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Question

The value of $$'a'$$ for which the quadratic equation $$2{x^2} - x\left( {{a^2} + 8a - 1} \right) + {a^2} - 4a = 0$$ has roots opposite signs, lie in the interval 


A
1<a<5
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B
0<a<4
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C
1<a<2
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D
2<a<6
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Solution

The correct option is B $$0 < a < 4$$
R.E.F image 
Given equation is : $$ 2x^{2}-x(a^{2}+8a-1)+a^{2}-4a = 0 $$
Comparing it with the equation : $$ ax^{2}+bx+c = 0 $$
we have $$ a = 2 $$
$$ b = -(a^{2}+8a-1) $$
$$ c = a^{2}-4a $$
For the roots to have opposite signs.
product of roots $$ < 0 $$
$$ \Rightarrow \dfrac{c}{a}< 0 $$
$$ \Rightarrow \dfrac{a^{2}-4a}{2}< 0 $$
$$ \Rightarrow a^{2}-4a< 0 $$
$$ \Rightarrow a(a-4)< 0 $$
$$ \Rightarrow (a-0)(a-4)< 0 $$
$$ \therefore aE(0,4) $$ 

1177711_1034389_ans_da1477a90d8e47c7aedafa3159ccd581.jpg

Mathematics

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