Solve z^2-(2+i)z+3+i=0

$$z ^ { 2 } - ( 2 + i ) z + 3 + i = 0$$

Solve for z

$$z=1+2i$$

$$z=1-i$$

Steps Using the Quadratic Formula

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.

$$z^{2}+\left(-2-i\right)z+\left(3+i\right)=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -2-i for b, and 3+i for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$z=\frac{2+i±\sqrt{\left(-2-i\right)^{2}-4\left(3+i\right)}}{2}$$

Square -2-i.

$$z=\frac{2+i±\sqrt{3+4i-4\left(3+i\right)}}{2}$$

Multiply -4 times 3+i.

$$z=\frac{2+i±\sqrt{3+4i+\left(-12-4i\right)}}{2}$$

Add 3+4i to -12-4i.

$$z=\frac{2+i±\sqrt{-9}}{2}$$

Take the square root of -9.

$$z=\frac{2+i±3i}{2}$$

Now solve the equation z=\frac{2+i±3i}{2} when ± is plus. Add 2+i to 3i.

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

Divide 2+4i by 2.

$$z=1+2i$$

Now solve the equation z=\frac{2+i±3i}{2} when ± is minus. Subtract 3i from 2+i.

$$z=\frac{2-2i}{2}$$

Divide 2-2i by 2.

$$z=1-i$$

The equation is now solved.

$$z=1+2i z=1-i$$

Steps for Completing the Square

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.

$$z^{2}+\left(-2-i\right)z+\left(3+i\right)=0$$

Subtract 3+i from both sides of the equation.

$$z^{2}+\left(-2-i\right)z+\left(3+i\right)-\left(3+i\right)=-3-i$$

Subtracting 3+i from itself leaves 0.

$$z^{2}+\left(-2-i\right)z=-3-i$$

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

$$z^{2}+\left(-2-i\right)z+\left(-1-\frac{1}{2}i\right)^{2}=-3-i+\left(-1-\frac{1}{2}i\right)^{2}$$

Square -1-\frac{1}{2}i.

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

Add -3-i to \frac{3}{4}+i.

$$z^{2}+\left(-2-i\right)z+\left(\frac{3}{4}+i\right)=-\frac{9}{4}$$

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

$$\left(z+\left(-1-\frac{1}{2}i\right)\right)^{2}=-\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(z+\left(-1-\frac{1}{2}i\right)\right)^{2}}=\sqrt{-\frac{9}{4}}$$

Simplify.

$$z+\left(-1-\frac{1}{2}i\right)=\frac{3}{2}i z+\left(-1-\frac{1}{2}i\right)=-\frac{3}{2}i$$

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

$$z=1+2i z=1-i$$