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:
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:
Updated On: Apr 10, 2026
- -135
- -567
- 135
- 567
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The Correct Option is B
Solution and Explanation
The given equation is \( x^2 - 3x + r = 0 \) with roots \( \alpha \) and \( \beta \), and the equation \( x^2 + 3x + r = 0 \) with roots \( \frac{\alpha}{2} \) and \( 2\beta \).
Using Vieta's formulas, we know:
\[
\alpha + \beta = 3 \quad \text{(sum of the roots of the first equation)}
\]
and
\[
\frac{\alpha}{2} + 2\beta = -3 \quad \text{(sum of the roots of the second equation)}.
\]
Now, solving these two equations:
1. \( \alpha + \beta = 3 \),
2. \( \frac{\alpha}{2} + 2\beta = -3 \).
Multiplying the second equation by 2 to eliminate the fraction:
\[
\alpha + 4\beta = -6.
\]
Now, subtract the first equation from this:
\[
(\alpha + 4\beta) - (\alpha + \beta) = -6 - 3,
\]
\[
3\beta = -9,
\]
\[
\beta = -3.
\]
Substitute \( \beta = -3 \) into the first equation:
\[
\alpha - 3 = 3,
\]
\[
\alpha = 6.
\]
Now, we have \( \alpha = 6 \) and \( \beta = -3 \).
Next, we use the quadratic equation \( x^2 + 6x = m \) with roots \( 2\alpha + \beta + 2r \) and \( \alpha - 2\beta - \frac{r}{2} \).
Substitute \( \alpha = 6 \) and \( \beta = -3 \):
\[
2\alpha + \beta + 2r = 2(6) + (-3) + 2r = 12 - 3 + 2r = 9 + 2r,
\]
\[
\alpha - 2\beta - \frac{r}{2} = 6 - 2(-3) - \frac{r}{2} = 6 + 6 - \frac{r}{2} = 12 - \frac{r}{2}.
\]
Now, applying the sum and product of the roots of the quadratic equation \( x^2 + 6x = m \):
\[
\text{Sum of the roots} = -6, \quad \text{Product of the roots} = m.
\]
The sum of the roots:
\[
(9 + 2r) + \left( 12 - \frac{r}{2} \right) = -6,
\]
\[
21 + \frac{3r}{2} = -6,
\]
\[
\frac{3r}{2} = -27,
\]
\[
r = -18.
\]
Now, substitute \( r = -18 \) into the product of the roots:
\[
(9 + 2r) \cdot \left( 12 - \frac{r}{2} \right) = m,
\]
\[
(9 + 2(-18)) \cdot \left( 12 - \frac{-18}{2} \right) = m,
\]
\[
(9 - 36) \cdot (12 + 9) = m,
\]
\[
-27 \cdot 21 = m,
\]
\[
m = -567.
\]
Final Answer: (B) -567
Final Answer: (B) -567
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