Solve 13/12-2x+4x^2+1/12=1

$$\frac { 13 } { 12 } - 2 x + 4 x ^ { 2 } + \frac { 1 } { 12 } = 1$$

Solve for x

$$x=\frac{\sqrt{3}}{12}+\frac{1}{4}\approx 0.394337567$$

$$x=-\frac{\sqrt{3}}{12}+\frac{1}{4}\approx 0.105662433$$

Steps Using the Quadratic Formula

Add \frac{13}{12} and \frac{1}{12} to get \frac{7}{6}.

$$\frac{7}{6}-2x+4x^{2}=1$$

Subtract 1 from both sides.

$$\frac{7}{6}-2x+4x^{2}-1=0$$

Subtract 1 from \frac{7}{6} to get \frac{1}{6}.

$$\frac{1}{6}-2x+4x^{2}=0$$

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.

$$4x^{2}-2x+\frac{1}{6}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -2 for b, and \frac{1}{6} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 4\times \frac{1}{6}}}{2\times 4}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 4\times \frac{1}{6}}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-2\right)±\sqrt{4-16\times \frac{1}{6}}}{2\times 4}$$

Multiply -16 times \frac{1}{6}.

$$x=\frac{-\left(-2\right)±\sqrt{4-\frac{8}{3}}}{2\times 4}$$

Add 4 to -\frac{8}{3}.

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

Take the square root of \frac{4}{3}.

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

The opposite of -2 is 2.

$$x=\frac{2±\frac{2\sqrt{3}}{3}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{2±\frac{2\sqrt{3}}{3}}{8}$$

Now solve the equation x=\frac{2±\frac{2\sqrt{3}}{3}}{8} when ± is plus. Add 2 to \frac{2\sqrt{3}}{3}.

$$x=\frac{\frac{2\sqrt{3}}{3}+2}{8}$$

Divide 2+\frac{2\sqrt{3}}{3} by 8.

$$x=\frac{\sqrt{3}}{12}+\frac{1}{4}$$

Now solve the equation x=\frac{2±\frac{2\sqrt{3}}{3}}{8} when ± is minus. Subtract \frac{2\sqrt{3}}{3} from 2.

$$x=\frac{-\frac{2\sqrt{3}}{3}+2}{8}$$

Divide 2-\frac{2\sqrt{3}}{3} by 8.

$$x=-\frac{\sqrt{3}}{12}+\frac{1}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{3}}{12}+\frac{1}{4} x=-\frac{\sqrt{3}}{12}+\frac{1}{4}$$

Steps for Completing the Square

Add \frac{13}{12} and \frac{1}{12} to get \frac{7}{6}.

$$\frac{7}{6}-2x+4x^{2}=1$$

Subtract \frac{7}{6} from both sides.

$$-2x+4x^{2}=1-\frac{7}{6}$$

Subtract \frac{7}{6} from 1 to get -\frac{1}{6}.

$$-2x+4x^{2}=-\frac{1}{6}$$

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.

$$4x^{2}-2x=-\frac{1}{6}$$

Divide both sides by 4.

$$\frac{4x^{2}-2x}{4}=-\frac{\frac{1}{6}}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{2}{4}\right)x=-\frac{\frac{1}{6}}{4}$$

Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{1}{2}x=-\frac{\frac{1}{6}}{4}$$

Divide -\frac{1}{6} by 4.

$$x^{2}-\frac{1}{2}x=-\frac{1}{24}$$

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=-\frac{1}{24}+\left(-\frac{1}{4}\right)^{2}$$

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{1}{24}+\frac{1}{16}$$

Add -\frac{1}{24} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{1}{48}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{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}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{1}{48}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{1}{48}}$$

Simplify.

$$x-\frac{1}{4}=\frac{\sqrt{3}}{12} x-\frac{1}{4}=-\frac{\sqrt{3}}{12}$$

Add \frac{1}{4} to both sides of the equation.

$$x=\frac{\sqrt{3}}{12}+\frac{1}{4} x=-\frac{\sqrt{3}}{12}+\frac{1}{4}$$