Solve z^2=4z+1

$${ z }^{ 2 } =4z+1$$

Solve for z

$$z=\sqrt{5}+2\approx 4.236067977$$

$$z=2-\sqrt{5}\approx -0.236067977$$

Steps Using the Quadratic Formula

Subtract 4z from both sides.

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

Subtract 1 from both sides.

$$z^{2}-4z-1=0$$

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

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

Square -4.

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

Multiply -4 times -1.

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

Add 16 to 4.

$$z=\frac{-\left(-4\right)±\sqrt{20}}{2}$$

Take the square root of 20.

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

The opposite of -4 is 4.

$$z=\frac{4±2\sqrt{5}}{2}$$

Now solve the equation z=\frac{4±2\sqrt{5}}{2} when ± is plus. Add 4 to 2\sqrt{5}.

$$z=\frac{2\sqrt{5}+4}{2}$$

Divide 4+2\sqrt{5} by 2.

$$z=\sqrt{5}+2$$

Now solve the equation z=\frac{4±2\sqrt{5}}{2} when ± is minus. Subtract 2\sqrt{5} from 4.

$$z=\frac{4-2\sqrt{5}}{2}$$

Divide 4-2\sqrt{5} by 2.

$$z=2-\sqrt{5}$$

The equation is now solved.

$$z=\sqrt{5}+2 z=2-\sqrt{5}$$

Steps for Completing the Square

Subtract 4z from both sides.

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

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -2.

$$z^{2}-4z+4=1+4$$

Add 1 to 4.

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

Factor z^{2}-4z+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}.

$$\left(z-2\right)^{2}=5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(z-2\right)^{2}}=\sqrt{5}$$

Simplify.

$$z-2=\sqrt{5} z-2=-\sqrt{5}$$

Add 2 to both sides of the equation.

$$z=\sqrt{5}+2 z=2-\sqrt{5}$$