Question

Let \( \alpha, \beta \) be the roots of the equation \( x^2 - 3x + r = 0 \), and \( \frac{\alpha}{2}, 2\beta \) 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 \( \alpha - 2\beta - \frac{r}{2} \), then \( m \) is equal to:

• A. -135
• B. -567
• C. 135
• D. 567

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The sum and product of the roots of a quadratic equation \(ax^2 + bx + c = 0\) are given by \(-b/a\) and \(c/a\), respectively.
We apply these relations to the given quadratic equations to find the values of \(\alpha, \beta\), and \(r\).
Step 2: Key Formula or Approach:
For the first equation \(x^2 - 3x + r = 0\), the sum of roots is \(\alpha + \beta = 3\) and the product is \(\alpha\beta = r\).
For the second equation \(x^2 + 3x + r = 0\), the sum of roots is \(\frac{\alpha}{2} + 2\beta = -3\) and the product is \(\left(\frac{\alpha}{2}\right)(2\beta) = r\).
Step 3: Detailed Explanation:
The product of roots condition for the second equation yields \(\alpha\beta = r\), which is consistent with the first equation.
Now we solve the system of linear equations for the sum of roots.
\[ \alpha + \beta = 3 \] \[ \frac{\alpha}{2} + 2\beta = -3 \implies \alpha + 4\beta = -6 \] Subtracting the first equation from the second yields a new relation.
\[ 3\beta = -9 \implies \beta = -3 \] Substituting \(\beta = -3\) into the first equation gives the value of \(\alpha\).
\[ \alpha - 3 = 3 \implies \alpha = 6 \] Now we find \(r\) by multiplying the roots.
\[ r = \alpha\beta = (6)(-3) = -18 \] Next, we calculate the roots \(R_1\) and \(R_2\) for the final equation \(x^2 + 6x - m = 0\).
\[ R_1 = 2\alpha + \beta + 2r = 2(6) + (-3) + 2(-18) = 12 - 3 - 36 = -27 \] \[ R_2 = \alpha - 2\beta - \frac{r}{2} = 6 - 2(-3) - \left(\frac{-18}{2}\right) = 6 + 6 + 9 = 21 \] The sum of these roots is \(R_1 + R_2 = -27 + 21 = -6\), which matches the coefficient of \(x\) in \(x^2 + 6x - m = 0\).
The product of the roots is given by the constant term \(-m\).
\[ R_1 R_2 = -m \] \[ (-27)(21) = -m \] \[ -567 = -m \implies m = 567 \] Step 4: Final Answer:
The value of \(m\) is \(567\).