What Is a Quadratic Equation?
A quadratic equation is a polynomial equation whose highest power of the variable is $2$. The standard form is:
$$ax^2 + bx + c = 0, \quad a \neq 0$$
Here, $a$, $b$, and $c$ are constants and $x$ is the variable. The requirement that $a \neq 0$ is what makes the equation quadratic — if $a = 0$, the $x^2$ term vanishes and it becomes a linear equation.
The two solutions (called roots or zeros) are given by the quadratic formula:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
A quadratic equation has up to two solutions. The graph of the corresponding function $f(x) = ax^2 + bx + c$ is a parabola — a U-shaped curve that opens upward when $a > 0$ and downward when $a < 0$.
The Three Methods for Solving Quadratic Equations
Three methods solve every quadratic equation. Each one works for different situations.
Method 1: Factoring
Factoring works when the quadratic can be split into two linear factors with integer coefficients. Solve $x^2 - 5x + 6 = 0$.
Step 1. Find two numbers that multiply to $c = 6$ and add to $b = -5$. They are $-2$ and $-3$ (since $(-2)(-3) = 6$ and $-2 + -3 = -5$).
Step 2. Rewrite the quadratic as a product:
$$x^2 - 5x + 6 = (x - 2)(x - 3) = 0$$
Step 3. A product equals zero only when at least one factor is zero. So $x - 2 = 0$ or $x - 3 = 0$, giving:
$$\boxed{x = 2 \quad \text{or} \quad x = 3}$$
Check: $2^2 - 5(2) + 6 = 0$ ✓ and $3^2 - 5(3) + 6 = 0$ ✓.
Factoring is the fastest method when it works — but it only works when the roots are nice rational numbers. For most quadratics in real applications, factoring fails.
Method 2: Completing the Square
Completing the square is al-Khwarizmi's original method. It works for every quadratic. Solve $x^2 + 6x - 7 = 0$.
Step 1. Move the constant to the RHS:
$$x^2 + 6x = 7$$
Step 2. Take half the coefficient of $x$ — half of $6$ is $3$ — and square it: $3^2 = 9$. Add this to both sides:
$$x^2 + 6x + 9 = 7 + 9$$
Step 3. The LHS is now a perfect square trinomial:
$$(x + 3)^2 = 16$$
Step 4. Take square roots of both sides — remembering the $\pm$:
$$x + 3 = \pm 4$$
Step 5. Solve each case: $x = -3 + 4 = 1$ or $x = -3 - 4 = -7$.
$$\boxed{x = 1 \quad \text{or} \quad x = -7}$$
Method 3: The Quadratic Formula
The quadratic formula is the most general method — derived once, by completing the square on the general form $ax^2 + bx + c = 0$, and then memorised:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Solve $2x^2 - 7x + 3 = 0$. Identify $a = 2$, $b = -7$, $c = 3$.
Step 1. Compute the discriminant:
$$b^2 - 4ac = (-7)^2 - 4(2)(3) = 49 - 24 = 25$$
Step 2. Substitute into the formula:
$$x = \frac{-(-7) \pm \sqrt{25}}{2(2)} = \frac{7 \pm 5}{4}$$
Step 3. Compute both solutions:
$$x = \frac{7 + 5}{4} = 3 \quad \text{or} \quad x = \frac{7 - 5}{4} = \frac{1}{2}$$
$$\boxed{x = 3 \quad \text{or} \quad x = \tfrac{1}{2}}$$
The Discriminant — Why It Matters
The piece under the square root in the quadratic formula has a special name: the discriminant, written $\Delta$ (Greek capital delta) or $D$:
$$\Delta = b^2 - 4ac$$
The discriminant tells you how many solutions the quadratic equation has — before you finish solving. Three cases:
Discriminant | Number of real solutions | What the graph looks like |
|---|---|---|
$\Delta > 0$ | Two distinct real solutions | The parabola crosses the x-axis at two points |
$\Delta = 0$ | One repeated real solution | The parabola just touches the x-axis at one point (vertex) |
$\Delta < 0$ | No real solutions (two complex) | The parabola doesn't touch the x-axis |
Example. For $x^2 - 5x + 6 = 0$, the discriminant is $\Delta = (-5)^2 - 4(1)(6) = 25 - 24 = 1 > 0$ — so there are two distinct real solutions (which we found above: $x = 2$ and $x = 3$).
Example. For $x^2 + 4 = 0$, the discriminant is $\Delta = 0^2 - 4(1)(4) = -16 < 0$ — so there are no real solutions. The graph $y = x^2 + 4$ is a parabola entirely above the x-axis.
What Are the Sum and Product of Roots? (Vieta's Formulas)
For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, two relationships hold without solving the equation:
$$\alpha + \beta = -\frac{b}{a} \qquad \alpha \beta = \frac{c}{a}$$
These are Vieta's formulas, named for François Viète (1540–1603), the French mathematician who first wrote them down explicitly.
Why they're true. If $\alpha$ and $\beta$ are the roots, the quadratic factors as $a(x - \alpha)(x - \beta) = 0$. Expanding:
$$a(x - \alpha)(x - \beta) = a\bigl[x^2 - (\alpha + \beta)x + \alpha\beta\bigr] = ax^2 - a(\alpha + \beta)x + a\alpha\beta$$
Matching coefficients with $ax^2 + bx + c$: the $x$-coefficient gives $-a(\alpha + \beta) = b$, so $\alpha + \beta = -b/a$. The constant gives $a\alpha\beta = c$, so $\alpha\beta = c/a$.
Worked example. For $x^2 - 5x + 6 = 0$ (where $a = 1$, $b = -5$, $c = 6$):
Sum of roots: $\alpha + \beta = -(-5)/1 = 5$
Product of roots: $\alpha\beta = 6/1 = 6$
Two numbers that sum to 5 and multiply to 6 are 2 and 3 — and indeed the roots are $x = 2$ and $x = 3$ ✓.
Why this is useful.
Check your work. After solving by formula or factoring, plug your roots into Vieta's to verify.
Build a quadratic from roots. Given desired roots 4 and $-1$: sum $= 3$, product $= -4$. The monic quadratic is $x^2 - 3x - 4 = 0$.
Symmetric functions of roots. Quantities like $\alpha^2 + \beta^2$, $\frac{1}{\alpha} + \frac{1}{\beta}$, $(\alpha - \beta)^2$ can be computed without finding the roots, using only $\alpha+\beta$ and $\alpha\beta$.
Example — symmetric function shortcut. Find $\alpha^2 + \beta^2$ for $2x^2 - 7x + 3 = 0$ without solving.
Sum: $\alpha + \beta = 7/2$. Product: $\alpha\beta = 3/2$. Use the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$:
$$\alpha^2 + \beta^2 = (7/2)^2 - 2(3/2) = 49/4 - 3 = 37/4$$
No roots needed.
Where Quadratics Show Up in the Real World
Quadratic equations describe an entire class of physical situations:
Projectile motion. A ball thrown into the air follows the path $y = -\frac{1}{2} g t^2 + v_0 t + h_0$ — a quadratic in $t$. Solving $y = 0$ tells you when it lands.
Area problems. A rectangle with length $x + 5$ and width $x$ has area $x(x + 5) = x^2 + 5x$. Setting this equal to a target area gives a quadratic.
Optimisation. The maximum or minimum value of a quadratic $ax^2 + bx + c$ occurs at the vertex, $x = -\frac{b}{2a}$. This is used everywhere from minimising cost to maximising area.
Physics — kinetic energy. $E = \frac{1}{2} m v^2$ is quadratic in $v$, so doubling the speed quadruples the energy. This is why car crashes at 60 mph cause four times the damage of crashes at 30 mph.
Engineering — parabolic reflectors. Satellite dishes and car headlights use the parabolic shape because every ray entering parallel to the axis reflects to a single focal point.
At Bhanzu, our trainers always tie the quadratic formula back to one of these real-world situations early in the teaching. The reason is simple: the formula is hard to remember in isolation, but it sticks once a student has seen what it actually computes.
A Worked Example — Wrong Path First
Solve $x^2 + 4x - 21 = 0$ using the quadratic formula.
The intuitive (wrong) approach. A student in a hurry plugs in $a = 1$, $b = 4$, $c = -21$ and writes:
$$x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-21)}}{2(1)} = \frac{-4 \pm \sqrt{16 + 84}}{2}$$
So far so good — but then writes $\sqrt{100} = 10$ and forgets the $\pm$:
$$x = \frac{-4 + 10}{2} = 3$$
One solution found. The student moves on — and misses the second root.
Why it fails. The $\pm$ in the formula is doing real work — it produces two solutions. Skipping it costs you exactly half the answer.
The correct method.
$$x = \frac{-4 \pm \sqrt{100}}{2} = \frac{-4 \pm 10}{2}$$
Two cases:
$x = \frac{-4 + 10}{2} = \frac{6}{2} = 3$
$x = \frac{-4 - 10}{2} = \frac{-14}{2} = -7$
$$\boxed{x = 3 \quad \text{or} \quad x = -7}$$
Check: $3^2 + 4(3) - 21 = 9 + 12 - 21 = 0$ ✓ and $(-7)^2 + 4(-7) - 21 = 49 - 28 - 21 = 0$ ✓.
The rusher who skips the $\pm$ is one of the most common archetypes in quadratic problem sets — the answer they get isn't wrong, exactly; it's just missing the other one.
Common Mistakes with Quadratic Equations
Three failure modes account for most of the wrong answers on quadratic problems.
Mistake 1: Forgetting the $\pm$ in the quadratic formula
Where it slips in: Plugging into the formula and computing only one root.
Don't do this: $x = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$ as your final answer.
The correct way: Always write both roots: $x = \frac{-b + \sqrt{\Delta}}{2a}$ AND $x = \frac{-b - \sqrt{\Delta}}{2a}$. A quadratic has up to two roots; if you only report one, you've found half the answer.
Mistake 2: Sign error with $-b$ when $b$ is already negative
Where it slips in: $-b$ in the formula when the coefficient $b$ is itself negative.
Don't do this: For $b = -7$, writing $-b = -7$. (You've forgotten that $-b$ means the opposite of $b$.)
The correct way: $-b = -(-7) = 7$. The negative-of-a-negative is positive. The memorizer who learned "$-b$ means add a minus sign" without the underlying logic hits this constantly. Double-check the sign explicitly when $b$ is negative.
Mistake 3: Dividing both sides by a variable
Where it slips in: $x^2 = 5x$ tempts students to "divide both sides by $x$" and get $x = 5$.
Don't do this: Dividing by $x$ — because if $x = 0$ is a solution (which it is here), you've just lost it.
The correct way: Move everything to one side: $x^2 - 5x = 0$. Factor: $x(x - 5) = 0$. Apply zero-product: $x = 0$ or $x = 5$. Both are solutions. The second-guesser who instinctively distrusts "divide by $x$" is right — and should trust that instinct.
The Mathematicians Who Shaped Quadratic Equations
Brahmagupta (598–670 CE, India) — First to give a systematic, written rule for solving quadratic equations in his 628 CE treatise Brāhmasphuṭasiddhānta. Also the first mathematician to acknowledge that quadratics have two solutions — including negative ones, which earlier mathematicians had refused to accept.
Muhammad ibn Musa al-Khwarizmi (c. 780–c. 850, Persia/Baghdad) — Developed the geometric method of completing the square for quadratics around 820 CE in Al-Jabr. His procedure remains the foundation of how completing-the-square is taught today.
Bhaskara II (1114–1185, India) — In his 1150 CE book Bījaganita ("Seed Counting"), he gave essentially the modern quadratic formula with both positive and negative roots, and explained the role of the discriminant in determining the nature of the solutions.
Three mathematicians — separated by 600 years and three civilisations — built the formula on every high-school algebra test today.
A Practical Next Step
Try these three problems before going further into polynomial equations.
Solve $x^2 - 4x - 5 = 0$ by factoring. (Find two numbers that multiply to $-5$ and add to $-4$.)
Solve $2x^2 + 5x - 3 = 0$ using the quadratic formula. (Compute the discriminant first.)
Find the discriminant of $3x^2 - 4x + 2 = 0$. How many real solutions does it have?
If problem 2 felt tough, go back to the worked example using $2x^2 - 7x + 3 = 0$ above and follow the steps in parallel. Want to build this confidence with a live Bhanzu trainer? Book a free demo class — online globally.
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