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# Quadratic Formula and Complex Sums

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Practice Quadratic Formula and Complex Sums
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The quadratic function $y=x^2-2x+3$ (shown below) does not intersect the x-axis and therefore has no real roots. What are the complex roots of the function?

### Guidance

Recall that the imaginary number, $i$ , is a number whose square is –1:

${\color{red}i^2 = -1}$ and ${\color{red}i=\sqrt{-1}}$

The sum of a real number and an imaginary number is called a complex number . Examples of complex numbers are $5+4i$ and $3-2i$ . All complex numbers can be written in the form $a+bi$ where $a$ and $b$ are real numbers. Two important points:

• The set of real numbers is a subset of the set of complex numbers where $b=0$ . Examples of real numbers are $2, 7, \frac{1}{2}, -4.2$ .
• The set of imaginary numbers is a subset of the set of complex numbers where $a=0$ . Examples of imaginary numbers are $i, -4i, \sqrt{2}i$ .

This means that the set of complex numbers includes real numbers, imaginary numbers, and combinations of real and imaginary numbers.

When a quadratic function does not intersect the x-axis, it has complex roots. When solving for the roots of a function algebraically using the quadratic formula, you will end up with a negative under the square root symbol. With your knowledge of complex numbers, you can still state the complex roots of a function just like you would state the real roots of a function.

#### Example A

Solve the following quadratic equation for $x$ .

$\boxed{m^2-2m+5=0}$

Solution: You can use the quadratic formula to solve. For this quadratic equation, $a=1, b=-2, c=5$ .

$& m= \frac{-b \pm \sqrt{b^2-4ac}}{2a} \\& m= \frac{-({\color{red}-2}) \pm \sqrt{({\color{red}-2})^2-4({\color{red}1})({\color{red}5})}}{2({\color{red}1})} \\& m= \frac{2 \pm \sqrt{4-20}}{2} \\& m= \frac{2 \pm \sqrt{-16}}{2} \qquad \quad \sqrt{-16} = \sqrt{16} \times i =4i \\& m= \frac{2 \pm 4i}{2} \\& m=1 \pm 2i \\& m=1+2i \ or \ m=1-2i$

There are no real solutions to the equation. The solutions to the quadratic equation are $1+2i \ and \ 1-2i$ .

#### Example B

Solve the following equation by rewriting it as a quadratic and using the quadratic formula.

$\frac{3}{e+3} - \frac{2}{e+2} =1$

Solution: To rewrite as a quadratic equation, multiply each term by $(e + 3) (e + 2)$ .

$& \frac{3}{e+3} {\color{red}(e+3)(e+2)} - \frac{2}{e+2} {\color{red}(e+3)(e+2)} = 1 {\color{red}(e+3)(e+2)} \\& 3(e+2)-2(e+3) = (e+3)(e+2)$

Expand and simplify.

$& 3e+6-2e-6=e^2+2e+3e+6 \\& e^2+4e+6=0$

Solve using the quadratic formula. For this quadratic equation, $a=1, b=4, c=6$ .

$& e= \frac{-b \pm \sqrt{b^2-4ac}}{2a} \\& e= \frac{-({\color{red}4}) \pm \sqrt{({\color{red}4})^2-4({\color{red}1})({\color{red}6})}}{2({\color{red}1})} \\& e= \frac{-4 \pm \sqrt{16-24}}{2} \\& e= \frac{-4 \pm \sqrt{-8}}{2} \qquad \quad \sqrt{-8} = \sqrt{8} \times i = \sqrt{4 \cdot 2} \times i = 2i \sqrt{2} \\& e= \frac{-4 \pm 2i\sqrt{2}}{2}\\& e= -2 \pm i \sqrt{2} \\& e= -2 + i \sqrt{2} \ or \ e=-2-i\sqrt{2}$

There are no real solutions to the equation. The solutions to the equation are $-2+i\sqrt{2} \ and \ -2-i\sqrt{2}$

#### Example C

Sketch the graph of the following quadratic function. What are the roots of the function?

$y=x^2-4x+5$

Solution: Use your calculator or a table to make a sketch of the function. You should get the following:

As you can see, the quadratic function has no $x$ -intercepts; therefore, the function has no real roots. To find the roots (which will be complex), you must use the quadratic formula.

For this quadratic function, $a=1, b=-4, c=5$ .

$& x= \frac{-b \pm \sqrt{b^2-4ac}}{2a} \\& x= \frac{-({\color{red}-4}) \pm \sqrt{({\color{red}-4})^2-4({\color{red}1})({\color{red}5})}}{2({\color{red}1})} \\& x= \frac{4 \pm \sqrt{16-20}}{2} \\& x= \frac{4 \pm \sqrt{-4}}{2} \qquad \quad \sqrt{-4} = \sqrt{4} \times i = 2i \\& x= \frac{4 \pm 2i}{2} \\& x= 2 \pm i \\& x= 2 + i \ or \ x=2-i$

The complex roots of the quadratic function are $2+i \ and \ 2-i$ .

#### Concept Problem Revisited

To find the complex roots of the function $y=x^2-2x+3$ , you must use the quadratic formula.

For this quadratic function, $a=1, b=-2, c=3$ .

$& x= \frac{-b \pm \sqrt{b^2-4ac}}{2a} \\& x= \frac{-({\color{red}-2}) \pm \sqrt{({\color{red}-2})^2-4({\color{red}1})({\color{red}3})}}{2({\color{red}1})} \\& x= \frac{2 \pm \sqrt{4-12}}{2} \\& x= \frac{2 \pm \sqrt{-8}}{2} \qquad \quad \sqrt{-8} = \sqrt{8} \times i = 2\sqrt{2}i \\& x= \frac{2 \pm 2\sqrt{2}i}{2} \\& x= 1 \pm \sqrt{2}i$

### Vocabulary

Complex Number
A complex number is a number in the form $a + bi$ where $a$ and $b$ are real numbers and $i^2=-1$ .
Imaginary Number
An imaginary number is a number such that its square is a negative number.
$\sqrt{-16}$ is an imaginary number beacause its square is –16.
$\sqrt{-16}=i\sqrt{16} = 4i$

### Guided Practice

1. Solve the following quadratic equation. Express all solutions in simplest radical form.

$2n^2+n=-4$

2. Solve the following quadratic equation. Express all solutions in simplest radical form.

$m^2+(m+1)^2+(m+2)^2=-1$

3. Is it possible for a quadratic function to have exactly one complex root?

1. $2n^2+n=-4$

Set the equation equal to zero.

$2n^2+n+4=0$

$& x= \frac{-b \pm \sqrt{b^2-4ac}}{2a} \\& n= \frac{-({\color{red}1}) \pm \sqrt{({\color{red}1})^2-4({\color{red}2})({\color{red}4})}}{2({\color{red}2})} \\& n= \frac{-1 \pm \sqrt{1-32}}{4} \\& n= \frac{-1 \pm \sqrt{-31}}{4} \\& n= \frac{-1 \pm i\sqrt{31}}{4}$

2. $m^2+(m+1)^2+(m+2)^2=-1$

Expand and simplify.

$& m^2 + (m+1)(m+1) + (m+2)(m+2) = -1 \\& m^2 + m^2 + m + m + 1+ m^2 +2m +2m +4 = -1 \\& 3m^2+6m+5=-1$

Write the equation in general form.

$3m^2 +6m+6=0$

Divide by 3 to simplify the equation.

$& \frac{3m^2}{{\color{red}3}} + \frac{6m}{{\color{red}3}} + \frac{6}{{\color{red}3}} = \frac{0}{{\color{red}3}} \\& m^2+2m+2=0$

$& m= \frac{-b \pm \sqrt{b^2-4ac}}{2a} \\& m= \frac{-({\color{red}2}) \pm \sqrt{({\color{red}2})^2-4({\color{red}1})({\color{red}2})}}{2({\color{red}1})} \\& m= \frac{-2 \pm \sqrt{4-8}}{2} \\& m= \frac{-2 \pm \sqrt{-4}}{2} \\& m= \frac{-2 \pm 2i}{2} \\& m= -1 \pm i$

3. No, even in higher degree polynomials, complex roots will always come in pairs. Consider when you use the quadratic formula-- if you have a negative under the square root symbol, both the + version and the - version of the two answers will end up being complex.

### Practice

1. If a quadratic function has 2 x-intercepts, how many complex roots does it have? Explain.
2. If a quadratic function has no x-intercepts, how many complex roots does it have? Explain.
3. If a quadratic function has 1 x-intercept, how many complex roots does it have? Explain.
4. If you want to know whether a function has complex roots, which part of the quadratic formula is it important to focus on?
5. You solve a quadratic equation and get 2 complex solutions. How can you check your solutions?
6. In general, you can attempt to solve a quadratic equation by graphing, factoring, completing the square, or using the quadratic formula. If a quadratic equation has complex solutions, what methods do you have for solving the equation?

Solve the following quadratic equations. Express all solutions in simplest radical form.

1. $x^2+x+1=0$
2. $5y^2-8y=-6$
3. $2m^2-12m+19=0$
4. $-3x^2-2x=2$
5. $2x^2+4x=-11$
6. $-x^2+x-23=0$
7. $-3x^2+2x=14$
8. $x^2+5=-x$
9. $\frac{1}{2}d^2+4d=-12$