Solve x=1x^2-4

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

Solve for x

$$x = \frac{\sqrt{17} + 1}{2} \approx 2.561552813$$

$$x=\frac{1-\sqrt{17}}{2}\approx -1.561552813$$

Steps Using the Quadratic Formula

Subtract 1x^{2} from both sides.

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

Add 4 to both sides.

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

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.

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

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

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

Square 1.

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

Multiply -4 times -1.

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

Multiply 4 times 4.

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

Add 1 to 16.

$$x=\frac{-1±\sqrt{17}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-1±\sqrt{17}}{-2}$$

Now solve the equation x=\frac{-1±\sqrt{17}}{-2} when ± is plus. Add -1 to \sqrt{17}.

$$x=\frac{\sqrt{17}-1}{-2}$$

Divide -1+\sqrt{17} by -2.

$$x=\frac{1-\sqrt{17}}{2}$$

Now solve the equation x=\frac{-1±\sqrt{17}}{-2} when ± is minus. Subtract \sqrt{17} from -1.

$$x=\frac{-\sqrt{17}-1}{-2}$$

Divide -1-\sqrt{17} by -2.

$$x=\frac{\sqrt{17}+1}{2}$$

The equation is now solved.

$$x=\frac{1-\sqrt{17}}{2} x=\frac{\sqrt{17}+1}{2}$$

Steps for Completing the Square

Subtract 1x^{2} from both sides.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 1 by -1.

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

Divide -4 by -1.

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

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.

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

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

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

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

$$x^{2}-x+\frac{1}{4}=\frac{17}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{17}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{17}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{17}}{2} x-\frac{1}{2}=-\frac{\sqrt{17}}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{\sqrt{17}+1}{2} x=\frac{1-\sqrt{17}}{2}$$