Quadratic Equations
Cheat sheetClass 10 · Mathematics
The standard form
$ax^2 + bx + c = 0$, where $a$, $b$, $c$ are real numbers and $a \neq 0$. A value of $x$ that satisfies the equation is a root.
Two ways to solve
- Factorisation — split the middle term so that the two parts multiply to $ac$ and add to $b$, then set each factor to zero.
- Quadratic formula — works every time:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Nature of the roots
The discriminant $D = b^2 - 4ac$ tells you what kind of roots to expect before you solve.
| Discriminant | Roots |
|---|---|
| $D > 0$ | Two distinct real roots |
| $D = 0$ | Two equal real roots, $x = -\frac{b}{2a}$ |
| $D < 0$ | No real roots |
Worked example
Solve $2x^2 - 5x + 3 = 0$.
$D = (-5)^2 - 4(2)(3) = 25 - 24 = 1$, so there are two distinct real roots.
$x = \frac{5 \pm 1}{4}$, which gives $x = \frac{3}{2}$ or $x = 1$.
Word problems
Turn the sentence into an equation first. The product of two consecutive positive integers is 306:
$x(x + 1) = 306 \Rightarrow x^2 + x - 306 = 0 \Rightarrow (x + 18)(x - 17) = 0$
The integers are 17 and 18 — reject $x = -18$, because the question asks for positive integers.
Mistakes that cost marks
- Forgetting that $a \neq 0$ — without the $x^2$ term it is not a quadratic.
- Dropping the $\pm$ sign and finding only one root.
- Dividing both sides by $x$, which loses the root $x = 0$.
- Not checking whether a root makes sense in a word problem, such as a negative age or length.