The given quadratic equation is $ (\lambda + 2)x^2 - 3\lambda x + 4\lambda = 0 $. We are given that $ \lambda \neq -2 $ and the equation has two positive roots.
For a quadratic equation $ax^2 + bx + c = 0$ to have two positive roots, three conditions must be satisfied:
- Real Roots: The discriminant $ \Delta = b^2 - 4ac \ge 0 $.
- Positive Sum of Roots: $ -b/a > 0 $.
- Positive Product of Roots: $ c/a > 0 $.
For the given equation $ (\lambda + 2)x^2 - 3\lambda x + 4\lambda = 0 $, we have $ a = \lambda + 2 $, $ b = -3\lambda $, and $ c = 4\lambda $.
1. Discriminant $ \Delta \ge 0 $
Calculate the discriminant:
$ \Delta = (-3\lambda)^2 - 4(\lambda + 2)(4\lambda) $ $ \Delta = 9\lambda^2 - 16\lambda(\lambda + 2) $ $ \Delta = 9\lambda^2 - 16\lambda^2 - 32\lambda $ $ \Delta = -7\lambda^2 - 32\lambda $Set $ \Delta \ge 0 $:
$ -7\lambda^2 - 32\lambda \ge 0 $Multiply by -1 and reverse the inequality:
$ 7\lambda^2 + 32\lambda \le 0 $Factor the expression:
$ \lambda(7\lambda + 32) \le 0 $The inequality holds for $ \lambda $ between the roots $ \lambda = 0 $ and $ \lambda = -32/7 $. So, $ -32/7 \le \lambda \le 0 $. Since $ -32/7 \approx -4.57 $, this interval is approximately $ [-4.57, 0] $.
2. Sum of Roots $ -b/a > 0 $
Calculate the sum of the roots:
$ -\frac{b}{a} = -\frac{-3\lambda}{\lambda + 2} = \frac{3\lambda}{\lambda + 2} $Set the sum of roots greater than 0:
$ \frac{3\lambda}{\lambda + 2} > 0 $This inequality holds when $ \lambda $ and $ \lambda + 2 $ have the same sign.
- Case 1: $ 3\lambda > 0 $ and $ \lambda + 2 > 0 \implies \lambda > 0 $ and $ \lambda > -2 \implies \lambda > 0 $.
- Case 2: $ 3\lambda < 0 $ and $ \lambda + 2 < 0 \implies \lambda < 0 $ and $ \lambda < -2 \implies \lambda < -2 $.
Therefore, the sum of roots is positive when $ \lambda \in (-\infty, -2) \cup (0, \infty) $.
3. Product of Roots $ c/a > 0 $
Calculate the product of the roots:
$ \frac{c}{a} = \frac{4\lambda}{\lambda + 2} $Set the product of roots greater than 0:
$ \frac{4\lambda}{\lambda + 2} > 0 $This inequality holds when $ \lambda $ and $ \lambda + 2 $ have the same sign.
- Case 1: $ 4\lambda > 0 $ and $ \lambda + 2 > 0 \implies \lambda > 0 $ and $ \lambda > -2 \implies \lambda > 0 $.
- Case 2: $ 4\lambda < 0 $ and $ \lambda + 2 < 0 \implies \lambda < 0 $ and $ \lambda < -2 \implies \lambda < -2 $.
Therefore, the product of roots is positive when $ \lambda \in (-\infty, -2) \cup (0, \infty) $.
4. Combine Conditions for $ \lambda $
We need to find the values of $ \lambda $ that satisfy all three conditions:
- $ \lambda \in [-32/7, 0] $ (from discriminant)
- $ \lambda \in (-\infty, -2) \cup (0, \infty) $ (from sum of roots)
- $ \lambda \in (-\infty, -2) \cup (0, \infty) $ (from product of roots)
The intersection of these intervals is $ \lambda \in [-32/7, -2) $. Approximately, $ -4.57 \le \lambda < -2 $.
5. Identify Integral Values of $ \lambda $
We need to find the integral values of $ \lambda $ within the interval $ [-32/7, -2) $. The integers in this range are $ -4 $ and $ -3 $.
The constraint $ \lambda \neq -2 $ is satisfied.
Thus, there are 2 possible integral values for $ \lambda $.
Similar Questions
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Important Questions from Algebra
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