Quadratic

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complex roots

Zeros of Quadratic Functions

To obtain the zeros of a quadratic function, we solve quadratic equations and we have already seen three methods of solving quadratic equations.

Factoring
Completing the Square
Quadratic Formula

All three methods rely on first rewriting the equation, so that one side equals $0$.

When we first investigated quadratic equations, we were only interested in real roots or real solutions. We are now in a position to fill-in the rest of the picture, because we can now work with square roots of negative numbers.

Complex Solutions

Solve $3t^2 + 2t + 5 = 0$

Using the quadratic formula, we get two solutions

\[ t = \frac {-2 \pm \sqrt {2^2 - 4 \cdot 3 \cdot 5}}{2 \cdot 3} = \frac {-2 \pm \sqrt {-56}}{6} = \frac {-2 \pm \sqrt {56}\sqrt {-1}}{6} = \frac {-2 \pm \sqrt {56} \, i}{6} = \frac {-2 \pm i \,\sqrt {56}}{6} \]

We can simplify this, if you would like.

\[ t = \frac {-2 \pm \sqrt {4 \cdot 14} \, i}{6} = \frac {-2 \pm i \, \sqrt {4 \cdot 14}}{6} = \frac {-2 \pm i \,\sqrt {4} \sqrt {14}}{6} = \frac {-2 \pm 2 i \, \sqrt {14}}{6} = \frac {2(-1 \pm i \,\sqrt {14} \, i)}{6} = \frac {-1 \pm i \,\sqrt {14}}{3} \]

Complex Roots

Factor $3 t^2 + 2 t + 5$

In the previous example we found the roots of this polynomial to be

\[ \frac {-1 + \sqrt {14} \, i}{3} \, \text { and } \, \frac {-1 - \sqrt {14} \, i}{3} \]

Therefore, the factored form of this polynomial looks like

\[ A \left (t - \frac {-1 + \sqrt {14} \, i}{3} \right ) \left (t - \frac {-1 - \sqrt {14} \, i}{3} \right ) \]

If we were to mulitply out the two factors, we would end up with $1$ as a leading coefficient. Therefore, $A = 3$.

\[ 3 \left (t - \frac {-1 + \sqrt {14} \, i}{3} \right ) \left (t - \frac {-1 - \sqrt {14} \, i}{3} \right ) \]

Let’s multiply out the factorization in the last example and see if it works.

\[ 3 \left (t - \frac {-1 + \sqrt {14} \, i}{3} \right ) \left (t - \frac {-1 - \sqrt {14} \, i}{3} \right ) \]

\[ 3 \left ( t^2 - \left (\frac {-1 + \sqrt {14} \, i}{3} + \frac {-1 - \sqrt {14} \, i}{3}\right ) t + \left (\frac {-1 + \sqrt {14} \, i}{3}\right ) \left (\frac {-1 - \sqrt {14} \, i}{3}\right ) \right ) \]

\[ 3 \left ( t^2 - \left (\frac {-2}{3} \right ) t + \frac {(-1 + \sqrt {14} \, i)(-1 - \sqrt {14} \, i)}{9}\right ) \]

\[ 3 \left ( t^2 - \left (\frac {-2}{3} \right ) t + \frac {1 - 14 \, i^2}{9}\right ) \]

\[ 3 \left ( t^2 - \left (\frac {-2}{3} \right ) t + \frac {15}{9} \right ) \]

\[ 3 \left ( t^2 - \left (\frac {-2}{3} \right ) t + \frac {5}{3}\right ) \]

\[ 3 t^2 + 2 t + 5 \]

Analyze $g(k) = \frac {1}{10} (k+6)(k+2)(k-4)$

Domain: This is a polynomial funciton, so its domain is all real numbers.

Zeros: $g$ is given in factored form and the zeros of $g$ are the zeros of these factors: $-6$, $-2$, and $4$.

Continuity: This is a polynomial function, so it is continuous on its domain. There are not discontinuities or singularities.

End-Behavior: This is a polynomial function of odd degree with a positive leading coefficient.

\[ \lim \limits _{x \to -\infty } g(k) = -\infty \, \text { and } \, \lim \limits _{x \to \infty } g(k) = \infty \]

Range: This also tells us that the range is $(-\infty , \infty )$.

Behavior:

Algebra really has no way of finding the critical numbers of $g$ and obtaining exact intervals where $g$ increases and decreases. Caluclus will provide a method of obtaining the derivative, which will help with this problem.

A DESMOS graph suggests that there are two critical numbers. DESMOS approximates these to be around $-4.239$ and $1.573$.

Note: These are not the middle points between the zeros.

These critical numbers are the locations of a local maximum and minimum for our polynomial function. And, a graph is about the best we can do for reasoning on local extrema.

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Calculus will show us how to get the derivative:

\[ g'(k) = \frac {3}{10} k^2 + \frac {4}{5} x - 2 \]

The critical numbers of $g(k)$ are the zeros of the derivative, which is a quadratic function. We can use the quadratic formula.

To make things a little eaier to write and read, let’s factor out $\frac {1}{10}$.

\[ g'(k) = \frac {3}{10} k^2 + \frac {4}{5} x - 2 = \frac {1}{10} ( 3 k^2 + 8 k - 20 ) \]

The zeros of $g'(k)$ are the zeros of $3 k^2 + 8 k - 20$.

We need to solve $3 k^2 + 8 k - 20 = 0$

\[ k = \frac {-8 \pm \sqrt {64 + 4 \cdot 3 \cdot 20}}{6} = \frac {-8 \pm \sqrt {304}}{6} \]

\[ = \frac {-8 \pm 4 \sqrt {19}}{6} = \frac {-4 \pm 2 \sqrt {19}}{3} \]

\[ \frac {-4 - 2 \sqrt {19}}{3} \approx 1.573 \]

\[ \frac {-4 + 2 \sqrt {19}}{3} \approx -4.239 \]

And, that agrees with our graph.

$g$ increases on $\left ( -\infty , \frac {-4 - 2 \sqrt {19}}{3} \right )$
$g$ decreases on $\left ( \frac {-4 - 2 \sqrt {19}}{3}, \frac {-4 + 2 \sqrt {19}}{3} \right )$
$g$ increases on $\left ( \frac {-4 + 2 \sqrt {19}}{3}, \infty \right )$

Extrema: (Maximums and minimums)

$\blacktriangleright $ $g\left ( \frac {-4 - 2 \sqrt {19}}{3} \right )$ is a local maximum of $g$ occuring at $\frac {-4 - 2 \sqrt {19}}{3}$

$\blacktriangleright $ $g\left ( \frac {-4 + 2 \sqrt {19}}{3} \right )$ is a local minimum of $g$ occuring at $\frac {-4 + 2 \sqrt {19}}{3}$

The end-behavior of $g$ tells us that there is no global maximum or minimum.

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more examples can be found by following this link
More Examples of Zeros

2025-05-18 02:00:23