Discriminant

Video lesson · from 0:093 worked examples

One lesson video covers all of Polynomials & Quadratics, so it opens at 0:09 for this topic — not from the beginning.

Theory

Given a quadratic equation of the form, $ax^2+bx+c=0$ , the discriminant is defined by $b^2-4ac$ .

It is worth seeing where it comes from. Any quadratic equation can be solved with the quadratic formula:

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

The discriminant is exactly the expression under the square root. That is why it controls the roots: you cannot take the square root of a negative number, and a square root of zero adds nothing, so the sign of $b^2-4ac$ decides how many roots there are before you calculate anything.

b² − 4ac > 0

2 real and distinct roots

b² − 4ac = 0

1 real and equal root
i.e. repeated root

b² − 4ac < 0

No real roots

This gives three cases, which you should be able to state and recognise:

\begin{aligned} b^2 - 4ac &\gt 0 &&\Rightarrow \text{two real and distinct roots} \\ b^2 - 4ac &= 0 &&\Rightarrow \text{two real and equal roots (a repeated root)} \\ b^2 - 4ac &\lt 0 &&\Rightarrow \text{no real roots} \end{aligned}

When $b^2-4ac = 0$ the parabola touches the $x$ -axis at exactly one point — its turning point sits on the axis. This is the case examiners use most, because “equal roots” and “tangent to” mean the same thing.

If $b^2-4ac \gt 0$ and is a perfect square, the roots are rational and the quadratic factorises; if it is positive but not a perfect square, the roots are irrational.

The discriminant has many uses including finding unknown terms in a quadratic equation.

The Golden Rule: before substituting, write the equation in the form $ax^2+bx+c=0$ with everything on one side — reading $a$ , $b$ and $c$ off an equation that has not been rearranged is the most common way to lose these marks.

⚠️ Common Examiner Traps

State the condition you are using. The most frequently repeated fault on discriminant questions is using an incorrect condition, or never stating the inequality being solved at all. Write down $b^2-4ac \lt 0$ (or whichever applies) before you solve it.
Brackets when substituting: marks are routinely lost for not using brackets when substituting into $b^2-4ac$ . With $b$ negative or a coefficient in terms of $k$ , brackets are essential.
Two regions need two inequalities: a single inequality is often wrongly used to describe two separate regions. If the answer is “ $k \lt 0$ or $k \gt 4$ ”, write both — it cannot be compressed into one chain.
Match the condition to the wording: “two distinct real roots” is $\gt 0$ ; “equal roots” and “is a tangent to” are both $= 0$ ; “no real roots” and “does not intersect” are both $\lt 0$ .

Worked examples

Example 1

Find the nature of the roots of $9x^2+21x+16=0$

We use the discriminant $b^2-4ac$ with $a=9, b=21, c=16$ .

b^2-4ac = 21^2 - 4(9)(16)

\phantom{b^2-4ac} = 441 - 576

\phantom{b^2-4ac} = -135

Since $b^2-4ac < 0$ , there are no real roots.

Example 2

Find the value of $q$ such that $6x^2+12x+q=0$ has real roots.

For real roots (either repeated or distinct), we need $b^2-4ac \ge 0$ .

$a = 6$
$b = 12$
$c = q$

12^2 - 4(6)(q) \ge 0

144 - 24q \ge 0

144 \ge 24q

q \le 6

Example 3

Show that $(2k+4)x^2+(3k+2)x+(k-2)=0$ always has real roots.

Here $a=(2k+4), b=(3k+2), c=(k-2)$ .

b^2-4ac = (3k+2)^2 - 4(2k+4)(k-2)

\phantom{b^2-4ac} = (9k^2+12k+4) - 4(2k^2-4k+4k-8)

\phantom{b^2-4ac} = 9k^2+12k+4 - 8k^2 + 32

\phantom{b^2-4ac} = k^2+12k+36

\phantom{b^2-4ac} = (k+6)^2

Since any real number squared is greater than or equal to zero, $(k+6)^2 \ge 0$ .

Therefore, the roots are always real.