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Difficulty: MediumQuadratic Equations and Factoring

If mm and nn are positive integers such that the quadratic equation x2−mx+n=0x^2 - mx + n = 0 has two roots that are distinct prime numbers, and the discriminant of the equation is equal to 11, what is the value of m+nm + n?

  1. A
    5
  2. B
    7
  3. 11Answer
  4. D
    13
  5. E
    1

Answer

11
For the quadratic equation x2−mx+n=0x^2 - mx + n = 0 with roots p1p_1 and p2p_2, the discriminant is D=(p1−p2)2D = (p_1 - p_2)^2. Given D=1D = 1, the difference between the two prime roots is ∣p1−p2∣=1|p_1 - p_2| = 1. The only prime numbers with a difference of 1 are 2 and 3. Thus, m=2+3=5m = 2 + 3 = 5 and n=2×3=6n = 2 \times 3 = 6, giving m+n=11m + n = 11.

Step-by-Step Solution

1
Express the discriminant of the quadratic equation in terms of its roots.
If p1p_1 and p2p_2 are the roots of x2−mx+n=0x^2 - mx + n = 0, Vieta's formulas give p1+p2=mp_1 + p_2 = m and p1p2=np_1 p_2 = n. The discriminant is D=m2−4n=(p1+p2)2−4p1p2=(p1−p2)2D = m^2 - 4n = (p_1 + p_2)^2 - 4p_1 p_2 = (p_1 - p_2)^2.
Relating the discriminant directly to the difference of the roots simplifies the constraint.
2
Determine the roots using the given discriminant value.
Since D=1D = 1, we have (p1−p2)2=1(p_1 - p_2)^2 = 1, which implies ∣p1−p2∣=1|p_1 - p_2| = 1.
Taking the square root of both sides indicates the two roots differ by 1.
3
Identify the prime numbers that satisfy this condition.
The only pair of prime numbers that differ by 1 is 2 and 3, because 2 is the only even prime and all other primes are odd.
Consecutive integers that are both prime must be 2 and 3.
4
Calculate mm, nn, and their sum m+nm + n.
m=2+3=5m = 2 + 3 = 5 and n=2×3=6n = 2 \times 3 = 6. Therefore, m+n=5+6=11m + n = 5 + 6 = 11.
Substitute the root values into the sum and product formulas.

Key Concept

Quadratic discriminant and root relationships (Vieta's Formulas)
Estimated Time:1m 30s
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Topics

Quadratic Equations and Factoring

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