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Solving Quadratic Equations

The purpose of this section is to learn how to solve quadratic equations.

Definition

Definition Quadratic Equation
Given numbers \(a\), \(b\), and \(c\) with \(a \neq 0\), a quadratic equation is any equation of the form$$ax^2 + bx + c = 0.$$
Example
Is \(3x^2 + 5x + 4\) a quadratic polynomial? If yes, identify the coefficients \(a\), \(b\), and \(c\).

Yes: \(a=3\), \(b=5\), \(c=4\).

Definition Root
A root of the equation \(ax^2 + bx + c = 0\) is any number that, when substituted for \(x\), makes the equation true.
Example
Are \(1\) and \(3\) roots of the equation \(x^2 - 3x + 2 = 0\)?

To check if \(1\) and \(3\) are roots, substitute each into the equation:
  • For \(x=1\), \(1^2 - 3 \cdot 1 + 2 = 1 - 3 + 2 = 0\).
    So \(1\) is a root.
  • For \(x=3\), \(3^2 - 3 \cdot 3 + 2 = 9 - 9 + 2 = 2 \neq 0\).
    So \(3\) is not a root

A quadratic equation may have no real solution. For example, \(x^2 =-1\) has no real solution because the square of a real number cannot be negative.

Solving by Factorization

To solve a quadratic equation of the form \( ax^2 + bx + c = 0 \) with \( a \neq 0 \), we can leverage our understanding of solving linear equations. The key idea is to transform the quadratic equation into a product of linear factors using factorization. This allows us to convert the problem into solving simpler linear equations, which we already know how to handle.
Proposition Null Factor Law
If \(ab = 0\), then \(a = 0\) or \(b = 0\).
Example
Solve \((x - 1)(x + 2) = 0\).

$$\begin{aligned}(x-1)(x+2) &= 0\\ x-1 = 0 &\quad \text{or} \quad x+2=0 &&\text{(null factor law)} \\ x = 1 &\quad \text{or} \quad x = -2 &&\text{(solve each equation)}\\ \end{aligned}$$

Method Solving by Factorization
To solve a quadratic equation:
  1. Factorize,
  2. Apply the null factor law,
  3. Solve the resulting linear equations.
This turns the problem of solving a quadratic equation into finding a factorization.

Factorization Techniques for Special Forms of Equations

Proposition Common Factor Law for equations of the form \(x^2 + ax\)
$$x^2 + ax = x(x + a)$$
Example
Find the roots of \(x^2 - 2x = 0\).

$$\begin{aligned}x^2 - 2x &= 0 \\ x(x-2) &= 0 &&\text{(factorizing)} \\ x = 0 &\quad \text{or} \quad x-2=0 &&\text{(null factor law)} \\ x = 0 &\quad \text{or} \quad x=2 &&\text{(solving linear equations)} \\ \end{aligned}$$

Proposition Perfect Square for equations of the form \(x^2 + 2ax + a^2\)
$$x^2 + 2ax + a^2 = (x + a)^2$$
Example
Solve \(x^2 + 2x + 1 = 0\).

$$\begin{aligned}x^2 + 2x + 1 &= 0 \\ (x + 1)^2 &= 0 &&\text{(perfect square)} \\ x + 1 &= 0 &&\text{(null factor law)} \\ x &= -1 &&\text{(solving linear equation)} \\ \end{aligned}$$So, \(-1\) is a double root.

Proposition Difference of Squares for equations of the form \(x^2 - a^2\)
$$x^2 - a^2 = (x - a)(x + a)$$
Example
Solve \(x^2 - 9 = 0\).

$$\begin{aligned}x^2 - 9 &= 0 \\ (x - 3)(x + 3) &= 0 &&\text{(difference of squares)} \\ x - 3 = 0 &\quad \text{or} \quad x + 3 = 0 &&\text{(null factor law)} \\ x = 3 &\quad \text{or} \quad x = -3 &&\text{(solving linear equations)} \\ \end{aligned}$$

Factorization by Completing the Square

Some quadratics like \(x^2 + 2x - 3\) cannot be factored easily. In this case, we use completing the square.
Proposition Completing the Square
$$x^2 + bx + c = \left(x + \frac{b}{2}\right)^2 + c - \left(\frac{b}{2}\right)^2.$$

$$\begin{aligned}\left(x + \frac{b}{2}\right)^2 &= x^2 + bx + \left(\frac{b}{2}\right)^2 &\quad &\text{(perfect square)}\\ \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c &= x^2 + bx + \left(\frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c &\quad &\text{(adding \(- \left(\frac{b}{2}\right)^2 + c\) to both sides)}\\ \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c &= x^2 + bx + c &\quad &\text{(simplifying)}.\end{aligned}$$

Example
Complete the square for \(x^2 + 10x + 24 = 0\).

We know that \((x + 5)^2 = x^2 + 10x + 25\). So$$\begin{aligned}x^2 + 10x + 24 &= 0 \\ x^2 + 10x + 25 - 1 &= 0 &&\text{(rewrite \(24\) as \(25 - 1\))}\\ (x + 5)^2 - 1 &= 0 &&\text{(complete the square)}.\end{aligned}$$

Method General method to solve quadratic equations
  • Step 1: Complete the square
  • Step 2: Use the difference of squares
  • Step 3: Apply the null factor law
  • Step 4: Solve the linear equations
Example
Solve \(x^2 + 10x + 24 = 0\).

We know that \((x + 5)^2 = x^2 + 10x + 25\). So$$\begin{aligned}x^2 + 10x + 24 &= 0 \\ x^2 + 10x + 25 - 1 &= 0 &&\text{(rewrite \(24\) as \(25 - 1\))}\\ (x + 5)^2 - 1 &= 0 &&\text{(complete the square)} \\ (x + 5)^2 - 1^2 &= 0 &&\text{(difference of squares)}\\ (x + 5 - 1)(x + 5 + 1) &= 0 &&\text{(factorize)}\\ (x + 4)(x + 6) &= 0 &&\text{(simplify)}\\ x + 4 = 0 &\text{ or } x + 6 = 0 &&\text{(null factor law)} \\ x = -4 &\text{ or } x = -6 &&\text{(solve)}.\end{aligned}$$

Quadratic Formula

Definition Discriminant
Given a quadratic equation \(ax^2 + bx + c = 0\), the discriminant, denoted \(\Delta\), is defined as $$\Delta = b^2 - 4ac.$$
Proposition Quadratic formula
For any quadratic equation \(ax^2 + bx + c = 0\):
  • If \(\Delta > 0\), there are two real roots:$$x = \frac{-b - \sqrt{\Delta}}{2a}\text{ or }x = \frac{-b + \sqrt{\Delta}}{2a}$$
  • If \(\Delta = 0\), there is one real root:$$x = \frac{-b}{2a}.$$
  • If \(\Delta < 0\), there are no real roots.

Suppose \(ax^2 + bx + c = 0\), where \(a \neq 0\).$$\begin{aligned}ax^2 + bx + c &= 0 \\ \therefore x^2 + \frac{b}{a}x + \frac{c}{a} &= 0 \quad \text{(divide each term by \(a\), since \(a \neq 0\))} \\ \therefore x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 + \frac{c}{a} &= 0 \quad \text{(complete the square)} \\ \therefore \left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a^2} &= 0 \quad \text{(simplify)} \\ \therefore \left(x + \frac{b}{2a}\right)^2 - \frac{\Delta}{4a^2} &= 0 \quad \text{(where \(\Delta = b^2 - 4ac\))}.\end{aligned}$$Now, consider the cases based on the discriminant \(\Delta\):
  • Case \(\Delta \geq 0\): Since \(\frac{\Delta}{4a^2} \geq 0\), a real square root exists.$$\begin{aligned}\left(x + \frac{b}{2a}\right)^2 - \left(\sqrt{\frac{\Delta}{4a^2}}\right)^2 &= 0 \\ \therefore \left(x + \frac{b}{2a} - \sqrt{\frac{\Delta}{4a^2}}\right) \left(x + \frac{b}{2a} + \sqrt{\frac{\Delta}{4a^2}}\right) &= 0 \quad \text{(difference of squares)}.\end{aligned}$$Applying the null factor law:$$x + \frac{b}{2a} - \sqrt{\frac{\Delta}{4a^2}} = 0 \quad \text{or} \quad x + \frac{b}{2a} + \sqrt{\frac{\Delta}{4a^2}} = 0.$$Solving these linear equations:$$x = -\frac{b}{2a} + \sqrt{\frac{\Delta}{4a^2}} \quad \text{or} \quad x = -\frac{b}{2a} - \sqrt{\frac{\Delta}{4a^2}}.$$Simplifying:$$x = \frac{-b \pm \sqrt{\Delta}}{2a}.$$
    • If \(\Delta > 0\), there are two distinct real roots.
    • If \(\Delta = 0\), there is one real root (double root): \(x = -\frac{b}{2a}\).
  • Case \(\Delta < 0\): Then \(\frac{\Delta}{4a^2} < 0\), so$$\left(x + \frac{b}{2a}\right)^2 = \frac{\Delta}{4a^2} < 0.$$Since the square of a real number is non-negative, there are no real solutions.

Example
Consider the quadratic equation \(x^2 + 2x - 3 = 0\).
  1. Find the discriminant.
  2. Hence, state the nature of the roots of the equation.
  3. Solve the equation.

\(x^2 + 2x - 3 = 0\) has \(a=1\), \(b=2\), \(c=-3\).
  1. \(\begin{aligned}[t]\Delta &= b^2 - 4ac \\&= (2)^2 - 4(1)(-3) \\&= 4 + 12 \\&= 16\end{aligned}\)
  2. As \(\Delta > 0\), there are 2 distinct roots.
  3. \(\begin{aligned}[t]x &= \frac{-b - \sqrt{\Delta}}{2a} &\text{ or } x &= \frac{-b + \sqrt{\Delta}}{2a} \\x &= \frac{-2 - \sqrt{16}}{2 \cdot 1} &\text{ or } x &= \frac{-2 + \sqrt{16}}{2 \cdot 1} \\x &= \frac{-2 - 4}{2} &\text{ or } x &= \frac{-2 + 4}{2} \\x &= -3 &\text{ or } x &= 1\end{aligned}\)