DE-BOOK Quadratic Equations

Section Quadratic Equations

Quadratic equations appear frequently in differential equations, especially when solving second-order equations with constant coefficients. For example, the characteristic equation associated with a linear differential equation like \(ay'' + by' + cy = 0\) is a quadratic. Understanding how to solve quadratics is essential for finding solutions to these types of problems.

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A quadratic equation has the general form \(ax^2 + bx + c = 0\text{,}\) where \(a \ne 0\text{.}\) There are several common techniques to solve such equations:

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Factoring: Try to express the quadratic as a product of two binomials.

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Quadratic formula: Use the general formula
\begin{equation*} x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\text{.} \end{equation*}

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Completing the square: Rewrite the quadratic to form a perfect square trinomial.

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Each method has its advantages. Factoring is often the quickest when possible. The quadratic formula works for all cases and is especially useful when the equation doesn’t factor easily. Completing the square is less commonly used early on, but becomes more important in later topics, such as solving differential equations by variation of parameters or Laplace transforms.

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Let’s look at a few examples that illustrate all three methods.

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🌌 Example 303. Repeated Root.

Solve \(4x^2 + 12x + 9 = 0\text{.}\)

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Solution.

Factoring: Try factoring directly:

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\begin{align*} 4x^2 + 12x + 9 \amp = 0\\ (2x + 3)(2x + 3) \amp = 0\\ 2x + 3 = 0 \amp \Rightarrow x = -\frac{3}{2} \end{align*}

This gives a repeated root, also called a double root.

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Quadratic formula:

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\begin{align*} x \amp = \frac{-12 \pm \sqrt{144 - 144}}{8}\\ x \amp = \frac{-12}{8} = -\frac{3}{2} \end{align*}

Completing the square:

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\begin{align*} 4x^2 + 12x + 9 \amp = 0\\ 4x^2 + 12x \amp = -9\\ x^2 + 3x \amp = -\frac{9}{4}\\ x^2 + 3x + \frac{9}{4} \amp = 0\\ \left(x + \frac{3}{2}\right)^2 \amp = 0\\ x \amp = -\frac{3}{2} \end{align*}

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Answer.

\begin{equation*} x = -\frac{3}{2} \text{ (double root)} \end{equation*}

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🌌 Example 304. Two Real Roots.

Solve \(2x^2 - 9x - 35 = 0\text{.}\)

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Solution.

Factoring:

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\begin{align*} 2x^2 - 9x - 35 \amp = 0\\ (2x + 5)(x - 7) \amp = 0\\ x = -\frac{5}{2}, \quad x = 7 \end{align*}

Quadratic formula:

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\begin{align*} x \amp = \frac{9 \pm \sqrt{361}}{4}\\ x \amp = \frac{9 \pm 19}{4} = 7, -\frac{5}{2} \end{align*}

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Answer.

\begin{equation*} x = -\frac{5}{2},\quad x = 7 \end{equation*}

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🌌 Example 305. Complex Roots.

Solve \(x^2 - 4x + 13 = 0\text{.}\)

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Solution.

Quadratic formula:

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\begin{align*} x \amp = \frac{4 \pm \sqrt{-36}}{2}\\ x \amp = 2 \pm 3i \end{align*}

Completing the square:

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\begin{align*} x^2 - 4x + 13 \amp = 0\\ x^2 - 4x + 4 \amp = -9\\ (x - 2)^2 \amp = -9\\ x \amp = 2 \pm 3i \end{align*}

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Answer.

\begin{equation*} x = 2 \pm 3i \end{equation*}

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Name at least two methods for solving quadratic equations.
Answer.

Factoring, using the quadratic formula, completing the square.
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How many solutions does a quadratic equation have?
Answer.
A quadratic equation has two solutions. These may be:
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1. Two distinct real roots
  
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2. One repeated real root (double root)
  
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3. Two complex conjugate roots
  
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✳️ Solving Quadratic Equations.

The solution to the quadratic equation

\begin{equation} a x^2 + b x + c = 0\tag{47} \end{equation}

is given by the quadratic formula:

\begin{equation} x = \frac{-b \pm \sqrt{\DLBb b^2 - 4ac}}{2a}\text{.}\tag{48} \end{equation}

Notes:

The \(\pm\) gives two solutions, say \(x_1\) and \(x_2\text{.}\)
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\(x_1\) and \(x_2\) are also known as the roots of \(\ a x^2 + b x + c\text{.}\)
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The value, \({\DLBb b^2 - 4ac} \ \text{,}\) under the root in is called the discriminant.
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Equation (47) can be written as \(\quad\ds (x - x_1)(x - x_2) = 0 \text{.}\)
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If \({\DLBb b^2 - 4ac > 0}\text{,}\) then \(x_1\) and \(x_2\) are different real numbers.
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If \({\DLBb b^2 - 4ac = 0}\text{,}\) then \(x_1\) and \(x_2\) are the same real number (repeated).
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If \({\DLBb b^2 - 4ac < 0}\text{,}\) then \(x_1\) and \(x_2\) are complex and can be written as

\begin{equation*} x_1 = \alpha + \beta i, \quad x_2 = \alpha - \beta i\text{.} \end{equation*}

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