Solve 2x-1=4x^2

$$2 x - 1 = 4 x ^ { 2 }$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{3}i+1}{4}\approx 0.25-0.433012702i$$

$$x=\frac{1+\sqrt{3}i}{4}\approx 0.25+0.433012702i$$

Steps Using the Quadratic Formula

Subtract 4x^{2} from both sides.

$$2x-1-4x^{2}=0$$

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.

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

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

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

Square 2.

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

Multiply -4 times -4.

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

Multiply 16 times -1.

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

Add 4 to -16.

$$x=\frac{-2±\sqrt{-12}}{2\left(-4\right)}$$

Take the square root of -12.

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

Multiply 2 times -4.

$$x=\frac{-2±2\sqrt{3}i}{-8}$$

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

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

Divide -2+2i\sqrt{3} by -8.

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

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

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 4x^{2} from both sides.

$$2x-1-4x^{2}=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$2x-4x^{2}=1$$

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.

$$-4x^{2}+2x=1$$

Divide both sides by -4.

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

Dividing by -4 undoes the multiplication by -4.

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

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

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

Divide 1 by -4.

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

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.

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

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

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{1}{4}+\frac{1}{16}$$

Add -\frac{1}{4} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{3}{16}$$

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}.

$$\left(x-\frac{1}{4}\right)^{2}=-\frac{3}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{-\frac{3}{16}}$$

Simplify.

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

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

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