Solve 4+x=4x+4x^2

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

Solve for x

$$x=\frac{\sqrt{73}-3}{8}\approx 0.693000468$$

$$x=\frac{-\sqrt{73}-3}{8}\approx -1.443000468$$

Steps Using the Quadratic Formula

Subtract 4x from both sides.

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

Combine x and -4x to get -3x.

$$4-3x=4x^{2}$$

Subtract 4x^{2} from both sides.

$$4-3x-4x^{2}=0$$

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.

$$-4x^{2}-3x+4=0$$

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

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

Square -3.

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

Multiply -4 times -4.

$$x=\frac{-\left(-3\right)±\sqrt{9+16\times 4}}{2\left(-4\right)}$$

Multiply 16 times 4.

$$x=\frac{-\left(-3\right)±\sqrt{9+64}}{2\left(-4\right)}$$

Add 9 to 64.

$$x=\frac{-\left(-3\right)±\sqrt{73}}{2\left(-4\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{73}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{3±\sqrt{73}}{-8}$$

Now solve the equation x=\frac{3±\sqrt{73}}{-8} when ± is plus. Add 3 to \sqrt{73}.

$$x=\frac{\sqrt{73}+3}{-8}$$

Divide 3+\sqrt{73} by -8.

$$x=\frac{-\sqrt{73}-3}{8}$$

Now solve the equation x=\frac{3±\sqrt{73}}{-8} when ± is minus. Subtract \sqrt{73} from 3.

$$x=\frac{3-\sqrt{73}}{-8}$$

Divide 3-\sqrt{73} by -8.

$$x=\frac{\sqrt{73}-3}{8}$$

The equation is now solved.

$$x=\frac{-\sqrt{73}-3}{8} x=\frac{\sqrt{73}-3}{8}$$

Steps for Completing the Square

Subtract 4x from both sides.

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

Combine x and -4x to get -3x.

$$4-3x=4x^{2}$$

Subtract 4x^{2} from both sides.

$$4-3x-4x^{2}=0$$

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

$$-3x-4x^{2}=-4$$

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.

$$-4x^{2}-3x=-4$$

Divide both sides by -4.

$$\frac{-4x^{2}-3x}{-4}=-\frac{4}{-4}$$

Dividing by -4 undoes the multiplication by -4.

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

Divide -3 by -4.

$$x^{2}+\frac{3}{4}x=-\frac{4}{-4}$$

Divide -4 by -4.

$$x^{2}+\frac{3}{4}x=1$$

Divide \frac{3}{4}, the coefficient of the x term, by 2 to get \frac{3}{8}. Then add the square of \frac{3}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{4}x+\left(\frac{3}{8}\right)^{2}=1+\left(\frac{3}{8}\right)^{2}$$

Square \frac{3}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{4}x+\frac{9}{64}=1+\frac{9}{64}$$

Add 1 to \frac{9}{64}.

$$x^{2}+\frac{3}{4}x+\frac{9}{64}=\frac{73}{64}$$

Factor x^{2}+\frac{3}{4}x+\frac{9}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{8}\right)^{2}=\frac{73}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{8}\right)^{2}}=\sqrt{\frac{73}{64}}$$

Simplify.

$$x+\frac{3}{8}=\frac{\sqrt{73}}{8} x+\frac{3}{8}=-\frac{\sqrt{73}}{8}$$

Subtract \frac{3}{8} from both sides of the equation.

$$x=\frac{\sqrt{73}-3}{8} x=\frac{-\sqrt{73}-3}{8}$$