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Solving Quadratic Equations

Quadratic equations are second-degree equations that model many real situations—from areas and trajectories to profit and optimization. In this chapter you will learn to recognize the standard form \(ax^2+bx+c=0\), identify the coefficients, and choose an appropriate solving method: factoring (including special products), completing the square, and the quadratic formula. You will also use the discriminant to predict the number of real solutions.

Quadratic Equation

Definition Quadratic Equation
A quadratic equation is an equation that can be written in the standard form:$$ax^2 + bx + c = 0$$where \(a, b,\) and \(c\) are known coefficients, and \(a \neq 0\). The condition \(a \neq 0\) is what makes the equation quadratic.
A solution or root of the equation is a value of \(x\) that makes the statement true.
Example
Consider the equation \(3x^2 + 5x + 4 = 0\). Is it a quadratic equation? If yes, identify the coefficients \(a\), \(b\), and \(c\).

Yes. It is in the form \(ax^2+bx+c=0\) with \(a=3\), \(b=5\), \(c=4\), and \(a\neq 0\), so it is a quadratic equation.

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

A primary strategy for solving quadratic equations is to use the Null Factor Law. This law allows us to break a single quadratic equation into simpler linear equations. To use it, we must first factorize the quadratic expression.
Method Solving by Factorization
The primary strategy for solving quadratic equations is to use the Null Factor Law.
  1. Write the equation in standard form, \(ax^2+bx+c=0\).
  2. Factorize the quadratic expression completely.
  3. Apply the Null Factor Law: set each factor equal to zero.
  4. Solve each resulting linear equation.
Proposition Null Factor Law
If the product of two or more factors is equal to zero, then at least one of the factors must be equal to zero.
If \(\textcolor{colordef}{A}\,\textcolor{colorprop}{B} = 0\) then \(\textcolor{colordef}{A=0}\) or \(\textcolor{colorprop}{B=0}\).
Note: one or both of the factors can be zero.
Example
Solve \((x - 1)(x + 2) = 0\).

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

Factorization Techniques for Special Forms of Equations

Before learning a general method, we first master solving equations that can be factored using familiar patterns.
Proposition Common Factor (\(c\equal 0\))
For equations of the form \(ax^2+bx=0\), the common factor is \(x\):$$x(ax+b)=0 \Leftrightarrow x=0 \text{ or } ax+b=0$$
Example
Find the roots of \(x^2 - 2x = 0\).

$$\begin{aligned}x^2 - 2x &= 0 \\ x(x-2) &= 0 &&\color{gray}\text{(Factor out the common factor }x\color{gray}\text{)} \\ x = 0 &\quad \text{or} \quad x-2=0 &&\color{gray}\text{(Null factor law)} \\ x = 0 &\quad \text{or} \quad x=2\end{aligned}$$

Proposition Difference of Squares (\(b\equal 0\))
For equations of the form \(x^2-k=0\) (where \(k>0\)):$$x^2 - (\sqrt{k})^2 = 0 \Leftrightarrow (x - \sqrt{k})(x + \sqrt{k})=0$$
Example
Solve \(x^2 - 9 = 0\).

$$\begin{aligned}x^2 - 9 &= 0 \\ x^2 - 3^2 &= 0 \\ (x - 3)(x + 3) &= 0 &&\color{gray}\text{(Difference of squares)} \\ x - 3 = 0 &\quad \text{or} \quad x + 3 = 0 &&\color{gray}\text{(Null factor law)} \\ x = 3 &\quad \text{or} \quad x = -3\end{aligned}$$

Proposition Perfect Squares
For equations of the form \(x^2 \pm 2ax + a^2=0\):$$\begin{aligned}x^2 + 2ax + a^2 &= (x + a)^2 = 0 \Leftrightarrow x + a = 0,\\ x^2 - 2ax + a^2 &= (x - a)^2 = 0 \Leftrightarrow x - a = 0.\end{aligned}$$
Example
Solve \(x^2 + 2x + 1 = 0\).

$$\begin{aligned}x^2 + 2x + 1 &= 0 \\ (x + 1)^2 &= 0 &&\color{gray}\text{(Perfect square factorization)} \\ x + 1 &= 0 &&\color{gray}\text{(Null factor law)} \\ x &= -1\end{aligned}$$This is a repeated root (double root).

The General Method: Completing the Square

When a quadratic expression is not one of the special forms, we need a general method. This method is called completing the square. The idea is to rewrite the quadratic as a perfect square plus (or minus) a constant. When solving an equation, this eventually leads to a difference of two squares or to taking square roots.
Proposition Completing the Square
Any quadratic expression of the form \(x^2 + bx + c\) (leading coefficient \(1\)) can be put into vertex form by completing the square:$$x^2 + bx + c = \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c.$$For a general quadratic \(ax^2 + bx + c\) with \(a\neq 0\), we first factor out \(a\):$$ax^2 + bx + c = a\left(x^2 + \frac{b}{a}x + \frac{c}{a}\right)$$and then complete the square inside the parentheses.

We start with the expression \(x^2 + bx + c\). Our goal is to manipulate the first two terms to create a perfect square trinomial of the form \((x+k)^2 = x^2 + 2kx + k^2\).$$ \begin{aligned}x^2 + bx + c = & (x^2 + bx) + c && \color{gray}\text{(Group the \(x\)-terms)} \\ \end{aligned} $$Comparing \(x^2+bx\) to \(x^2+2kx\), we see that we need \(b=2k\), so \(k=\frac{b}{2}\). The term needed to complete the square is \(k^2 = \left(\frac{b}{2}\right)^2\). We add and subtract this term inside the parentheses.$$ \begin{aligned}x^2 + bx + c = & \left(x^2 + bx + \left(\frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2\right) + c && \color{gray}\text{(Add and subtract the “magic” term)} \\ = & \left(x^2 + bx + \left(\frac{b}{2}\right)^2\right) - \left(\frac{b}{2}\right)^2 + c && \color{gray}\text{(Group the perfect square trinomial)} \\ = & \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c && \color{gray}\text{(Factor the trinomial)}\end{aligned} $$This completes the proof.

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

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

Method General method to solve quadratic equations
Given an equation \(ax^2+bx+c=0\) with \(a\neq 0\):
  • Step 1: Complete the square to rewrite the left-hand side as a perfect square plus/minus a constant.
  • Step 2: Use the difference of squares (or take square roots) to isolate \(x\).
  • Step 3: Apply the Null Factor Law if the expression has been written as a product.
  • Step 4: Solve the linear equations obtained.
Example
Solve \(x^2 + 10x + 24 = 0\).

We know that \((x + 5)^2 = x^2 + 10x + 25\). So$$\begin{aligned}x^2 + 10x + 24 &= 0 \\ x^2 + 10x + 25 -25 +24 &= 0 \\ (x + 5)^2 - 1 &= 0 &&\color{gray}\text{(Complete the square)} \\ (x + 5)^2 - 1^2 &= 0 &&\color{gray}\text{(Difference of squares)}\\ (x + 5 - 1)(x + 5 + 1) &= 0 &&\color{gray}\text{(Factorize)}\\ (x + 4)(x + 6) &= 0 &&\color{gray}\text{(Simplify)}\\ x + 4 = 0 &\text{ or } x + 6 = 0 &&\color{gray}\text{(Null factor law)} \\ x = -4 &\text{ or } x = -6 &&\color{gray}\text{(Solve)}.\end{aligned}$$

Quadratic Formula

Applying the method of completing the square to the general equation \(ax^2 + bx + c = 0\) gives a formula that solves any quadratic equation. This is the quadratic formula.
Proposition Quadratic formula
For any quadratic equation \(ax^2 + bx + c = 0\), the discriminant, denoted \(\Delta\), is defined as $$\Delta = b^2 - 4ac.$$Its sign determines the number of real solutions:
  • 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 (a double 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 \\ x^2 + \frac{b}{a}x + \frac{c}{a} &= 0 \quad \color{gray}\text{(Divide each term by \(a\), since \(a \neq 0\))} \\ x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 + \frac{c}{a} &= 0 \quad \color{gray}\text{(Complete the square)} \\ \left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a^2} &= 0 \quad \color{gray}\text{(Simplify)} \\ \left(x + \frac{b}{2a}\right)^2 - \frac{\Delta}{4a^2} &= 0 \quad \color{gray}\text{(Where \(\Delta = b^2 - 4ac\))}.\end{aligned}$$Now, consider the cases based on the discriminant \(\Delta\):
  • Case \(\Delta \geq 0\): Since \(\dfrac{\Delta}{4a^2} \geq 0\), a real square root exists.$$\begin{aligned}\left(x + \frac{b}{2a}\right)^2 &= \frac{\Delta}{4a^2} \\ \left(x + \frac{b}{2a}\right)^2 - \left(\sqrt{\frac{\Delta}{4a^2}}\right)^2 &= 0 \\ \left(x + \frac{b}{2a} - \sqrt{\frac{\Delta}{4a^2}}\right) \left(x + \frac{b}{2a} + \sqrt{\frac{\Delta}{4a^2}}\right) &= 0 \quad \color{gray}\text{(Difference of squares)}.\end{aligned}$$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): \(x = -\dfrac{b}{2a}\).
  • Case \(\Delta < 0\): Then \(\dfrac{\Delta}{4a^2} < 0\), 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 two distinct real solutions.
  3. \(\begin{aligned}[t]x &= \frac{-b - \sqrt{\Delta}}{2a} &&\text{ or } && x = \frac{-b + \sqrt{\Delta}}{2a} \\x &= \frac{-2 - \sqrt{16}}{2 \cdot 1} &&\text{ or } && x = \frac{-2 + \sqrt{16}}{2 \cdot 1} \\x &= \frac{-2 - 4}{2} &&\text{ or } && x = \frac{-2 + 4}{2} \\x &= -3 &&\text{ or }&& x = 1\end{aligned}\)