Solve x=4x^2+5x-4

$$x=4 { x }^{ 2 } +5x-4$$

Solve for x

$$x=\frac{\sqrt{5}-1}{2}\approx 0.618033989$$

$$x=\frac{-\sqrt{5}-1}{2}\approx -1.618033989$$

Steps Using the Quadratic Formula

Subtract 4x^{2} from both sides.

$$x-4x^{2}=5x-4$$

Subtract 5x from both sides.

$$x-4x^{2}-5x=-4$$

Combine x and -5x to get -4x.

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

Add 4 to both sides.

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

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.

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

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

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

Square -4.

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

Multiply -4 times -4.

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

Multiply 16 times 4.

$$x=\frac{-\left(-4\right)±\sqrt{16+64}}{2\left(-4\right)}$$

Add 16 to 64.

$$x=\frac{-\left(-4\right)±\sqrt{80}}{2\left(-4\right)}$$

Take the square root of 80.

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

The opposite of -4 is 4.

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

Multiply 2 times -4.

$$x=\frac{4±4\sqrt{5}}{-8}$$

Now solve the equation x=\frac{4±4\sqrt{5}}{-8} when ± is plus. Add 4 to 4\sqrt{5}.

$$x=\frac{4\sqrt{5}+4}{-8}$$

Divide 4+4\sqrt{5} by -8.

$$x=\frac{-\sqrt{5}-1}{2}$$

Now solve the equation x=\frac{4±4\sqrt{5}}{-8} when ± is minus. Subtract 4\sqrt{5} from 4.

$$x=\frac{4-4\sqrt{5}}{-8}$$

Divide 4-4\sqrt{5} by -8.

$$x=\frac{\sqrt{5}-1}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{5}-1}{2} x=\frac{\sqrt{5}-1}{2}$$

Steps for Completing the Square

Subtract 4x^{2} from both sides.

$$x-4x^{2}=5x-4$$

Subtract 5x from both sides.

$$x-4x^{2}-5x=-4$$

Combine x and -5x to get -4x.

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

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.

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

Divide both sides by -4.

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

Dividing by -4 undoes the multiplication by -4.

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

Divide -4 by -4.

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

Divide -4 by -4.

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

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

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add 1 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{5}{4}$$

Factor x^{2}+x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{5}{4}$$

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{\sqrt{5}-1}{2} x=\frac{-\sqrt{5}-1}{2}$$