If \(\alpha\) and \(\beta\) are the roots of \(2x^{2} - 5x + 6 = 0\), find the equation whose roots are \((\alpha + 1)\) and \((\beta + 1)\).
Answer Details
To find the equation whose roots are \((\alpha + 1)\) and \((\beta + 1)\), we can use the relationship between the roots and coefficients of a quadratic equation.
Let's start by finding the sum and product of the roots of the original equation, \(2x^{2} - 5x + 6 = 0\). We can use the formulae:
Sum of roots, \(S = -\frac{b}{a} = -\frac{-5}{2} = \frac{5}{2}\)
Product of roots, \(P = \frac{c}{a} = \frac{6}{2} = 3\)
Now, let's add 1 to both roots:
\(\alpha + 1\) and \(\beta + 1\)
The sum of these roots would be:
\((\alpha + 1) + (\beta + 1) = \alpha + \beta + 2\)
And the product would be:
\((\alpha + 1)(\beta + 1) = \alpha \beta + \alpha + \beta + 1\)
Now, we want to find the equation whose roots are \((\alpha + 1)\) and \((\beta + 1)\). Let's call this equation \(ax^{2} + bx + c = 0\).
According to the relationship between roots and coefficients, we know that:
\(\frac{-b}{a} = \frac{5}{2}\) and \(\frac{c}{a} = 3\)
Solving for \(b\) and \(c\) in terms of \(a\), we get:
\(b = -\frac{5a}{2}\) and \(c = 3a\)
Now, we can use the sum and product of the new roots to form two equations:
\(\frac{-b}{a} = \frac{5}{2} \implies -\frac{5a}{2a} = \frac{5}{2} \implies -5 = 5\)
This equation is clearly not true, which means that the first option, \(2x^{2} - 9x + 15 = 0\), is not the correct answer.
Let's move on to the second option:
\((\alpha + \beta) + 2 = \frac{5}{2} + 2 = \frac{9}{2}\)
\((\alpha \beta + \alpha + \beta + 1) = 3 + \frac{5}{2} + \frac{5}{2} + 1 = 8\)
Substituting the values of \(b\) and \(c\) in terms of \(a\), we get:
\(a(x^{2} - \frac{9}{2}x + \frac{13}{2}) = 0\)
Simplifying, we get:
\(2x^{2} - 9x + 13 = 0\)
Therefore, the answer is the second option, \(2x^{2} - 9x + 13 = 0\).