Discriminant — nature of roots (real - equal - complex): Exercises

The one tool used everywhere is the discriminant: $Δ = b^{2} - 4 a c for a x^{2} + b x + c = 0, a \neq = 0.$ Its sign is the gatekeeper. Positive → two real roots. Zero → one repeated real root. Negative → two complex conjugate roots. Everything below is just that idea, pushed harder. See Quadratic Formula and Completing the Square for where $Δ$ comes from.

Figure — Discriminant — nature of roots (real - equal - complex)

The picture above is your compass: it shows the three parabola positions the sign of $Δ$ predicts. Keep it in mind for every geometric exercise.

Level 1 — Recognition

Read off the nature of the roots. No solving required.

Exercise 1.1

State the nature of the roots of $x^{2} + 7 x + 10 = 0$ .

Recall Solution

Identify $a = 1, b = 7, c = 10$ . $Δ = 7^{2} - 4 (1) (10) = 49 - 40 = 9.$ What we did: plugged coefficients into $b^{2} - 4 a c$ . Why: we only need the sign, not the roots. Since $9 > 0$ , the roots are two distinct real roots. (They happen to be $- 2$ and $- 5$ , but L1 does not require finding them.)

Exercise 1.2

State the nature of the roots of $4 x^{2} - 12 x + 9 = 0$ .

Recall Solution

$a = 4, b = - 12, c = 9$ . Note $b = - 12$ keeps its sign; squaring removes it. $Δ = (- 12)^{2} - 4 (4) (9) = 144 - 144 = 0.$ $Δ = 0$ → one repeated real root (multiplicity 2). The equation is a perfect square $(2 x - 3)^{2} = 0$ .

Exercise 1.3

State the nature of the roots of $x^{2} + x + 1 = 0$ .

Recall Solution

$a = 1, b = 1, c = 1$ . $Δ = 1^{2} - 4 (1) (1) = 1 - 4 = - 3.$ $Δ = - 3 < 0$ → two complex conjugate roots. There is no real solution, but two complex ones exist — never say "no solution."

Level 2 — Application

Compute $Δ$ , then find the actual roots.

Exercise 2.1

Find the nature of the roots of $2 x^{2} - 7 x + 3 = 0$ and solve.

Recall Solution

$a = 2, b = - 7, c = 3$ . $Δ = (- 7)^{2} - 4 (2) (3) = 49 - 24 = 25 > 0.$ Two distinct real roots. Now use Quadratic Formula with $- b = + 7$ : $x = \frac{7 \pm 25}{2 ( 2 )} = \frac{7 \pm 5}{4} .$ $x_{1} = \frac{12}{4} = 3, x_{2} = \frac{2}{4} = \frac{1}{2} .$ Check: $2 (3)^{2} - 7 (3) + 3 = 18 - 21 + 3 = 0$ ✓.

Exercise 2.2

Solve $x^{2} + 4 x + 13 = 0$ over the complex numbers.

Recall Solution

$a = 1, b = 4, c = 13$ . $Δ = 4^{2} - 4 (1) (13) = 16 - 52 = - 36 < 0.$ Complex conjugate roots. Since $- 36 = i 36 = 6 i$ : $x = \frac{- 4 \pm 6 i}{2} = - 2 \pm 3 i .$ So $α = - 2$ , $β = 3$ . By the Conjugate Root Theorem, the two roots must be a conjugate pair because the coefficients are all real. See Complex Numbers.

Exercise 2.3

Solve $9 x^{2} - 6 x + 1 = 0$ .

Recall Solution

$a = 9, b = - 6, c = 1$ . $Δ = (- 6)^{2} - 4 (9) (1) = 36 - 36 = 0.$ One repeated root. With $Δ = 0$ the formula collapses to $x = \frac{- b}{2 a}$ : $x = \frac{- ( - 6 )}{2 ( 9 )} = \frac{6}{18} = \frac{1}{3} .$ The root $\frac{1}{3}$ has multiplicity 2: $9 x^{2} - 6 x + 1 = (3 x - 1)^{2}$ .

Level 3 — Analysis

Now $Δ$ becomes an equation or inequality to solve for a parameter.

Exercise 3.1

For what value(s) of $k$ does $x^{2} - k x + 4 = 0$ have equal roots?

Recall Solution

Equal roots ⟺ $Δ = 0$ . Here $a = 1, b = - k, c = 4$ . $Δ = (- k)^{2} - 4 (1) (4) = k^{2} - 16.$ Set to zero: $k^{2} - 16 = 0 \Rightarrow k^{2} = 16 \Rightarrow k = \pm 4.$ Two answers, $k = 4$ and $k = - 4$ , because $k^{2} = 16$ is satisfied by both signs. (For $k = 4$ : $(x - 2)^{2}$ ; for $k = - 4$ : $(x + 2)^{2}$ .)

Exercise 3.2

Find all $m$ for which $x^{2} + m x + m = 0$ has complex (non-real) roots.

Recall Solution

Complex ⟺ $Δ < 0$ . With $a = 1, b = m, c = m$ : $Δ = m^{2} - 4 m < 0 ⟹ m (m - 4) < 0.$ A product of two factors is negative when they have opposite signs. That happens between the roots $0$ and $4$ : $0 < m < 4.$ Answer: $m \in (0, 4)$ (open interval — the endpoints give $Δ = 0$ , i.e. equal real roots, which we exclude).

Exercise 3.3

For which $p$ does $p x^{2} + 4 x + 2 = 0$ have two distinct real roots?

Recall Solution

Two distinct real roots need $Δ > 0$ and $p \neq = 0$ (else it isn't quadratic — see Factoring Quadratics discussion of degree). $a = p, b = 4, c = 2$ : $Δ = 16 - 8 p > 0 ⟹ 16 > 8 p ⟹ p < 2.$ Combine with $p \neq = 0$ : $p \in (- \infty, 0) \cup (0, 2) .$

Level 4 — Synthesis

Combine the discriminant with other facts (Vieta, geometry, structure).

Exercise 4.1

The equation $x^{2} - 6 x + c = 0$ has two real roots whose difference is $4$ . Find $c$ .

Recall Solution

For $a x^{2} + b x + c = 0$ the roots are $\frac{- b \pm Δ}{2 a}$ , so their difference is $x_{1} - x_{2} = \frac{Δ}{a} (a > 0) .$ Here $a = 1$ , so difference $= Δ$ . We are told this equals $4$ : $Δ = 4 \Rightarrow Δ = 16.$ And $Δ = (- 6)^{2} - 4 (1) (c) = 36 - 4 c$ . So: $36 - 4 c = 16 \Rightarrow 4 c = 20 \Rightarrow c = 5.$ Cross-check with Vieta's Formulas: sum $= 6$ , product $= 5$ → roots $5, 1$ , difference $= 4$ ✓.

Exercise 4.2

A parabola $y = x^{2} + b x + 9$ is tangent to the x-axis. Find all possible $b$ and the point of tangency for each.

Recall Solution

"Tangent to x-axis" means it touches at exactly one point ⟺ $Δ = 0$ (see the middle parabola in the figure and Parabola and Axis of Symmetry). $a = 1, b = b, c = 9$ : $Δ = b^{2} - 36 = 0 \Rightarrow b = \pm 6.$ The repeated root (touch point) is $x = \frac{- b}{2 a}$ :

$b = 6$ : touch at $x = - 3$ , i.e. $(- 3, 0)$ . Curve is $(x + 3)^{2}$ .
$b = - 6$ : touch at $x = 3$ , i.e. $(3, 0)$ . Curve is $(x - 3)^{2}$ .

Exercise 4.3

Show that $(a - b) x^{2} + (b - c) x + (c - a) = 0$ always has real roots, for real $a, b, c$ with $a \neq = b$ .

Recall Solution

Real roots ⟺ $Δ \geq 0$ . Write $A = a - b, B = b - c, C = c - a$ . Notice a hidden fact: $A + B + C = (a - b) + (b - c) + (c - a) = 0 \Rightarrow C = - (A + B) .$ Now $Δ = B^{2} - 4 A C = B^{2} - 4 A (- (A + B)) = B^{2} + 4 A^{2} + 4 A B = (B + 2 A)^{2} .$ What we did: substituted $C$ using the sum-is-zero identity, then recognised a perfect square. Why it matters: a square of a real number is $\geq 0$ , so $Δ = (B + 2 A)^{2} \geq 0$ always. Hence the roots are always real. (One is $x = 1$ , since plugging $x = 1$ gives $A + B + C = 0$ .)

Level 5 — Mastery

Multiple simultaneous conditions; watch every boundary and every degenerate case.

Exercise 5.1

Find all $k$ for which both equations $x^{2} + k x + 1 = 0 and x^{2} + x + k = 0$ have real roots.

Recall Solution

Impose $Δ \geq 0$ on each, then intersect.

First: $Δ_{1} = k^{2} - 4 \geq 0 \Rightarrow k \leq - 2 or k \geq 2.$
Second: $Δ_{2} = 1 - 4 k \geq 0 \Rightarrow k \leq \frac{1}{4} .$ Intersect $(k \leq - 2 or k \geq 2)$ with $(k \leq \frac{1}{4})$ : the $k \geq 2$ branch dies (since $2 > \frac{1}{4}$ ), leaving $k \leq - 2, i.e. k \in (- \infty, - 2] .$ Endpoint $k = - 2$ included (equal roots still count as real).

Exercise 5.2

For $m x^{2} + (m + 3) x + 4 = 0$ , find all $m$ giving two distinct real roots, being careful about the leading coefficient.

Recall Solution

Need $m \neq = 0$ (else linear) and $Δ > 0$ . $Δ = (m + 3)^{2} - 4 \cdot m \cdot 4 = m^{2} + 6 m + 9 - 16 m = m^{2} - 10 m + 9.$ Factor: $m^{2} - 10 m + 9 = (m - 1) (m - 9) > 0$ . Product positive ⟺ both factors same sign ⟺ $m < 1$ or $m > 9$ . Now remove $m = 0$ : $m \in (- \infty, 0) \cup (0, 1) \cup (9, \infty) .$ Boundaries: at $m = 1$ and $m = 9$ , $Δ = 0$ (equal roots, excluded by "distinct"); at $m = 0$ , not quadratic (excluded).

Exercise 5.3

The quadratic $x^{2} - 2 (k + 1) x + (k^{2} + k - 8) = 0$ has roots that are real and of opposite sign. Find the range of $k$ .

Recall Solution

Two conditions must hold together. (1) Real roots: $Δ \geq 0$ . $Δ = [- 2 (k + 1)]^{2} - 4 (k^{2} + k - 8) = 4 (k + 1)^{2} - 4 (k^{2} + k - 8) .$ $= 4 (k^{2} + 2 k + 1) - 4 k^{2} - 4 k + 32 = 4 k^{2} + 8 k + 4 - 4 k^{2} - 4 k + 32 = 4 k + 36.$ So $Δ = 4 k + 36 \geq 0 \Rightarrow k \geq - 9.$ (2) Opposite-sign roots: by Vieta's Formulas, the product of roots is $\frac{c}{a} = k^{2} + k - 8$ . Roots have opposite signs ⟺ product $< 0$ : $k^{2} + k - 8 < 0 \Rightarrow k \in (\frac{- 1 - 33}{2}, \frac{- 1 + 33}{2}) .$ Numerically $33 \approx 5.745$ , so this is roughly $(- 3.372, 2.372)$ . Note: a negative product automatically forces $Δ > 0$ (real and distinct), so condition (2) already implies (1) here. Intersect with $k \geq - 9$ — no new restriction. $k \in (\frac{- 1 - 33}{2}, \frac{- 1 + 33}{2})$