Q21
2 marksVery Short AnswerSection B

If α,β\alpha, \beta are the zeroes of the polynomial p(x)=x2−3x−1p(x) = x^2 - 3x - 1, then find the value of 1α+1β\frac{1}{\alpha} + \frac{1}{\beta}.

Polynomials
Relationship between Zeroes and Coefficients
Official Answer

1α+1β=−3\frac{1}{\alpha} + \frac{1}{\beta} = -3, obtained directly from α+β=3\alpha+\beta = 3 and αβ=−1\alpha\beta = -1 without solving for the individual (irrational) roots.

sum of zeroesproduct of zeroesquadratic polynomialcoefficientsreciprocal of zeroes-b/ac/a

Marking Scheme

  • 11 mark: correctly finding α+β=3 and αβ=−1\alpha + \beta = 3 \text{ and } \alpha\beta = -1 using the coefficient relationships.
  • 21 mark: correctly combining as α+βαβ\frac{\alpha+\beta}{\alpha\beta} and arriving at the final answer -3.

Hint

Rewrite 1α+1β\frac{1}{\alpha} + \frac{1}{\beta} as α+βαβ\frac{\alpha+\beta}{\alpha\beta} and use α+β=−b/a,αβ=c/a\alpha+\beta = -b/a, \alpha\beta = c/a — no need to find α,β\alpha, \beta individually.

Quick Oral Answer

Using α+β=−b/a=3 and αβ=c/a=−1\alpha+\beta = -b/a = 3 \text{ and } \alpha\beta = c/a = -1 for p(x)=x2−3x−1p(x) = x^2-3x-1, I get 1α+1β=α+βαβ=3−1=−3\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{3}{-1} = -3.

Analysis & Explanation

Uses the sum-product relationship between zeroes and coefficients — no need to find the actual roots.


Concept

  • For ax2+bx+cax^2 + bx + c: sum of zeroes α+β=−b/a\alpha+\beta = -b/a, product αβ=c/a\alpha\beta = c/a.
  • Rewrite 1α+1β=α+βαβ\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} to avoid solving for irrational roots.

Key points

  • For p(x)=x2−3x−1p(x) = x^2 - 3x - 1: a=1,b=−3,c=−1a=1, b=-3, c=-1 ⟹ α+β=3,αβ=−1\alpha+\beta = 3, \alpha\beta = -1.
  • 1α+1β=3−1=−3\frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{-1} = -3.

Common mistakes

  • Solving the quadratic via the formula to get surd roots, then adding reciprocals — correct but slow and error-prone; the identity method is the intended 2-mark approach.

Common Mistakes

  1. 1Trying to find the actual irrational values of α\alpha and β\beta using the quadratic formula instead of using the sum-product shortcut, wasting time and risking arithmetic slips.
  2. 2Sign errors when reading b=−3b = -3 as +3 while computing −b/a-b/a, leading to a wrong sum of zeroes.
  3. 3Forgetting to divide by a (assuming a=1a = 1 always) in polynomials where the leading coefficient is not 1.

Interesting Facts

The relations α+β=−b/a\alpha + \beta = -b/a and αβ=c/a\alpha\beta = c/a come directly from expanding a(x−α)(x−β)=ax2−a(α+β)x+aαβa(x-\alpha)(x-\beta) = ax^2 - a(\alpha+\beta)x + a\alpha\beta and comparing coefficients with ax2+bx+cax^2+bx+c.

This 'reciprocal of zeroes' trick — expressing 1α+1β\frac{1}{\alpha} + \frac{1}{\beta} as α+βαβ\frac{\alpha+\beta}{\alpha\beta} — is a standard technique that also extends to finding α2+β2\alpha^2+\beta^2, α3+β3\alpha^3+\beta^3, or (α−β)2(\alpha-\beta)^2 without solving for individual roots.

x2−3x−1=0x^2 - 3x - 1 = 0 actually has irrational roots 3±132\frac{3\pm\sqrt{13}}{2}, which is precisely why the coefficient-relationship shortcut is essential here rather than direct computation.

Spotted a mistake or something unclear?

Tell us — we fix reported answers fast.

Frequently Asked Questions

Why not just solve for α\alpha and β\beta directly?

The roots of x2−3x−1=0x^2-3x-1=0 are irrational (3±132\frac{3\pm\sqrt{13}}{2}), so directly computing 1α+1β\frac{1}{\alpha}+\frac{1}{\beta} would involve messy surd arithmetic; using α+βαβ\frac{\alpha+\beta}{\alpha\beta} avoids this entirely.

What is the general formula used here?

For ax2+bx+cax^2+bx+c with zeroes α,β\alpha, \beta: α+β=−b/a\alpha+\beta = -b/a and αβ=c/a\alpha\beta = c/a; then 1α+1β=α+βαβ\frac{1}{\alpha}+\frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta}.