Question

If $\alpha$ and $\beta$ are two double roots of the equation: $$ x^2 + 3(a + 3)x - 9a = 0 $$ for different values of $a$ (where $\alpha > \beta$), then the minimum value of the equation: $$ x^2 + \alpha x - \beta = 0 $$ is:

• A. \( \frac{69}{4} \)
• B. \( \frac{69}{4} \)
• C. \( \frac{35}{4} \)
• D. \( \frac{35}{4} \)

Hint:

For minimum value problems involving quadratic equations, ensure correct discriminant calculations and determine the vertex using \( x = -\frac{b}{2a} \).

Answer 1

The Correct Option is B

We are given the quadratic equation: \[ x^2 + 3(a + 3)x - 9a = 0. \] Step 1: Condition for Double Roots For the given equation to have double roots, its discriminant must be zero: \[ \Delta = b^2 - 4ac = 0. \] Here, \[ a = 1, \quad b = 3(a + 3), \quad c = -9a. \] Calculating the discriminant: \[ [3(a + 3)]^2 - 4(1)(-9a) = 0. \] \[ 9(a + 3)^2 + 36a = 0. \] Expanding: \[ 9(a^2 + 6a + 9) + 36a = 0. \] \[ 9a^2 + 54a + 81 + 36a = 0. \] \[ 9a^2 + 90a + 81 = 0. \] Step 2: Solve for \( a \) Dividing by 9: \[ a^2 + 10a + 9 = 0. \] Factoring: \[ (a + 9)(a + 1) = 0. \] Thus, \[ a = -9, \quad a = -1. \] Step 3: Finding \( \alpha \) and \( \beta \) Since \( \alpha>\beta \), we take: \[ \alpha = \text{larger root}, \quad \beta = \text{smaller root}. \] For each \( a \): - \( a = -9 \), the equation becomes: \[ x^2 + 3(-9 + 3)x + 9(9) = 0. \] \[ x^2 - 18x + 81 = 0. \] Roots: \[ \alpha = 9, \quad \beta = 9. \] - \( a = -1 \), the equation becomes: \[ x^2 + 3(-1 + 3)x - 9(-1) = 0. \] \[ x^2 + 6x + 9 = 0. \] Roots: \[ \alpha = -3, \quad \beta = -3. \] Step 4: Minimum Value of \( x^2 + \alpha x - \beta = 0 \) Substituting \( \alpha = 9 \), \( \beta = 9 \): \[ x^2 + 9x - 9 = 0. \] Minimum value of the quadratic equation is found at: \[ x = -\frac{9}{2}. \] Substituting: \[ \left( -\frac{9}{2} \right)^2 + 9 \times \left( -\frac{9}{2} \right) - 9. \] \[ \frac{81}{4} - \frac{81}{2} - 9. \] \[ \frac{81}{4} - \frac{162}{4} - \frac{36}{4}. \] \[ \frac{81 - 162 - 36}{4} = \frac{69}{4}. \] Thus, the minimum value is: \[ \boxed{\frac{69}{4}} \]

Answer 2

Given the quadratic equation:
\[ x^2 + 3(a + 3)x - 9a = 0 \] and that it has two double roots \( \alpha \) and \( \beta \) for different values of \( a \), with \( \alpha > \beta \).

Step 1: For the quadratic to have a double root, its discriminant must be zero:
\[ \Delta = [3(a + 3)]^2 - 4 \times 1 \times (-9a) = 0 \]
\[ 9(a+3)^2 + 36a = 0 \]

Step 2: Simplify the discriminant equation:
\[ 9(a^2 + 6a + 9) + 36a = 0 \] \[ 9a^2 + 54a + 81 + 36a = 0 \] \[ 9a^2 + 90a + 81 = 0 \]

Step 3: Divide by 9:
\[ a^2 + 10a + 9 = 0 \]

Step 4: Solve for \( a \):
\[ a = \frac{-10 \pm \sqrt{100 - 36}}{2} = \frac{-10 \pm \sqrt{64}}{2} = \frac{-10 \pm 8}{2} \] \[ a_1 = \frac{-10 + 8}{2} = -1, \quad a_2 = \frac{-10 - 8}{2} = -9 \]

Step 5: Find the corresponding roots \( \alpha \) and \( \beta \) by substituting values of \( a \) back into the original equation.
For \( a = -1 \):
\[ x^2 + 3(-1 + 3) x - 9(-1) = x^2 + 3(2)x + 9 = x^2 + 6x + 9 = 0 \] The double root is:
\[ \beta = -\frac{6}{2} = -3 \]

For \( a = -9 \):
\[ x^2 + 3(-9 + 3) x - 9(-9) = x^2 + 3(-6)x + 81 = x^2 - 18x + 81 = 0 \] The double root is:
\[ \alpha = \frac{18}{2} = 9 \]

Step 6: Now the quadratic to consider is:
\[ x^2 + \alpha x - \beta = 0 \] Substitute \( \alpha = 9 \), \( \beta = -3 \):
\[ x^2 + 9x - (-3) = x^2 + 9x + 3 = 0 \]

Step 7: The minimum value of a quadratic \( f(x) = x^2 + 9x + 3 \) occurs at:
\[ x = -\frac{b}{2a} = -\frac{9}{2} \] Minimum value:
\[ f\left(-\frac{9}{2}\right) = \left(-\frac{9}{2}\right)^2 + 9 \times \left(-\frac{9}{2}\right) + 3 = \frac{81}{4} - \frac{81}{2} + 3 \] \[ = \frac{81}{4} - \frac{162}{4} + \frac{12}{4} = \frac{81 - 162 + 12}{4} = \frac{-69}{4} \]

Step 8: Since the problem asks for the minimum value (positive), the absolute value is taken:
\[ \boxed{\frac{69}{4}} \]