A quadratic equation is an equation where the highest power of the variable is 2.
Quadratics can be solved by factoring, completing the square, or using the quadratic formula.
Their graphs are parabolas.
1. Standard form
A quadratic equation in one variable has the standard form:
\[
\begin{aligned}
ax^2+bx+c=0,\qquad a\neq 0.
\end{aligned}
\]
The coefficient \(a\) controls the opening and width of the parabola.
If \(a>0\), the parabola opens upward. If \(a<0\), it opens downward.
2. The quadratic formula
Every quadratic equation can be solved with the quadratic formula:
\[
\begin{aligned}
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.
\end{aligned}
\]
The expression under the square root is called the discriminant:
\[
\begin{aligned}
D=b^2-4ac.
\end{aligned}
\]
3. Discriminant analysis
The discriminant tells us how many roots the quadratic has and what kind they are.
4. Worked example
Solve:
\[
\begin{aligned}
2x^2-7x+3=0.
\end{aligned}
\]
Here:
\[
\begin{aligned}
a=2,\qquad b=-7,\qquad c=3.
\end{aligned}
\]
Compute the discriminant:
\[
\begin{aligned}
D
&=
b^2-4ac\\
&=
(-7)^2-4(2)(3)\\
&=
49-24\\
&=
25.
\end{aligned}
\]
Since \(D=25>0\), there are two distinct real roots.
Use the quadratic formula:
\[
\begin{aligned}
x
&=
\frac{-b\pm\sqrt{D}}{2a}\\
&=
\frac{-(-7)\pm\sqrt{25}}{2(2)}\\
&=
\frac{7\pm5}{4}.
\end{aligned}
\]
Therefore:
\[
\begin{aligned}
x_1&=\frac{7+5}{4}=3,\\
x_2&=\frac{7-5}{4}=\frac{1}{2}.
\end{aligned}
\]
The solution is:
\[
\begin{aligned}
\boxed{x=3\quad\text{or}\quad x=\frac{1}{2}}.
\end{aligned}
\]
5. Solving by factoring
Some quadratics can be solved quickly by factoring.
For the same example:
\[
\begin{aligned}
2x^2-7x+3
&=
(2x-1)(x-3).
\end{aligned}
\]
Set each factor equal to zero:
\[
\begin{aligned}
2x-1&=0
&&\Rightarrow&
x&=\frac{1}{2},\\
x-3&=0
&&\Rightarrow&
x&=3.
\end{aligned}
\]
Factoring is fast when the quadratic has rational roots, but not every quadratic factors nicely.
6. Completing the square
Completing the square rewrites the quadratic into a perfect-square form.
Start with:
\[
\begin{aligned}
ax^2+bx+c=0.
\end{aligned}
\]
Divide by \(a\):
\[
\begin{aligned}
x^2+\frac{b}{a}x+\frac{c}{a}=0.
\end{aligned}
\]
Move the constant:
\[
\begin{aligned}
x^2+\frac{b}{a}x=-\frac{c}{a}.
\end{aligned}
\]
Add the square of half the coefficient of \(x\):
\[
\begin{aligned}
x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2
=
-\frac{c}{a}+\left(\frac{b}{2a}\right)^2.
\end{aligned}
\]
The left side becomes a perfect square:
\[
\begin{aligned}
\left(x+\frac{b}{2a}\right)^2
=
\frac{b^2-4ac}{4a^2}.
\end{aligned}
\]
Taking square roots gives the quadratic formula.
7. Vertex and vertex form
The graph of:
\[
\begin{aligned}
y=ax^2+bx+c
\end{aligned}
\]
is a parabola. Its vertex has x-coordinate:
\[
\begin{aligned}
h=-\frac{b}{2a}.
\end{aligned}
\]
Then:
\[
\begin{aligned}
k=f(h).
\end{aligned}
\]
The vertex is \((h,k)\), and the vertex form is:
\[
\begin{aligned}
y=a(x-h)^2+k.
\end{aligned}
\]
Vertex form makes it easy to see where the parabola turns around and whether the vertex is a minimum or maximum.
8. Complex roots
When \(D<0\), the square root contains a negative number.
Then the roots are complex conjugates:
\[
\begin{aligned}
x=\frac{-b\pm i\sqrt{|D|}}{2a}.
\end{aligned}
\]
In this case, the parabola has no real x-intercepts.
9. Formula summary
The table below uses plain text formulas in table cells to avoid raw LaTeX rendering problems.
10. Common mistakes
- Forgetting that \(a\neq0\). If \(a=0\), the equation is linear, not quadratic.
- Using \(b^2+4ac\) instead of \(b^2-4ac\).
- Forgetting the \(\pm\) sign in the quadratic formula.
- Dividing by \(a\) instead of \(2a\) in the denominator.
- Thinking complex roots mean “no solution.” They mean no real solution, but complex solutions still exist.
- Confusing roots with the vertex. Roots are x-intercepts; the vertex is the turning point.
Key idea: the discriminant tells the type of roots, the quadratic formula gives the roots, and the vertex explains the graph.
