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WAEC 2022 · Paper 1 · Q36

If α\alpha and β\beta are the roots of x2+mx−n=0x^2 + mx - n = 0, where mm and nn are constants, form the equation whose roots are 1α\dfrac{1}{\alpha} and 1β\dfrac{1}{\beta}.

Worked solution (try it first)
  1. For x2+mx−n=0x^2 + mx - n = 0: α+β=−m\alpha + \beta = -m and αβ=−n\alpha\beta = -n.
  2. The new sum is 1α+1β=α+βαβ\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta}
    =−m−n= \dfrac{-m}{-n}
    =mn= \dfrac{m}{n}, and the new product is 1αβ=−1n\dfrac{1}{\alpha\beta} = -\dfrac{1}{n}.
  3. The equation is x2−mnx−1n=0x^2 - \dfrac{m}{n}x - \dfrac{1}{n} = 0.
  4. Multiply by nn to get nx2−mx−1=0nx^2 - mx - 1 = 0, option D.

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