Question
If the quadratic equation $(\lambda + 2)x^{2} - 3\lambda x + 4\lambda = 0$, $\lambda \neq - 2$, has two positive roots, then the number of possible integral values of $\lambda$ is :
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Solution
Sol. [Key Concept: Two positive roots need $D \geq 0$, product $> 0$, sum $> 0$ and $af(0) > 0$; count the integer $\lambda$.]
$$f(x) = (\lambda + 2)x^{2} - 3\lambda x + 4\lambda $$(C-1) $\ af(0) > 0$
$$(\lambda + 2)4\lambda > 0 $$$\Rightarrow \lambda < - 2$ or $\lambda > 0$
(C-2) $- \frac{b}{2a} > 0$
$$\frac{3\lambda}{2(\lambda + 2)} > 0 $$$\Rightarrow \lambda < - 2$ or $\lambda > 0$
(C-3) $D \geq 0$
$${( - 3\lambda)^{2} - 4(\lambda + 2) \times 4\lambda \geq 0 }{\lambda(7\lambda + 32) \leq 0 }{\Rightarrow \lambda \in \left\lbrack \frac{- 32}{7},0 \right\rbrack }$$Intersection of $C - 1,C - 2$ and $C - 3$
$${\Rightarrow \lambda \in \left\lbrack \frac{- 32}{7}, - 2 \right) }{\lambda \in \lbrack - 4.57, - 2) }{\Rightarrow \lambda = - 4, - 3 }$$∴ Number of values of $\lambda = 2$
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