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The set of all values of 'a' for which the roots of the equation $(1+a)x^2-3ax+4a=0$ exceeds 1

$\begin{array}{1 1}(A)\;(-1,2)&(B)\;(-\Large\frac{16}{7},\normalsize -1)\\(C)\;(-1,3)&(D)\;(-\Large\frac{16}{7},\normalsize 0)\end{array} $

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$a\neq 0,-1$
For both roots to exceed 1
$9a^2-4(4a)(1+a)\geq 0$
$\large\frac{3a}{1+a}$$ > 1$
$(1+a)[(1+a)-3a+4a] > 0$
$\Rightarrow a(7a+16)\leq 0$
$(2a-1)(1+a) >0$ and $(2a+1)(1+a) > 0$
$\Rightarrow -\large\frac{16}{7} $$\leq a\leq 0;a < -1$ or $a > \large\frac{1}{2}$
and $a < -1$ or $a > -\large\frac{1}{2}$
$\Rightarrow -\large\frac{16}{7}$$ \leq a \leq -1$
$\Rightarrow a\in [-\large\frac{16}{7},$$-1]$
Hence (B) is the correct answer.
answered Apr 15, 2014 by sreemathi.v

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