Solve 4+x=2x^2

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

Solve for x

$$x = \frac{\sqrt{33} + 1}{4} \approx 1.686140662$$

$$x=\frac{1-\sqrt{33}}{4}\approx -1.186140662$$

Steps Using the Quadratic Formula

Subtract 2x^{2} from both sides.

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

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.

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

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

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

Square 1.

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

Multiply -4 times -2.

$$x=\frac{-1±\sqrt{1+8\times 4}}{2\left(-2\right)}$$

Multiply 8 times 4.

$$x=\frac{-1±\sqrt{1+32}}{2\left(-2\right)}$$

Add 1 to 32.

$$x=\frac{-1±\sqrt{33}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-1±\sqrt{33}}{-4}$$

Now solve the equation x=\frac{-1±\sqrt{33}}{-4} when ± is plus. Add -1 to \sqrt{33}.

$$x=\frac{\sqrt{33}-1}{-4}$$

Divide -1+\sqrt{33} by -4.

$$x=\frac{1-\sqrt{33}}{4}$$

Now solve the equation x=\frac{-1±\sqrt{33}}{-4} when ± is minus. Subtract \sqrt{33} from -1.

$$x=\frac{-\sqrt{33}-1}{-4}$$

Divide -1-\sqrt{33} by -4.

$$x=\frac{\sqrt{33}+1}{4}$$

The equation is now solved.

$$x=\frac{1-\sqrt{33}}{4} x=\frac{\sqrt{33}+1}{4}$$

Steps for Completing the Square

Subtract 2x^{2} from both sides.

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

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

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

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.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide 1 by -2.

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

Divide -4 by -2.

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

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.

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

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

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

Add 2 to \frac{1}{16}.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{33}{16}$$

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}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{33}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{33}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{\sqrt{33}}{4} x-\frac{1}{4}=-\frac{\sqrt{33}}{4}$$

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

$$x=\frac{\sqrt{33}+1}{4} x=\frac{1-\sqrt{33}}{4}$$