Quadratic Equations Question - None

Question

Let a, b are two real roots of equation $x^2 + px + q = 0$, p, q ∈ ℝ, q ≠ 0. If the quadratic equation g(x) = 0 has two roots $a + \frac{1}{a}$, $b + \frac{1}{b}$ such that sum of its roots is equal to product of roots, then find the number of integral values q can attain.

Accepted Answer

Step 1: Apply Vieta's formulas to the original equation.

For x² + px + q = 0 with roots a and b:

a + b = −p

ab = q

Step 2: Find sum and product of new roots.

New roots are a + 1/a and b + 1/b.

Sum S = a + 1/a + b + 1/b = (a + b) + (1/a + 1/b) = (a + b) + (a + b)/(ab)

S = −p + (−p)/q = −p(1 + 1/q) = −p(q + 1)/q

Product P = (a + 1/a)(b + 1/b) = ab + a/b + b/a + 1/(ab)

P = ab + (a² + b²)/(ab) + 1/(ab) = q + ((a + b)² − 2ab)/(ab) + 1/q

P = q + (p² − 2q)/q + 1/q = q + (p² − 2q + 1)/q = (q² + p² − 2q + 1)/q

Step 3: Apply the condition S = P.

−p(q + 1)/q = (q² + p² − 2q + 1)/q

Multiply both sides by q:

−p(q + 1) = q² + p² − 2q + 1

−pq − p = q² + p² − 2q + 1

0 = q² + p² + pq − 2q + p + 1

0 = q² + pq + p² + p − 2q + 1

Step 4: Rearrange as a quadratic in p.

p² + p(q + 1) + (q² − 2q + 1) = 0

p² + p(q + 1) + (q − 1)² = 0

Step 5: Apply the discriminant condition.

For real p, discriminant ≥ 0:

Δ = (q + 1)² − 4(q − 1)² ≥ 0

(q + 1)² − 4(q − 1)² ≥ 0

Let u = q + 1 and v = q − 1, so u = v + 2:

(v + 2)² − 4v² ≥ 0

v² + 4v + 4 − 4v² ≥ 0

−3v² + 4v + 4 ≥ 0

3v² − 4v − 4 ≤ 0

3(q − 1)² − 4(q − 1) − 4 ≤ 0

Let w = q − 1:

3w² − 4w − 4 ≤ 0

(3w + 2)(w − 2) ≤ 0

−2/3 ≤ w ≤ 2

−2/3 ≤ q − 1 ≤ 2

1/3 ≤ q ≤ 3

Step 6: Find integral values.

Since q ∈ ℝ with 1/3 ≤ q ≤ 3 and q ≠ 0, the integral values are q ∈ {1, 2, 3}.

∴ Answer: 3

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