If \(\alpha\) and \(\beta\) are the roots of \(2x^{2} – 5x + 6 = 0\), find the equation whose roots are \((\alpha + 1)\) and \((\beta + 1)\).
- \(2x^{2} - 9x + 15 = 0\)
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\(2x^{2} - 9x + 13 = 0\)
- \(2x^{2} - 9x - 13 = 0\)
- \(2x^{2} - 9x - 15 = 0\)
The correct answer is: 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: \(x^{2} - (\frac{-b}{a})x + (\frac{c}{a}) = 0\)
= \(x^{2} - (\frac{9}{2})x + \frac{13}{2} = 2x^{2} - 9x + 13 = 0\)