Quadratic equations and Vieta's formulas

The quadratic equation is a “pillar” of the exam: it appears both as a question in its own right and, after a substitution, inside higher-degree, rational, irrational, exponential and trigonometric equations. That is why the discriminant and Vieta’s theorem must become automatic. In this lesson we see where the root formula comes from, how to tell the number of roots, how to learn about the roots without solving, and how to handle questions with a parameter.

§1. Definition and incomplete equations

A quadratic equation is an equation of the form ax² + bx + c = 0 with a ≠ 0. a is the leading coefficient, b the second coefficient and c the constant term. If a = 1, the equation is monic (reduced): x² + px + q = 0. If a = 0, the equation becomes linear — this matters in parameter problems.
If b or c is zero, the equation is incomplete and is solved without the discriminant:

• c = 0: ax² + bx = 0 → x(ax + b) = 0 → x = 0 or x = −b/a.
• b = 0: ax² + c = 0 → x² = −c/a; if −c/a > 0, then x = ±√(−c/a), otherwise there are no real roots.
• b = c = 0: ax² = 0 → x = 0.

For example, 3x² − 12 = 0 → x² = 4 → x = ±2; 5x² + 2x = 0 → x = 0 or x = −2/5; x² + 9 = 0 has no real roots.
A frequent mistake: dividing x² = 4x by x and getting x = 4. The root x = 0 is lost. The correct way is x(x − 4) = 0.

§2. The discriminant

The formula comes from completing the square. Multiply ax² + bx + c = 0 by 4a: 4a²x² + 4abx + 4ac = 0, then add and subtract b²: (2ax + b)² = b² − 4ac. The right-hand side is called the discriminant:

• D > 0 — two distinct real roots;
• D = 0 — one (repeated) root x = −b/(2a);
• D < 0 — no real roots, because a square cannot be negative.

If the second coefficient is even, the working is simpler:

For example, x² − 6x + 5 = 0: k = −3, D/4 = 9 − 5 = 4, x = 3 ± 2 → 1; 5.
Another useful sign: if a + b + c = 0, the roots are 1 and c/a; if a − b + c = 0, the roots are −1 and −c/a.

Worked example

Solve 2x² − 3x − 2 = 0.

1. a = 2, b = −3, c = −2.
2. D = 9 − 4 · 2 · (−2) = 9 + 16 = 25, √D = 5.
3. x = (3 ± 5)/4.
4. Answer: x₁ = 2, x₂ = −1/2.

§3. Vieta’s theorem

If we add and multiply the two root formulas, √D cancels:

For a monic equation: x₁ + x₂ = −p, x₁x₂ = q.
Converse: numbers with sum s and product q are the roots of x² − sx + q = 0. This is handy for mental solving: x² − 5x + 6 = 0 → product 6, sum 5 → 2 and 3.
Without finding the roots, we can compute their symmetric expressions:

Signs of the roots: c/a < 0 — roots of opposite signs; c/a > 0 and −b/a > 0 — both positive; c/a > 0 and −b/a < 0 — both negative (provided D ≥ 0).
Before using Vieta’s theorem, check that D ≥ 0; otherwise there are no real roots.

Worked example (official sample)

For the roots of 2x² − 5x − 3 = 0, find 1/x₁ + 1/x₂.

1. D = 25 + 24 = 49 > 0; x₁ + x₂ = 5/2, x₁x₂ = −3/2.
2. 1/x₁ + 1/x₂ = (x₁ + x₂)/(x₁x₂) = (5/2) : (−3/2) = −5/3.

§4. Factorising a quadratic trinomial

A quadratic trinomial with roots x₁, x₂ can be written as:

This is needed for simplifying fractions and solving inequalities. For example, the roots of 2x² + 5x − 3 are 1/2 and −3, so 2x² + 5x − 3 = 2(x − 1/2)(x + 3) = (2x − 1)(x + 3).
Graphical meaning: the parabola y = ax² + bx + c crosses the x-axis at the roots. D > 0 — at two points, D = 0 — the vertex touches the axis, D < 0 — it does not meet the axis. The x-coordinate of the vertex, x₀ = −b/(2a), is the mean of the roots.

§5. Equations that reduce to quadratics

• Biquadratic ax⁴ + bx² + c = 0: substitute t = x² ≥ 0.
• A repeated expression: t = (x − 1)², t = x² + x and so on.
• A rational equation: bring to a common denominator and discard roots that make a denominator zero.

Worked example (official sample)

(x − 1)⁴ + 2x = x² + 73. How many real roots are there, and what is their product?

1. x² − 2x = (x − 1)² − 1. Let t = (x − 1)² ≥ 0: t² − (t − 1) = 73.
2. t² − t − 72 = 0 → t = 9 or t = −8 (rejected, t ≥ 0).
3. (x − 1)² = 9 → x = 4 or x = −2.
4. 2 roots, product −8.

§6. Questions with a parameter

The number of roots is determined by the sign of the discriminant:

• One root: D = 0 (and a ≠ 0; if a depends on the parameter, check the case a = 0 separately — the equation becomes linear and may have one root).
• No roots: D < 0; two roots: D > 0.
• Roots of opposite signs: c/a < 0 (then D > 0 holds automatically).

Worked example

For which k does x² + kx + 9 = 0 have exactly one root?

1. The leading coefficient is 1 ≠ 0, so the condition is D = 0.
2. D = k² − 36 = 0.
3. Answer: k = ±6.

Check yourself

1. Solve x² − 5x + 6 = 0.

2. Solve x² + 6x + 9 = 0.

3. For x² − 4x − 1 = 0, find x₁² + x₂².

4. Solve x⁴ − 5x² + 4 = 0.

5. Write the monic quadratic equation with roots −2 and 5.

6. For which a does ax² − 4x + 1 = 0 have exactly one root?

7. The absolute difference of the roots of x² − 7x + 10 = 0.

8. Solve 3x² − 7x + 2 = 0.

Common mistakes

• Writing x₁ + x₂ = b/a in Vieta’s theorem and forgetting the sign (correct: −b/a).
• Writing the “sum of the real roots” by Vieta even when D < 0.
• Not rejecting a root t < 0 in a biquadratic equation.
• Ignoring the case a = 0 when the leading coefficient is a parameter.
• Writing only x = 2 from x² = 4 (it should be x = ±2).
• Dividing x² = 4x by x and losing the root x = 0.