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Difficulty: HardQuadratic Equations and Polynomial Factoring

If mm and nn are non-zero real numbers such that mm and nn are the roots of the quadratic equation x2+mx+n=0x^2 + mx + n = 0, what is the value of m−nm - n?

  1. A
    -3
  2. B
    -1
  3. C
    0
  4. D
    1
  5. 3Answer

Answer

3
The correct answer is 3. By Vieta's formulas, for x2+mx+n=0x^2 + mx + n = 0, the sum of roots m+n=−mm + n = -m gives 2m+n=02m + n = 0, and the product of roots m⋅n=nm \cdot n = n gives m=1m = 1 since n≠0n \neq 0. Substituting m=1m = 1 into 2m+n=02m + n = 0 yields n=−2n = -2. Thus, m−n=1−(−2)=3m - n = 1 - (-2) = 3.

Step-by-Step Solution

1
Apply Vieta's formulas to express the sum and product of roots in terms of coefficients.
For the quadratic equation x2+mx+n=0x^2 + mx + n = 0, the sum of roots is m+n=−mm + n = -m, and the product of roots is m⋅n=nm \cdot n = n.
By Vieta's relations for a standard quadratic equation ax2+bx+c=0ax^2 + bx + c = 0, the sum of roots is −ba-\frac{b}{a} and the product is ca\frac{c}{a}.
2
Solve the product relation for mm.
Since n≠0n \neq 0, divide both sides of m⋅n=nm \cdot n = n by nn to get m=1m = 1.
Because nn is specified to be non-zero, division by nn is valid and yields a unique non-zero value for mm.
3
Substitute m=1m = 1 into the sum relation to determine nn.
1+n=−1  ⟹  n=−21 + n = -1 \implies n = -2.
Rearranging m+n=−mm + n = -m gives 2m+n=02m + n = 0, so n=−2(1)=−2n = -2(1) = -2.
4
Calculate the target value m−nm - n.
m−n=1−(−2)=3m - n = 1 - (-2) = 3.
Subtracting −2-2 from 11 produces 1+2=31 + 2 = 3.

Key Concept

Vieta's Formulas and Root-Coefficient Relationships in Quadratic Equations
Estimated Time:2m 0s
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Topics

Quadratic Equations and Polynomial Factoring

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