Q21
2 marksVery Short AnswerSection B

  1. If α, β 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
Relation between zeroes and coefficients
Official Answer

1α+1β=−3\frac{1}{\alpha} + \frac{1}{\beta} = -3. From p(x)=x2−3x−1p(x) = x^2 - 3x - 1, sum of zeroes α+β=3\alpha + \beta = 3 and product αβ=−1\alpha\beta = -1; rewriting 1α+1β\frac{1}{\alpha} + \frac{1}{\beta} as α+βαβ\frac{\alpha + \beta}{\alpha\beta} gives 3−1=−3\frac{3}{-1} = -3.

sum of zeroes 3product of zeroes −11/α + 1/β = (α+β)/(αβ)−b/a and c/aanswer −3symmetric functions of zeroes

Marking Scheme

  • 11 mark: Correctly writing α+β=3\alpha + \beta = 3 and αβ=−1\alpha\beta = -1 from the coefficients (½ mark each).
  • 2½ mark: Expressing 1α+1β\frac{1}{\alpha} + \frac{1}{\beta} as α+βαβ\frac{\alpha + \beta}{\alpha\beta}.
  • 3½ mark: Correct final value −3-3 with proper sign.
  • 4Full 2 marks only if the negative sign on the product is handled correctly (answer −3-3, not +3+3).

Hint

Do not find α and β. Combine as a single fraction: 1α+1β=α+βαβ\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}, then substitute sum=−ba\text{sum} = -\frac{b}{a} and product=ca\text{product} = \frac{c}{a}.

Quick Oral Answer

Since α+β=−ba=3\alpha + \beta = -\frac{b}{a} = 3 and αβ=ca=−1\alpha\beta = \frac{c}{a} = -1, the expression 1α+1β\frac{1}{\alpha} + \frac{1}{\beta} equals α+βαβ\frac{\alpha + \beta}{\alpha\beta} = 33 divided by −1-1, which is −3-3 — found without ever solving the quadratic.

Analysis & Explanation

A direct 2-mark application of the sum/product of zeroes relationship, disguised as a reciprocal expression.


Concept

  • For ax2+bx+cax^2 + bx + c with zeroes α, β: α+β=−ba\alpha + \beta = -\frac{b}{a} and αβ=ca\alpha\beta = \frac{c}{a}.
  • Combine the required expression over a common denominator: 1α+1β=α+βαβ\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}, avoiding the need to solve for α, β individually.

Common mistakes

  • Sign error: since c=−1c = -1, the product αβ is negative; writing 31\frac{3}{1} instead of 3−1\frac{3}{-1} is a frequent slip.
  • This 'symmetric function' technique also applies to α2+β2\alpha^2 + \beta^2, αβ+βα\frac{\alpha}{\beta} + \frac{\beta}{\alpha}, etc.

Common Mistakes

  1. 1Sign error: writing the product αβ=ca\alpha\beta = \frac{c}{a} as +1+1 instead of −1-1 (since c = −1-1), giving a wrong answer of +3+3.
  2. 2Trying to solve x2−3x−1=0x^2 - 3x - 1 = 0 for α and β explicitly — the roots are irrational (involve 13\sqrt{13}), wasting time and inviting arithmetic errors.
  3. 3Confusing sum and product formulas: using α+β=ca\alpha + \beta = \frac{c}{a} or αβ=−ba\alpha\beta = -\frac{b}{a}.

Interesting Facts

Because 1α+1β\frac{1}{\alpha} + \frac{1}{\beta} and α+β\alpha + \beta share the same sign only when the product αβ\alpha\beta is positive, the sign of the product secretly controls the sign of this expression — a neat consistency check.

1α\frac{1}{\alpha} and 1β\frac{1}{\beta} are actually the zeroes of the 'reversed' polynomial −x2−3x+1-x^2 - 3x + 1 (coefficients reversed), so their sum can also be read directly as −−3−1=−3-\frac{-3}{-1} = -3, matching our answer.

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Frequently Asked Questions

Do I need to find α and β to solve 1α+1β\frac{1}{\alpha} + \frac{1}{\beta}?

No. Rewrite it as α+βαβ\frac{\alpha + \beta}{\alpha\beta}. Since α+β=−ba=3\alpha + \beta = -\frac{b}{a} = 3 and αβ=ca=−1\alpha\beta = \frac{c}{a} = -1 come straight from the coefficients, substitute to get 3−1=−3\frac{3}{-1} = -3 without solving the quadratic.

What is the sum and product of zeroes of x2−3x−1x^2 - 3x - 1?

With a=1,b=−3,c=−1a = 1, b = -3, c = -1: sum α+β=−ba=3\alpha + \beta = -\frac{b}{a} = 3 and product αβ=ca=−1\alpha\beta = \frac{c}{a} = -1.

Why is the answer negative?

Because the product of the zeroes αβ=−1\alpha\beta = -1 is negative. The expression equals (positive sum) ÷ (negative product) = 3÷(−1)=−33 \div (-1) = -3, so the sign of c drives the negative result.