10.3 Solve Quadratic Equations Using the Quadratic Formula

The quadratic formula lets you solve any quadratic equation, even when factoring or completing the square won't work cleanly. It's the most versatile method in your toolkit for this unit, and understanding it well will save you time on both homework and exams.

Solving Quadratic Equations

Application of the Quadratic Formula

Every quadratic equation can be written in standard form: $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants and $a \neq 0$. The quadratic formula solves for $x$ directly from those coefficients:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The $\pm$ symbol means you get two solutions: one where you add the square root term, and one where you subtract it.

Steps to apply the quadratic formula:

1. Write the equation in standard form ($ax^2 + bx + c = 0$) if it isn't already.
2. Identify $a$, $b$, and $c$. Be careful with signs, especially when terms are negative.
3. Substitute those values into the formula.
4. Simplify the expression under the square root (this part is called the discriminant).
5. Calculate both solutions by using $+$ and then $-$ with the square root term.

Example: Solve $2x^2 + 7x - 4 = 0$

• $a = 2$, $b = 7$, $c = -4$
• $x = \frac{-7 \pm \sqrt{7^2 - 4(2)(-4)}}{2(2)}$
• $x = \frac{-7 \pm \sqrt{49 + 32}}{4} = \frac{-7 \pm \sqrt{81}}{4} = \frac{-7 \pm 9}{4}$
• $x_1 = \frac{-7 + 9}{4} = \frac{2}{4} = \frac{1}{2}$
• $x_2 = \frac{-7 - 9}{4} = \frac{-16}{4} = -4$

A common mistake: when $c$ is negative, the $-4ac$ part becomes positive (negative times negative). Watch your signs in that step.

Interpretation of the Discriminant

The discriminant is the expression under the square root: $b^2 - 4ac$. Its value tells you what kind of solutions to expect before you finish solving.

• Positive discriminant ($b^2 - 4ac > 0$): two distinct real solutions. The parabola crosses the x-axis at two points.
• Zero discriminant ($b^2 - 4ac = 0$): one repeated real solution. The parabola just touches the x-axis at its vertex.
• Negative discriminant ($b^2 - 4ac < 0$): no real solutions. The parabola doesn't cross the x-axis at all. (At this level, you'll sometimes see these called complex solutions, which involve the imaginary unit $i$.)

Example: Determine the nature of solutions for $3x^2 - 5x + 2 = 0$

• $a = 3$, $b = -5$, $c = 2$
• Discriminant $= (-5)^2 - 4(3)(2) = 25 - 24 = 1$
• Since $1 > 0$, the equation has two distinct real solutions.

Checking the discriminant first is a quick way to know what you're dealing with, especially on multiple-choice problems where you just need to identify the number of solutions.

Choosing a Method for Quadratic Equations

You now have three methods for solving quadratics. Here's when each one works best:

• Factoring is the fastest option when:
  • The coefficients are integers
  • The leading coefficient ($a$) is 1 or factors out easily
  • The product of $a$ and $c$ is small enough to find factor pairs quickly
• Completing the square works well when:
  • The equation doesn't factor neatly
  • The leading coefficient is 1 (or you can divide it out)
  • You need the equation in vertex form ($y = a(x - h)^2 + k$)
• The quadratic formula is your go-to when:
  • Factoring isn't obvious
  • The leading coefficient isn't 1 and is hard to work with
  • The solutions turn out to be irrational (like $\sqrt{2}$) or complex

A solid general approach: try factoring first since it's quickest. If that doesn't work, use the quadratic formula. Save completing the square for when you specifically need vertex form.

Quadratic Equations and Their Graphs

When you graph a quadratic equation, you get a parabola. The solutions (roots) of $ax^2 + bx + c = 0$ correspond to the x-intercepts of that parabola. This connects the algebra to the geometry:

• Two real solutions = the parabola crosses the x-axis twice
• One repeated solution = the parabola touches the x-axis at exactly one point
• No real solutions = the parabola floats entirely above or below the x-axis

The quadratic formula finds these x-intercepts, whether they're nice whole numbers, messy fractions, or irrational values. If the discriminant is negative, the parabola has no x-intercepts, and the roots involve the imaginary unit $i$ (where $i = \sqrt{-1}$).