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If \(\alpha\) and \(\beta\) are the roots of \(2x^{2} - 5x + 6 = 0\),...

Further Mathematics
WAEC 2018

If \(\alpha\) and \(\beta\) are the roots of \(2x^{2} - 5x + 6 = 0\), find the equation whose roots are \((\alpha + 1)\) and \((\beta + 1)\).

  • A. \(2x^{2} - 9x + 15 = 0\)
  • B. \(2x^{2} - 9x + 13 = 0\)
  • C. \(2x^{2} - 9x - 13 = 0\)
  • D. \(2x^{2} - 9x - 15 = 0\)
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Correct Answer: Option B
Explanation

Note: Given the sum of the roots and its product, we can get the equation using the formula:

\(x^{2} - (\alpha + \beta)x + (\alpha\beta) = 0\). This will be used later on in the course of our solution.

Given equation: \(2x^{2} - 5x + 6 = 0; a = 2, b = -5, c = 6\).

\(\alpha + \beta = \frac{-b}{a} = \frac{-(-5)}{2} = \frac{5}{2}\)

\(\alpha\beta = \frac{c}{a} = \frac{6}{2} = 3\)

Given the roots of the new equation as \((\alpha + 1)\) and \((\beta + 1)\), their sum and product will be

\((\alpha + 1) + (\beta + 1) = \alpha + \beta + 2 = \frac{5}{2} + 2 = \frac{9}{2} = \frac{-b}{a}\)

\((\alpha + 1)(\beta + 1) = \alpha\beta + \alpha + \beta + 1 = 3 + \frac{5}{2} + 1 = \frac{13}{2} = \frac{c}{a}\)

The new equation is given by: \(x^{2} - (\frac{-b}{a})x + (\frac{c}{a}) = 0\)

= \(x^{2} - (\frac{9}{2})x + \frac{13}{2} = 2x^{2} - 9x + 13 = 0\)


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WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995
WAEC offline past questions - with all answers and explanations in one app - Download for free
WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
WAEC offline past questions - with all answers and explanations in one app - Download for free
WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995