Quadratic Equations

A quadratic equation is one whose standard form is

$ax^2 + bx + c = 0$

where $a \ne 0$. Here $a$ is the coefficient of the $x^2$ term, $b$ is the coefficient of the $x$ term, and $c$ is the constant term. The general solution, valid whenever the equation is in standard form with $a \ne 0$, is the quadratic formula:

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

Why the discriminant tells you what to expect

The expression under the square root has its own name and its own job. The discriminant is

$\Delta = b^2 - 4ac$

and it controls how many real solutions exist, because a square root of a negative number is not real. If you are working with real numbers:

• if $\Delta > 0$, there are two distinct real solutions;
• if $\Delta = 0$, there is one double real solution;
• if $\Delta < 0$, there are no real solutions.

The delta doesn't replace the calculation, but it tells you immediately what kind of result to expect, which is also useful for checking that your final answer makes sense. This is why $\Delta$ appears inside the formula: it is the part of $x = \frac{-b \pm \sqrt{\Delta}}{2a}$ that decides whether the $\pm$ produces two values, one value, or none.

Before any of this, the equation must actually be quadratic. Simply seeing a term with $x^2$ isn't enough. For example, $3x^2 + 5x = 2$ is quadratic, but it must first be rewritten as $3x^2 + 5x - 2 = 0$. If the $x^2$ term cancels out during the steps, the equation is no longer quadratic.

Worked example: from coefficients to solutions

Let's solve:

$2x^2 - x - 3 = 0$

First, identify the coefficients:

$a = 2,\quad b = -1,\quad c = -3$

Now calculate the discriminant:

$\Delta = (-1)^2 - 4 \cdot 2 \cdot (-3) = 1 + 24 = 25$

Since $\Delta > 0$, we expect two distinct real solutions. Apply the quadratic formula:

$x = \frac{-(-1) \pm \sqrt{25}}{2 \cdot 2}$

$x = \frac{1 \pm 5}{4}$

This gives:

$x_1 = \frac{1 + 5}{4} = \frac{3}{2}$

$x_2 = \frac{1 - 5}{4} = -1$

A quick check confirms both:

$2\left(\frac{3}{2}\right)^2 - \frac{3}{2} - 3 = 0$

$2(-1)^2 - (-1) - 3 = 0$

This highlights the key point: reading the signs correctly is just as important as the formula itself. If you get $b = -1$ wrong, the rest of the process will be incorrect.

Practice and verify

Try to solve:

$x^2 - 6x + 8 = 0$

First calculate $\Delta$, then decide whether to use the quadratic formula or factoring, and finish by substituting each solution back into the equation. If the trinomial factors immediately, factoring can be faster; when the factorization isn't obvious, the quadratic formula is the most reliable path, since it always produces the solutions from the coefficients $a$, $b$, and $c$. If you want to compare your steps afterward, use a math solver.

Calculation pitfalls

• Not setting the equation to zero before starting.
• Swapping the sign of $b$ or $c$ when copying the coefficients.
• Forgetting that the classification using $\Delta$ applies to real solutions.
• Using only one part of the formula and forgetting the $\pm$ symbol.
• Skipping the final verification.

When are they used?

Quadratic equations appear frequently in algebra, in the study of parabolas, and in problems where one quantity depends on the square of another. You'll also find them in exercises involving areas, intersections between graphs, and simple trajectory models. They aren't just for applying a memorized formula; they are used to describe situations where the relationship is non-linear.