10}$$

Now solve the equation x=\frac{-1±\sqrt{21}}{10} when ± is plus. Add -1 to \sqrt{21}.

$$x=\frac{\sqrt{21}-1}{10}$$

Now solve the equation x=\frac{-1±\sqrt{21}}{10} when ± is minus. Subtract \sqrt{21} from -1.

$$x=\frac{-\sqrt{21}-1}{10}$$

The equation is now solved.

$$x=\frac{\sqrt{21}-1}{10} x=\frac{-\sqrt{21}-1}{10}$$

Steps for Completing the Square

Add 6x^{2} to both sides.

$$x-x^{2}+6x^{2}=1$$

Combine -x^{2} and 6x^{2} to get 5x^{2}.

$$x+5x^{2}=1$$

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.

$$5x^{2}+x=1$$

Divide both sides by 5.

$$\frac{5x^{2}+x}{5}=\frac{1}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\frac{1}{5}x=\frac{1}{5}$$

Divide \frac{1}{5}, the coefficient of the x term, by 2 to get \frac{1}{10}. Then add the square of \frac{1}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{5}x+\frac{1}{100}=\frac{1}{5}+\frac{1}{100}$$

Add \frac{1}{5} to \frac{1}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{5}x+\frac{1}{100}=\frac{21}{100}$$

Factor x^{2}+\frac{1}{5}x+\frac{1}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{10}\right)^{2}=\frac{21}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{10}\right)^{2}}=\sqrt{\frac{21}{100}}$$

Simplify.

$$x+\frac{1}{10}=\frac{\sqrt{21}}{10} x+\frac{1}{10}=-\frac{\sqrt{21}}{10}$$

Subtract \frac{1}{10} from both sides of the equation.

$$x=\frac{\sqrt{21}-1}{10} x=\frac{-\sqrt{21}-1}{10}$$