Nature of roots and the discriminant

We've met the discriminant $D = b^2 - 4 a c$ several times. This topic devotes itself fully to interpreting it.

Statement

For a quadratic equation $a x^2 + b x + c = 0$ (with real coefficients and $a \ne 0$), define $D = b^2 - 4 a c$. Then the nature of the roots is:

• $D > 0$: two distinct real roots,
• $D = 0$: one repeated real root (also called a double root or equal roots),
• $D < 0$: no real roots (the roots are a pair of complex conjugates).

Additionally:

• $D$ is a perfect square (positive) $\Rightarrow$ roots are rational (when $a, b, c$ are rational).
• $D > 0$ but not a perfect square $\Rightarrow$ roots are irrational and conjugate ($p \pm q \sqrt{D}$).

Why this works

From the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2 a},$ the square root is the source of all branching. If $D < 0$, $\sqrt{D}$ is not real. If $D = 0$, both roots collapse to $-b/(2 a)$. If $D > 0$, we get two distinct real values.

The geometric picture (from Chapter 2): the parabola $y = a x^2 + b x + c$ meets the $x$-axis at the roots of the equation. So the discriminant is the number that decides "two intersections / one tangent / no intersection".

Applications

A common exam pattern: find the value(s) of $k$ for which the equation $a x^2 + b x + c = 0$ has equal roots. You set $D = 0$ and solve.

Another: find $k$ so that the equation has real and distinct roots. You set $D > 0$ and solve an inequality.

A third: show that for all real $k$ the equation has real roots. You show $D \ge 0$ for all real $k$ (typically by completing a square in $k$).

Worked examples

Example 1. Determine the nature of roots of $2 x^2 - 3 x + 5 = 0$.

$D = 9 - 40 = -31 < 0$. No real roots.

Example 2. Find $k$ so that $k x^2 - 2 k x + 6 = 0$ has equal roots.

$D = (2 k)^2 - 4 \cdot k \cdot 6 = 4 k^2 - 24 k$. Set $= 0$: $4 k (k - 6) = 0 \Rightarrow k = 0$ or $k = 6$. But $k = 0$ makes the equation non-quadratic. So $k = 6$.

Example 3. For what value of $p$ does $(p + 1) x^2 - 6 (p + 1) x + 3 (p + 9) = 0, p \ne -1$, have equal roots?

$D = 36 (p + 1)^2 - 12 (p + 1)(p + 9) = 12 (p + 1) [3 (p + 1) - (p + 9)] = 12 (p + 1) (2 p - 6)$.

Set $= 0$: $(p + 1)(p - 3) = 0$. Excluding $p = -1$, we get $p = 3$.

Example 4. Show that $x^2 + 2 x + 3 > 0$ for all real $x$, and hence that $x^2 + 2 x + 3 = 0$ has no real roots.

Complete the square: $x^2 + 2 x + 3 = (x + 1)^2 + 2 \ge 2 > 0$. So the LHS is strictly positive , the equation has no real roots.

Discriminant check: $D = 4 - 12 = -8 < 0$. ✓

Example 5. Find $k$ so that the equation $(k - 12) x^2 + 2 (k - 12) x + 2 = 0$ has equal roots (assume $k \ne 12$).

$D = 4 (k - 12)^2 - 8 (k - 12) = 4 (k - 12)[(k - 12) - 2] = 4 (k - 12)(k - 14)$.

Set $= 0$: $k = 12$ (excluded) or $k = 14$. So $k = 14$.

Try it yourself

1. Find the discriminant and describe the nature of roots: $2 x^2 - 4 x + 3 = 0$.
2. Find the discriminant and describe the nature of roots: $9 x^2 - 6 x + 1 = 0$.
3. For what value of $k$ does $k x (x - 2) + 6 = 0$ have equal roots?
4. Show that $x^2 + (a^2 + b^2) x + (a b)^2 = 0$ always has real roots.
5. Find values of $p$ for which $(p + 2) x^2 + (p + 5) x + 2 = 0$ has real and distinct roots.
6. Determine if $\sqrt{3} x^2 - 2 x - \sqrt{3} = 0$ has real roots; find them if so.
7. Find $m$ for which $m x^2 + (2 m - 1) x + (m - 2) = 0$ has equal roots.
8. Show that the equation $(b - c) x^2 + (c - a) x + (a - b) = 0$ always has real roots.
9. Find the values of $k$ for which the equation $x^2 - k x + 9 = 0$ has equal roots.
10. If the equation $(1 + m^2) x^2 + 2 m c x + (c^2 - a^2) = 0$ has equal roots, prove that $c^2 = a^2 (1 + m^2)$.

Pitfalls / Insight

• Don't exclude solutions automatically. Sometimes both $k$ values are valid; sometimes one is excluded (like the leading coefficient becoming zero).
• For "real roots", allow $D \ge 0$, not just $D > 0$.
• For inequalities in $k$, sketch a sign chart of the discriminant , it's a quadratic in $k$ in many problems.

Insight. The discriminant is a one-number diagnosis of a whole quadratic. It tells you what the formula will spit out , and often it lets you answer "nature of roots" questions without ever solving the equation.