Quadratic and cubic polynomials

4.2. Quadratic and cubic polynomials#

A polynomial of degree \(n\)

(4.1)#\[\begin{equation} a_n x^n + \cdots + a_1 x_1 + a_0 = 0 \end{equation}\]

has \(n\) roots, which may be real or complex. Real roots may be distinct or repeated. Complex roots always come in conjugate pairs, \(a \pm bi\), where a is called the real part, \(\pm b\) is called the imaginary part, and \(i = \sqrt{-1}\) is the imaginary unit.

4.2.1. Quadratic polynomials#

Quadratic polynomials can have three types of roots:

  • Two real roots

  • One (repeated) root

  • Two complex roots

Graphically, two real roots occur as two intersections of a parabola with the y-axis. For example, the roots of \(x^2 - 1 = 0\) occur at \(x = \pm 1\):

Two Real Solutions

One repeated root occurs as a single point of a parabola (its minimum or maximum) touching the y-axis. For example, the root of \((x-1)^2 = 0\) occurs at \(x = 1\):

One Duplicate Solution

Complex roots do not appear as intersections on a graph. For example, the roots of \(x^2 + 1 = 0\) are \(x = \pm i\):

Complex Solution

There are several strategies that can be used to find roots of quadratic polynomials:

  • Factoring

    (4.2)#\[\begin{equation} x^2 + 8 x - 9 = (x+9)(x-1) = 0 \end{equation}\]

    This equation will be true if either factor is zero, so the roots are \(x = -9\) or \(x = 1\).

  • Completing the square

    (4.3)#\[\begin{align} x^2 + 8x &= 9 \\ x^2 + 8x + 16 &= 9 + 16 \\ (x+4)^2 &= 25 \\ x + 4 &= \pm \sqrt{25} x = -4 \pm 5 \end{align}\]

    This gives the same roots!

  • Quadratic formula

    For the polynomial \(x^2 + 8x - 9 = 0\),

    (4.5)#\[\begin{align} x &= \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot -9}}{2 \cdot 1} \\ &= \frac{-8 \pm \sqrt{100}}{2} \\ &= -4 \pm 5 \end{align}\]

Example: Box optimization

An open-top box will be made from an 8.5” x 11” piece of paper by cutting out a square from each corner and folding the flaps. What size square should be cut to make the biggest box?

The volume of the box that will be obtained by cutting a square of edge length x is:

(4.6)#\[\begin{equation} V(x) = (8.5 - 2x)(11-2x)x = 4x^3 - 39x^2 + 93.5x \end{equation}\]

The volume will be at an extremum with respect to x if \(V'(x) = 0\):

(4.7)#\[\begin{equation} V'(x) = 12x^2 - 78x + 93.5 = 0 \end{equation}\]

The roots of this quadratic polynomial are:

(4.8)#\[\begin{align} x &= \frac{78 \pm \sqrt{(-78)^2 - 4 \cdot 12 \cdot -93.5}}{2 \cdot 12} \\ &= 1.59\, 4.91 \end{align}\]

Only the first root, 1.59”, is physical because there is not enough material to remove 4.91” from two corners!

Example: Complex roots

Find the roots of

(4.9)#\[\begin{equation} x^2 - 10x + 34 = 0 \end{equation}\]

I will use the technique of completing the square:

(4.10)#\[\begin{align} x^2 - 10x &= -34 \\ x^2 - 10x + 25 &= -34 + 25 \\ (x - 5)^2 &= -9 \\ x - 5 &= \pm \sqrt{-9} \\ x &= 5 \pm 3\sqrt{-1} \\ x &= 5 \pm 3i \end{align}\]

4.2.2. Cubic (and higher) polynomials#

General formulas for roots of cubic and quartic polynomials are known but complicated. No such formula exists for higher-order polynomials. In these cases, numerical methods are needed!

4.2.3. Skill builder problems#

Solve all roots to 3 significant figures.

  1. \(x^2 + 2x + 1 = 0\), by factoring.

    Solution
    (4.11)#\[\begin{align} x^2 + 2x + 1 &= 0 \\ (x + 1)^2 &= 0 \\ x &= -1 \end{align}\]

  2. \(x^2 - 8x + 5 = 0\), by completing the square.

    Solution
    (4.12)#\[\begin{align} x^2 - 8x &= -5 \\ x^2 - 8x + 16 &= -5 + 16 \\ (x - 4)^2 &= 11 \\ x &= 4 \pm \sqrt{11} \\ x &\approx 0.683,\, 7.32 \end{align}\]

  3. \(x^2 - 7x + 12 = 0\), by quadratic formula.

    Solution
    (4.13)#\[\begin{align} x &= \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 2 \cdot 12}}{2 \cdot 2} \\ x &= \frac{7 \pm \sqrt{49 - 96}}{4} \\ x &= \frac{7 \pm \sqrt{-47}}{4} \\ x &= \frac{7}{4} \pm \frac{\sqrt{47}}{4}i \\ x &\approx 1.75 \pm 1.71i \end{align}\]
Contents