Question

If \(\alpha, \beta\) are the roots of the equation \(x^2 + 3x + k = 0\) and \(\alpha + 1/\beta\), \(\beta + 1/\alpha\) are the roots of the equation \(4x^2 + px + 18 = 0\) then k satisfies the equation

• A. \(2x^2 - 13x + 20 = 0\)
• B. \(x^2 - 5x + 6 = 0\)
• C. \(2x^2 - 7x + 3 = 0\)
• D. \(x^2 - 8x + 15 = 0\)

Hint:

When relating the roots of two different polynomials, Vieta's formulas (\(sum = -b/a\), \(product = c/a\)) are your primary tools. Start by finding an expression that involves only one unknown (like \(k\) in this case), which often comes from the product of the new roots.

Answer 1

The Correct Option is B

For the first equation, \(x^2 + 3x + k = 0\), with roots \(\alpha\) and \(\beta\), we have from Vieta's formulas:
Sum of roots: \(\alpha + \beta = -3\).
Product of roots: \(\alpha \beta = k\).
For the second equation, \(4x^2 + px + 18 = 0\), the roots are \((\alpha + 1/\beta)\) and \((\beta + 1/\alpha)\).
Let's find the product of the roots for the second equation.
Product = \((\alpha + 1/\beta)(\beta + 1/\alpha) = \alpha\beta + \alpha(1/\alpha) + (1/\beta)\beta + (1/\beta)(1/\alpha)\).
Product = \(\alpha\beta + 1 + 1 + \frac{1}{\alpha\beta} = \alpha\beta + \frac{1}{\alpha\beta} + 2\).
From the second equation, the product of roots is also given by \(c/a = 18/4 = 9/2\).
Equating the two expressions for the product:
\(\alpha\beta + \frac{1}{\alpha\beta} + 2 = \frac{9}{2}\).
Substitute \(\alpha\beta = k\):
\(k + \frac{1}{k} + 2 = \frac{9}{2}\).
Subtract 2 from both sides: \(k + \frac{1}{k} = \frac{9}{2} - 2 = \frac{5}{2}\).
Multiply the entire equation by \(2k\) to eliminate fractions:
\(2k^2 + 2 = 5k\).
Rearranging gives a quadratic equation in \(k\): \(2k^2 - 5k + 2 = 0\).
Solving for k: \((2k-1)(k-2) = 0\), which gives \(k=1/2\) or \(k=2\).
The question asks which of the given equations is satisfied by \(k\). Let's test the value \(k=2\).
Checking option (B): \(x^2 - 5x + 6 = 0\).
Substituting \(x=k=2\), we get \((2)^2 - 5(2) + 6 = 4 - 10 + 6 = 0\).
The equation is satisfied. Therefore, \(k\) satisfies the equation \(x^2 - 5x + 6 = 0\).