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AdvancedGrades 9–1125 min41 / 82

Parameter problems: investigating the quadratic

What a parameter is and the special case a = 0, the number of roots, a quadratic that keeps its sign for all x, signs of the roots with Vieta’s theorem, reciprocal roots, the minimum of x₁² + x₂², both roots greater than a number, the position of the parabola’s vertex and linear systems with a parameter — answers written as intervals, with DİM-style closed and coded tasks.

Check yourself
In this lesson you will learn
  • Separate the case a = 0 in a quadratic with a parameter and investigate the number of roots with the discriminant
  • Apply the condition (a and D) for a quadratic to keep its sign for all x
  • Turn conditions on the roots (signs, reciprocal roots, x₁² + x₂², roots greater than k) into inequalities with Vieta’s theorem and D
  • Solve tasks on the position of the vertex and on the number of solutions of a linear system with a parameter

“For which values of a…” — DİM entrance exams contain such tasks every year: in 2026, 6 of the 60 tasks of groups I and II had a parameter (a quadratic negative for all x, reciprocal roots, a vertex in quadrant IV, a system with a given solution). A parameter is a fixed number whose value is not known; the task asks us to investigate all its possible values. This lesson builds on “Quadratic equations”, “The quadratic function and its graph” and “Quadratic inequalities and the interval method”.

The case a = 0 and the number of roots

  1. 1
    1. Leading coefficient

    If the coefficient of x² depends on the parameter, check separately the case where it is 0: the equation becomes linear.

  2. 2
    2. Discriminant

    For a ≠ 0, write D in terms of the parameter.

  3. 3
    3. Translate the condition

    Turn the condition in words into inequalities for D, x₁ + x₂, x₁x₂, the vertex or f(k).

  4. 4
    4. System

    Solve the system in the parameter and add the answer of the case a = 0.

  5. 5
    5. Answer

    Write the answer as an interval or a set and check the boundary values separately.

a ≠ 0: D = b² − 4ac; D > 0 — two distinct roots; D = 0 — one root; D < 0 — no real roots
where:
  • a, b, cthe coefficients of ax² + bx + c = 0 (may depend on the parameter)

For a = 0 the equation is bx + c = 0: one root if b ≠ 0; if b = 0, either no roots or every x is a root.

The number of roots

1) For which m does (m − 2)x² + 4x + 1 = 0 have exactly one root?
2) For which k does x² − 6x + k = 0 have two distinct roots?
3) For which k does kx² + 2x + 1 = 0 have no roots?

Show solution
1) m = 2: 4x + 1 = 0 ⇒ x = −1/4 — one root, fits.
m ≠ 2: D = 16 − 4(m − 2) = 24 − 4m = 0 ⇒ m = 6 (the equation is 4x² + 4x + 1 = (2x + 1)² = 0).
Answer: m ∈ {2, 6}. Forgetting the case a = 0 loses m = 2.
2) D = 36 − 4k > 0 ⇒ k < 9. Answer: (−∞, 9).
3) k = 0: 2x + 1 = 0 — has a root, does not fit.
k ≠ 0: D = 4 − 4k < 0 ⇒ k > 1. Answer: (1, +∞).

A quadratic that keeps its sign for all x

ax² + bx + c > 0 (all x) ⇔ a > 0 and D < 0; ax² + bx + c < 0 (all x) ⇔ a < 0 and D < 0
where:
  • adirection of the branches: a > 0 — up, a < 0 — down
  • D < 0the parabola does not meet the x-axis

For non-strict inequalities (≥ 0, ≤ 0) take D ≤ 0. If a depends on the parameter, check a = 0 separately: bx + c keeps its sign for all x only when b = 0.

Interactive
Loading simulation…
The parabola y = −x² + 4x + m. Move the slider m: for m < −4 the whole parabola is below the x-axis; at m = −4 it touches the axis.
Quadratics that keep their sign

1) For which m is −x² + 4x + m < 0 true for all x?
2) For which a is (a − 1)x² − 2(a − 1)x + 3 > 0 true for all x?
3) For which k is the domain of y = √(x² + 2x + k) the whole number line?

Show solution
1) a = −1 < 0 ✓. D = 16 + 4m < 0 ⇒ m < −4. Answer: (−∞, −4).
2) a = 1: 3 > 0 — true for all x, so a = 1 fits.
a ≠ 1: a − 1 > 0 and D/4 = (a − 1)² − 3(a − 1) = (a − 1)(a − 4) < 0 ⇒ 1 < a < 4.
Answer: [1, 4) — 1 is included, 4 is not (for a = 4, D = 0 and the expression is 0 at x = 1).
3) x² + 2x + k ≥ 0 for all x: a = 1 > 0, D = 4 − 4k ≤ 0 ⇒ k ≥ 1. Answer: [1, +∞).
DİM style: a closed task

For which k is kx² − 4x + k − 3 < 0 true for all x?
A) (−∞, −1) B) (−1, 4) C) (4, +∞) D) (−∞, 0) E) (−∞, −1) ∪ (4, +∞)

Show solution
k = 0: −4x − 3 < 0 holds only for x > −3/4 — does not fit.
k ≠ 0: { k < 0; D/4 = 4 − k(k − 3) < 0 } ⇒ { k < 0; k² − 3k − 4 > 0 } ⇒ { k < 0; k < −1 or k > 4 } ⇒ k < −1.
Answer: A.
Option E solves D < 0 alone and ignores k < 0: for k > 4 the branches point up and the expression is positive.

Conditions on the roots: Vieta’s theorem

x₁ + x₂ = −b/a; x₁ · x₂ = c/a; x₁² + x₂² = (x₁ + x₂)² − 2x₁x₂x₁ + x₂ = −b/a; x₁ · x₂ = c/a; x₁² + x₂² = (x₁ + x₂)² − 2x₁x₂
where:
  • x₁, x₂the roots of ax² + bx + c = 0

Vieta’s formulas work only when the roots exist: always add the condition D ≥ 0 (D > 0 for distinct roots).

ConditionSystem for the parameter
Roots of opposite signsc/a < 0 (then D > 0 automatically)
Both roots positiveD ≥ 0, x₁ + x₂ > 0, x₁x₂ > 0
Both roots negativeD ≥ 0, x₁ + x₂ < 0, x₁x₂ > 0
Reciprocal roots (x₁x₂ = 1)c/a = 1 and D ≥ 0
Opposite roots (x₁ = −x₂)b = 0 and c/a < 0
Both roots greater than kD ≥ 0, a · f(k) > 0, x₀ = −b/(2a) > k
f(x) = ax² + bx + c. Last row: the point k lies to the left of both roots — there f has the sign of a, and the vertex is to the right of k.

Each of the three conditions in the last row is needed. D ≥ 0 makes sure the roots exist. a · f(k) > 0 says that k is not between the roots: between the roots f has the sign opposite to a. But k can be to the left of both roots or to the right of both — x₀ > k tells these apart. For example, in x² − 2mx + m² − 1 = 0 with m = 0 we have D > 0 and f(2) = 3 > 0, yet the roots are −1 and 1, both less than 2: the condition x₀ > 2 fails (x₀ = 0).

Conditions with Vieta’s theorem

1) For which k are the roots of x² − 2x + k − 5 = 0 of opposite signs?
2) For which a are the roots of x² + (a − 1)x + a² − 3 = 0 reciprocal?
3) For which a is the sum of the squares of the roots of x² + (a − 2)x − a = 0 the least? What is this least value?
4) For which m are both roots of x² − 2mx + m² − 1 = 0 greater than 2?

Show solution
1) x₁x₂ = k − 5 < 0 ⇒ k < 5. Answer: (−∞, 5).
2) x₁x₂ = a² − 3 = 1 ⇒ a = ±2.
a = 2: x² + x + 1 = 0, D = 1 − 4 = −3 < 0 — no roots, a = 2 is rejected.
a = −2: x² − 3x + 1 = 0, D = 9 − 4 = 5 > 0 ✓.
Answer: −2.
3) x₁ + x₂ = 2 − a, x₁x₂ = −a; D = (a − 2)² + 4a = a² + 4 > 0 — the roots always exist.
x₁² + x₂² = (2 − a)² + 2a = a² − 2a + 4 = (a − 1)² + 3. The least value is 3, at a = 1.
4) D/4 = m² − (m² − 1) = 1 > 0 ✓; f(2) = 4 − 4m + m² − 1 = m² − 4m + 3 > 0 ⇒ m < 1 or m > 3; x₀ = m > 2.
System: m > 3. Check: the roots are m − 1 and m + 1, and m − 1 > 2 ⇔ m > 3. Answer: (3, +∞).
DİM style: a coded task

For which m are the roots of (m − 1)x² + 10x + 3m − 7 = 0 reciprocal?

Show solution
m ≠ 1 (otherwise the equation is linear). x₁x₂ = (3m − 7)/(m − 1) = 1 ⇒ 3m − 7 = m − 1 ⇒ m = 3.
Check: m = 3 ⇒ 2x² + 10x + 2 = 0, D = 100 − 16 = 84 > 0 — the roots exist and their product is 2/2 = 1.
Answer: 3.
Two methods: Vieta and the graph

For which a does x² − 6x + 5 = a have two distinct positive roots?

Show solution
With Vieta: x² − 6x + 5 − a = 0. D = 36 − 4(5 − a) = 16 + 4a > 0 ⇒ a > −4; x₁x₂ = 5 − a > 0 ⇒ a < 5; x₁ + x₂ = 6 > 0 ✓. Answer: (−4, 5).
With the graph: the parabola y = x² − 6x + 5 has vertex (3, −4) and meets the y-axis at (0, 5). The line y = a cuts the part with x > 0 at two points only for −4 < a < 5. For a = 5 the roots are 0 and 6 — 0 is not positive; for a = −4 there is one root. Both methods give the same answer.
Written task: a full investigation

For which a does ax² − (a + 1)x + 1 = 0 have two distinct positive roots? Write all stages of the solution.

Show solution
Solution. 1) a = 0: −x + 1 = 0 ⇒ x = 1 — one root, so a = 0 does not fit.
2) a ≠ 0: D = (a + 1)² − 4a = (a − 1)². Two distinct roots need D > 0 ⇒ a ≠ 1.
3) Roots: x = ((a + 1) ± |a − 1|)/(2a), i.e. x₁ = 1 and x₂ = 1/a (check: ax² − (a + 1)x + 1 = (ax − 1)(x − 1)).
4) x₁ = 1 > 0; x₂ = 1/a > 0 ⇒ a > 0.
5) Combine the conditions: a > 0 and a ≠ 1.
Answer: (0, 1) ∪ (1, +∞).
Criteria: the case a = 0 investigated — ⅓; the roots (or the D and Vieta conditions) found correctly — ⅔; a ≠ 1 taken into account and the answer written correctly — 1.

The vertex of the parabola and a parameter

x₀ = −b/(2a); y₀ = f(x₀) = −D/(4a)x₀ = −b/(2a); y₀ = f(x₀) = −D/(4a)
where:
  • (x₀; y₀)the vertex of y = ax² + bx + c

The vertex lies in quadrant I if x₀ > 0, y₀ > 0; II if x₀ < 0, y₀ > 0; III if x₀ < 0, y₀ < 0; IV if x₀ > 0, y₀ < 0. On the x-axis ⇔ D = 0; on the y-axis ⇔ b = 0.

Interactive
Loading simulation…
The vertex of y = x² − 2mx + 2m + 3 is (m, −m² + 2m + 3). Check with the slider m: the vertex is in quadrant IV only for m > 3.
Where the vertex lies

1) For which m is the vertex of y = x² − 2mx + 2m + 3 in quadrant IV?
2) For which c does the parabola y = x² + 4x + c touch the x-axis?
3) For which m does the vertex of y = x² + 2(m − 3)x + 5 lie on the y-axis?

Show solution
1) x₀ = m, y₀ = m² − 2m² + 2m + 3 = −m² + 2m + 3.
{ m > 0; −m² + 2m + 3 < 0 } ⇒ { m > 0; m² − 2m − 3 > 0 } ⇒ { m > 0; m < −1 or m > 3 } ⇒ m > 3.
Answer: (3, +∞).
2) Touching means the vertex is on the x-axis: D = 16 − 4c = 0 ⇒ c = 4 (y = (x + 2)²).
3) x₀ = −(m − 3) = 0 ⇒ m = 3.

Linear systems with a parameter

a₁/a₂ ≠ b₁/b₂ — one solution; a₁/a₂ = b₁/b₂ ≠ c₁/c₂ — no solution; a₁/a₂ = b₁/b₂ = c₁/c₂ — infinitely many solutionsa₁/a₂ ≠ b₁/b₂ — one solution; a₁/a₂ = b₁/b₂ ≠ c₁/c₂ — no solution; a₁/a₂ = b₁/b₂ = c₁/c₂ — infinitely many solutions
where:
  • a₁x + b₁y = c₁, a₂x + b₂y = c₂the equations of the system

Geometrically the lines intersect, are parallel or coincide. If a denominator is 0, check a₁b₂ − a₂b₁ = 0 instead of the ratios. More in “Systems of linear equations”.

Systems with a parameter

1) For which m does { mx + 4y = 8; x + my = 4 } have one solution, no solution, infinitely many solutions?
2) For which a and b is the pair (2, −3) a solution of { ax + 2y = 4; 3x − by = 15 }?
3) For which k does { 2x + ky = 5; x + 3y = 2 } have no solution?

Show solution
1) m/1 = 4/m ⇔ m² = 4 ⇔ m = ±2.
m ≠ ±2 — one solution.
m = 2: 2/1 = 4/2 = 8/4 = 2 — infinitely many solutions (the equations are the same).
m = −2: −2/1 = 4/(−2) = −2, but 8/4 = 2 ≠ −2 — no solution.
2) Substitute x = 2, y = −3: 2a − 6 = 4 ⇒ a = 5; 6 + 3b = 15 ⇒ b = 3.
3) 2/1 = k/3 ⇒ k = 6; 5/2 ≠ 2 ✓ — for k = 6 the lines are parallel and there is no solution.
Coded tasks: write the answer as a number
  1. 1.x² − 6x + k = 0 has exactly one root when k =
  2. 2.The greatest integer k for which x² − 6x + k = 0 has two distinct roots:
  3. 3.One root of x² + px + 12 = 0 is 3 ⇒ p =
  4. 4.In x² − 4x + m = 0, x₁² + x₂² = 10 ⇒ m =
  5. 5.The y-coordinate of the vertex of y = x² − 2mx + m² + 3:
  6. 6.{ 3x + ay = 6; x + 2y = 2 } has infinitely many solutions ⇒ a =

Key points

  • If the leading coefficient depends on the parameter, first investigate the case a = 0 separately.
  • ax² + bx + c > 0 for all x ⇔ a > 0, D < 0; < 0 ⇔ a < 0, D < 0.
  • Conditions on the roots are written with Vieta’s formulas, always together with D ≥ 0 (or D > 0).
  • The vertex is (−b/(2a), −D/(4a)): the quadrant is set by the signs of its coordinates.
  • A linear system: the ratios a₁/a₂, b₁/b₂, c₁/c₂ decide between one, no and infinitely many solutions.

Check yourself

12 questions. Every correct answer earns XP.

1 / 12
When is ax² + bx + c > 0 true for all x (a ≠ 0)?