Solve z/2+4=z^2

$$\frac { z } { 2 } + 4 = z ^ { 2 }$$

Solve for z

$$z = \frac{\sqrt{65} + 1}{4} \approx 2.265564437$$

$$z=\frac{1-\sqrt{65}}{4}\approx -1.765564437$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$z+8=2z^{2}$$

Subtract 2z^{2} from both sides.

$$z+8-2z^{2}=0$$

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.

$$-2z^{2}+z+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 1 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 1.

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

Multiply -4 times -2.

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

Multiply 8 times 8.

$$z=\frac{-1±\sqrt{1+64}}{2\left(-2\right)}$$

Add 1 to 64.

$$z=\frac{-1±\sqrt{65}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$z=\frac{-1±\sqrt{65}}{-4}$$

Now solve the equation z=\frac{-1±\sqrt{65}}{-4} when ± is plus. Add -1 to \sqrt{65}.

$$z=\frac{\sqrt{65}-1}{-4}$$

Divide -1+\sqrt{65} by -4.

$$z=\frac{1-\sqrt{65}}{4}$$

Now solve the equation z=\frac{-1±\sqrt{65}}{-4} when ± is minus. Subtract \sqrt{65} from -1.

$$z=\frac{-\sqrt{65}-1}{-4}$$

Divide -1-\sqrt{65} by -4.

$$z=\frac{\sqrt{65}+1}{4}$$

The equation is now solved.

$$z=\frac{1-\sqrt{65}}{4} z=\frac{\sqrt{65}+1}{4}$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$z+8=2z^{2}$$

Subtract 2z^{2} from both sides.

$$z+8-2z^{2}=0$$

Subtract 8 from both sides. Anything subtracted from zero gives its negation.

$$z-2z^{2}=-8$$

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.

$$-2z^{2}+z=-8$$

Divide both sides by -2.

$$\frac{-2z^{2}+z}{-2}=-\frac{8}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$z^{2}+\frac{1}{-2}z=-\frac{8}{-2}$$

Divide 1 by -2.

$$z^{2}-\frac{1}{2}z=-\frac{8}{-2}$$

Divide -8 by -2.

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

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.

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

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

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

Add 4 to \frac{1}{16}.

$$z^{2}-\frac{1}{2}z+\frac{1}{16}=\frac{65}{16}$$

Factor z^{2}-\frac{1}{2}z+\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}.

$$\left(z-\frac{1}{4}\right)^{2}=\frac{65}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(z-\frac{1}{4}\right)^{2}}=\sqrt{\frac{65}{16}}$$

Simplify.

$$z-\frac{1}{4}=\frac{\sqrt{65}}{4} z-\frac{1}{4}=-\frac{\sqrt{65}}{4}$$

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

$$z=\frac{\sqrt{65}+1}{4} z=\frac{1-\sqrt{65}}{4}$$