2.1.18 · D5Algebra — Introduction & Intermediate
Question bank — Discriminant — nature of roots (real - equal - complex)
This is a rapid-fire self-test on the ideas behind the discriminant, not the arithmetic. Every prompt below hides a common trap. Cover the answer, commit to a reason out loud, then reveal. If your reason was "just because," you haven't learned it yet.
True or false — justify
A quadratic with has no solutions.
False — it has no real solutions, but two complex conjugate solutions. "Nature of roots" and "existence of roots" are different questions.
If the equation has no roots because "zero means nothing."
False — gives exactly one real root repeated twice, . The parabola touches the axis at its vertex, so there IS an intersection.
only works when .
False — it works for every . The term specifically exists to handle a general leading coefficient; it drops out of the derivation from Completing the Square, not from assuming .
For a quadratic with real coefficients, if one root is then the other must be .
True — real coefficients force roots to come as conjugate pairs (the Conjugate Root Theorem). The in the formula splits symmetrically about .
If the two roots are always positive numbers.
False — only guarantees the roots are real and distinct, not their signs. Their signs depend on and (see Vieta's Formulas: sum , product ).
A larger discriminant means the roots are "more real."
False — "real" is not a matter of degree. A larger positive just means the two real roots are spread farther apart, since their gap is (the absolute value matters because may be negative).
is negative whenever is negative.
False — the sign of depends on both and . If then , pushing positive (real roots); but if then , which can drive negative depending on . So alone decides nothing.
The discriminant of equals the discriminant of .
True — replacing with gives . Multiplying an equation by doesn't change its roots, so mustn't change.
If two different quadratics have the same , they have the same roots.
False — fixes the nature and spread of roots, not their location. (the axis of symmetry) sets where they sit; equal with different gives different roots.
Spot the error
Student writes: "For , , and roots are ."
The line is fine since , but , so the numerator is , giving . Using in the numerator is the classic sign slip.
Student writes: " for , so ."
The formula is , not . Dropping the gives a wrong value; correct is . The comes from Completing the Square and is not optional.
Student writes: " has real roots for all ."
They forgot makes it linear, not quadratic, so it must be excluded: . Whenever the leading coefficient is a parameter, check first.
Student writes: ", so the equation is impossible and we stop."
Student writes: " so there is one root, hence the parabola never touches the x-axis."
Backwards — one repeated root means the parabola touches the axis at exactly that one point (its vertex). "Never touches" is the case.
Student writes: " has ... wait, there's no term so doesn't apply."
still applies: here , so , giving complex roots . A missing -term just means , not that is undefined.
Why questions
Why does the sign of , and not its exact value, decide the nature of the roots?
Because the nature depends only on whether is a positive real, zero, or imaginary — and that switch is controlled purely by whether is , , or .
Why does collapse the two roots into one?
The two roots are ; when the adds nothing, so both branches land on the same value .
Why must complex roots of a real-coefficient quadratic be conjugates and not, say, and ?
The formula forces both roots to share the same real part and equal-but-opposite imaginary parts . Only conjugate pairs fit this symmetric structure — see Conjugate Root Theorem.
Why does the term appear in at all?
It emerges when you complete the square: the right side becomes , and the numerator of that is the quantity under the root. It is not a memorized decoration.
Why can we tell the roots are real without computing them?
Because measures the argument of the square root before you ever divide; a non-negative guarantees a real , so the whole expression stays real. The discriminant is a "gatekeeper" you check first.
Why is (with , not ) the condition for "real roots"?
"Real roots" includes the equal-roots case , which is still real. Only if the problem said "two distinct real roots" would you require strict .
Why does the graph interpretation (crossings) match the algebra (sign of )?
Real roots are exactly the -values where hits , i.e. axis crossings. Two real roots ⇒ two crossings, one repeated ⇒ one tangent touch, complex ⇒ no crossings — see Graphing Quadratic Functions.
Edge cases
What is the discriminant when , e.g. ?
; its sign is just the sign of . If have opposite signs the roots are real (); same signs give complex roots.
What happens to "" if ?
The object stops being a quadratic — it's a linear equation with one root . The discriminant formula assumes and is meaningless otherwise.
If , what does tell you, and is anything special about the roots?
, always non-negative, so roots are always real; one of them is (factor ). Real roots are guaranteed whenever the constant term vanishes.
Can be zero while the roots are complex?
No — always gives a single real repeated root . Complex roots require strictly .
For a perfect-square trinomial like , what must be, and why?
, because a perfect square has a repeated root. Equivalently, factoring into a square is the algebraic signature of .
Recall Quick self-check
The three-way switch — what is it? ::: The sign of : positive ⇒ two distinct real, zero ⇒ one repeated real, negative ⇒ two complex conjugates. The one boundary condition students always forget when is a parameter? ::: Require , or the equation isn't quadratic at all.