Table of Contents
The quadratic formula is a universal method for solving any quadratic equation that can be written in the standard form
$$
ax^2 + bx + c = 0,\quad a \neq 0.
$$
In this chapter, we focus on:
- Writing a quadratic in the correct form for the formula.
- Applying the formula correctly.
- Understanding and using the discriminant.
- Interpreting the type and number of solutions from the formula.
- Choosing when the quadratic formula is a good method compared to others (like factoring or completing the square).
You should already be familiar with what a quadratic equation is and the basic idea of solving quadratics from the parent chapter.
The quadratic formula
For the quadratic equation
$$
ax^2 + bx + c = 0,\quad a \neq 0,
$$
the solutions for $x$ are given by the quadratic formula:
$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
$$
Here:
- $a$ is the coefficient of $x^2$,
- $b$ is the coefficient of $x$,
- $c$ is the constant term,
- the symbol $\pm$ means you get two values of $x$:
- one using $+$,
- one using $-$.
Using the formula is mainly about three steps:
- Make sure the equation is in standard form $ax^2 + bx + c = 0$.
- Identify $a$, $b$, and $c$ correctly.
- Substitute $a$, $b$, and $c$ into the formula and simplify.
Putting an equation into standard form
Before you can apply the formula, the equation must be written as $ax^2 + bx + c = 0$ on one side and $0$ on the other.
Typical adjustments:
- Move all terms to one side.
- Combine like terms.
- Make sure terms are in order $x^2$, $x$, constant (the order helps, but is not strictly required).
Example (setup only):
Solve $2x^2 - 3 = 5x$ using the quadratic formula.
- Move all terms to one side:
$$
2x^2 - 5x - 3 = 0.
$$ - Now you can read off:
$a = 2,\; b = -5,\; c = -3$.
The solving (plugging into the formula) will be done in later examples, but the key point here is: always rearrange to standard form first.
Step-by-step use of the quadratic formula
Once you have $a$, $b$, and $c$:
- Compute the discriminant: $D = b^2 - 4ac$.
- Take its square root (if possible): $\sqrt{D}$.
- Form the two solutions:
$$
x_1 = \frac{-b + \sqrt{D}}{2a},\quad
x_2 = \frac{-b - \sqrt{D}}{2a}.
$$
Example 1: Two real solutions
Solve $x^2 - 5x + 6 = 0$ using the quadratic formula.
- Identify coefficients: $a = 1,\; b = -5,\; c = 6$.
- Discriminant:
$$
D = b^2 - 4ac = (-5)^2 - 4(1)(6) = 25 - 24 = 1.
$$ - Square root of discriminant:
$$
\sqrt{D} = \sqrt{1} = 1.
$$ - Use the formula:
$$
x = \frac{-b \pm \sqrt{D}}{2a}
= \frac{-(-5) \pm 1}{2(1)}
= \frac{5 \pm 1}{2}.
$$ - Split into two answers:
$$
x_1 = \frac{5 + 1}{2} = 3,\quad
x_2 = \frac{5 - 1}{2} = 2.
$$
In this case the quadratic happened to factor, but the quadratic formula finds the same solutions without factoring.
Example 2: Fractions in the answer
Solve $2x^2 + 3x - 2 = 0$.
- $a = 2,\; b = 3,\; c = -2$.
- Discriminant:
$$
D = 3^2 - 4(2)(-2) = 9 + 16 = 25.
$$ - $\sqrt{D} = \sqrt{25} = 5$.
- Apply the formula:
$$
x = \frac{-3 \pm 5}{2\cdot 2}
= \frac{-3 \pm 5}{4}.
$$ - Solutions:
$$
x_1 = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2},\quad
x_2 = \frac{-3 - 5}{4} = \frac{-8}{4} = -2.
$$
The formula naturally gives fraction answers when needed.
The discriminant and types of solutions
The discriminant of a quadratic
$$
ax^2 + bx + c = 0
$$
is the expression
$$
D = b^2 - 4ac
$$
inside the square root in the quadratic formula.
The sign of $D$ tells you how many real solutions the quadratic has:
- If $D > 0$: two distinct real solutions.
- If $D = 0$: one real solution (a repeated root).
- If $D < 0$: no real solutions (the solutions are complex, involving $i$).
This is very useful when you only need to know the number or type of solutions, not their exact values.
Example 3: One real solution ($D = 0$)
Solve $x^2 - 4x + 4 = 0$.
- $a = 1,\; b = -4,\; c = 4$.
- Discriminant:
$$
D = (-4)^2 - 4(1)(4) = 16 - 16 = 0.
$$ - Apply the formula:
$$
x = \frac{-(-4) \pm \sqrt{0}}{2(1)}
= \frac{4 \pm 0}{2}
= \frac{4}{2}
= 2.
$$
There is just one real solution, $x = 2$, counted twice (a double root). The graph of this quadratic touches the $x$-axis at $x = 2$ and turns around there.
Example 4: No real solution ($D < 0$)
Solve $x^2 + 4x + 5 = 0$.
- $a = 1,\; b = 4,\; c = 5$.
- Discriminant:
$$
D = 4^2 - 4(1)(5) = 16 - 20 = -4.
$$ - Since $D < 0$, there is no real solution.
If you are working only with real numbers (as is typical in early Algebra I), you can stop here and simply say that the equation has no real solution. In a later chapter on complex numbers, you can express the solutions using $i$.
Common mistakes and how to avoid them
When using the quadratic formula, several simple errors happen frequently. Paying attention to these will save you a lot of rework.
- Not putting the equation into standard form first
For example, with $x^2 = 3x - 1$, if you just say $a = 1, b = 0, c = 0$ you are ignoring the other terms.
Correct approach:
$$
x^2 - 3x + 1 = 0,
$$
so $a = 1, b = -3, c = 1$.
- Sign mistakes in $b$
- If the equation is $x^2 - 7x + 10 = 0$, then $b = -7$ (not $7$).
- The formula has $-b$, so be careful: $-(-7) = 7$.
- Forgetting that $a$ might not be 1
For $2x^2 + x - 3 = 0$, $a = 2$ and the denominator is $2a = 4$, not $2$.
- Incorrectly computing the discriminant
- Remember to square the entire $b$ (including its sign): $b^2 = (-5)^2 = 25$, not $-25$.
- Multiply carefully: $4ac$ means $4 \cdot a \cdot c$.
- Forgetting the two solutions
After computing
$$
x = \frac{-b \pm \sqrt{D}}{2a},
$$
you must actually evaluate both:
$$
\frac{-b + \sqrt{D}}{2a} \quad\text{and}\quad \frac{-b - \sqrt{D}}{2a}.
$$
- Not simplifying answers when possible
While it is not always required, often you can simplify square roots or fractions:
- If $D = 36$, use $\sqrt{36} = 6$.
- Reduce common factors in numerator and denominator when appropriate.
When to use the quadratic formula
There are multiple methods to solve quadratic equations (factoring, completing the square, and the quadratic formula, among others). The quadratic formula is especially useful when:
- The quadratic does not factor easily with small integers.
- You want a general, guaranteed method.
- You need to understand the type of solutions (from the discriminant).
Some practical guidelines:
- If the quadratic factors quickly in your head (like $x^2 - 5x + 6 = 0$), factoring is often faster.
- If you are not sure whether it factors, or factoring looks messy, use the quadratic formula.
- In many word problems, once you form a quadratic equation, the quadratic formula is a reliable way to find the unknown quantity.
Example 5: A non-factorable quadratic (over integers)
Solve $3x^2 + 2x - 7 = 0$.
- $a = 3,\; b = 2,\; c = -7$.
- Discriminant:
$$
D = 2^2 - 4(3)(-7)
= 4 + 84
= 88.
$$ - $\sqrt{88}$ can be simplified:
$$
\sqrt{88} = \sqrt{4 \cdot 22} = 2\sqrt{22}.
$$ - Apply the formula:
$$
x = \frac{-2 \pm 2\sqrt{22}}{2\cdot 3}
= \frac{-2 \pm 2\sqrt{22}}{6}.
$$ - Factor $2$ from the numerator:
$$
x = \frac{2(-1 \pm \sqrt{22})}{6}
= \frac{-1 \pm \sqrt{22}}{3}.
$$
This quadratic does not factor nicely with integers, but the quadratic formula still gives exact solutions.
Practice structure for solving with the quadratic formula
When you solve problems using the quadratic formula, it is helpful to follow a consistent structure:
- Write the equation in standard form: $ax^2 + bx + c = 0$.
- State $a$, $b$, and $c$ clearly.
- Compute the discriminant:
$$
D = b^2 - 4ac.
$$ - Write the quadratic formula with your values substituted:
$$
x = \frac{-b \pm \sqrt{D}}{2a}.
$$ - Simplify step by step:
- Evaluate the square root if possible.
- Simplify the numerator and denominator.
- Reduce fractions and simplify radicals where reasonable.
- State the final answers clearly, usually as:
$$
x = \text{(first solution)} \quad\text{or}\quad x = \text{(second solution)}.
$$
Organizing your work this way makes it easier to check each part and to catch mistakes early.