JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
Let \(\alpha, \beta\) be the roots of the equation \(x^2 - 3x + r = 0\), and \(\dfrac{\alpha}{2}, 2\beta\) be the roots of the equation \(x^2 + 3x + r = 0\). If the roots of the equation \(x^2 + 6x = m\) are \(2\alpha + \beta + 2r\) and \(\alpha - 2\beta - \dfrac{r}{2}\), then \(m\) is equal to:
- A \(-135\)
- B \(-567\)
- C \(135\)
- D \(567\)
Answer & Solution
Correct Answer
(D) \(567\)
Step-by-step Solution
Detailed explanation
From the first equation \(x^2 - 3x + r = 0\), the sum and product of the roots are: \(\alpha + \beta = 3\) \(\alpha \beta = r\) From the second equation \(x^2 + 3x + r = 0\), the sum and product of the roots are: \(\dfrac{\alpha}{2} + 2\beta = -3\)…
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