Question

If the set of all $ a \in \mathbb{R} \setminus \{1\} $, for which the roots of the equation $ (1 - a)x^2 + 2(a - 3)x + 9 = 0 $ are positive is $ (-\infty, -\alpha] \cup [\beta, \gamma] $, then $ 2\alpha + \beta + \gamma $ is equal to ...........

Hint:

When solving quadratic inequalities, use both the discriminant condition and the sum and product of roots to narrow down the possible values for the variables.

Answer 1

Correct Answer: 7

Both the roots are positive.
For the quadratic equation \( (1 - a)x^2 + 2(a - 3)x + 9 = 0 \), we use the discriminant condition: \[ D \geq 0 \] This condition is satisfied for: \[ 4(a - 3)^2 - 4 \cdot (a - 3) \cdot 9 \geq 0 \] \[ a^2 - 6a + 9 + 9a + 9 \geq 0 \] \[ a^2 + 3a \geq 0 \] This gives: \[ a(a + 3) \geq 0 \quad \text{(Equation (i))} \] Now solving for \( a \): \[ a \in (-\infty, -3] \cup [0, \infty) \] Next, we apply the condition for the sum and product of the roots: \[ -\frac{b}{2a}>0 \] This gives: \[ \frac{2(a - 3)}{2(a - 1)}>0 \] Which implies: \[ a \in (-\infty, 1) \quad \text{(Equation (ii))} \] Therefore, combining these two conditions: \[ a \in (-\infty, -3] \cup [0, 1) \] Substituting into the given equation: \[ 2\alpha + \beta + \gamma = 7 \] Thus, the final value is: \[ 2\alpha + \beta + \gamma = 7 \]

Answer 2

We need the set of all real parameters \(a \ne 1\) for which both roots of the quadratic \( (1-a)x^2 + 2(a-3)x + 9 = 0 \) are positive. Then identify it as \( (-\infty,-\alpha] \cup [\beta,\gamma] \) and compute \( 2\alpha+\beta+\gamma \).

Concept Used:

For a quadratic \(Ax^2+Bx+C=0\) to have both roots positive, we require (i) real roots: \( \Delta=B^2-4AC\ge 0 \), (ii) sum of roots \( S=-\dfrac{B}{A}>0 \), and (iii) product of roots \( P=\dfrac{C}{A}>0 \).

Step-by-Step Solution:

Step 1: Identify coefficients.

\[ A=1-a,\qquad B=2(a-3),\qquad C=9. \]

Step 2: Product condition \(P>0\).

\[ P=\frac{C}{A}=\frac{9}{1-a}>0 \;\Rightarrow\; 1-a>0 \;\Rightarrow\; a<1. \]

Step 3: Sum condition \(S>0\).

\[ S=-\frac{B}{A}=\frac{2(a-3)}{a-1}. \] Since \(a<1\) (from Step 2), denominator \(a-1<0\). For \(S>0\) we need \(a-3<0\), i.e. \(a<3\), which is automatically true for \(a<1\). Hence the sum condition adds no new restriction beyond \(a<1\).

 

Step 4: Discriminant condition \(\Delta\ge 0\).

\[ \Delta=B^2-4AC=4(a-3)^2-36(1-a)=4\big[(a-3)^2-9(1-a)\big] =4\big(a^2+3a\big)=4a(a+3)\ge 0. \] \[ \Rightarrow\quad a\le -3 \;\; \text{or} \;\; a\ge 0. \]

Step 5: Combine with \(a<1\) (from Step 2):

\[ a\in(-\infty,-3]\ \cup\ [0,1). \] Given the problem states \(a\in\mathbb{R}\setminus\{1\}\), this can be written as \( (-\infty,-3] \cup [0,1] \) with the implicit exclusion of \(a=1\). Thus, \[ \alpha=3,\quad \beta=0,\quad \gamma=1. \]

 

Final Computation & Result

\[ 2\alpha+\beta+\gamma=2\cdot 3+0+1=\boxed{7}. \]