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Question

If the quadratic equation ax2+bx+c=0 with real coefficients has real and distinct roots in (1,2) then a and 5a+2b+c

A
have same sign.
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B
have opposite sign.
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C
are equal.
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D
are not real.
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Solution

The correct option is A have same sign.
Let α,β be the roots of ax2+bx+c=0
Given that 1<α,β<2


Case (1):a>0
f(2)>04a+2b+c>05a+2b+c>0

Case (2):a<0
f(2)<04a+2b+c<05a+2b+c<0

In both the cases a and 5a+2b+c has same sign.

Alternate solution:
a(5a+2b+c)=a2(5+2ba+ca)
=a2[52(α+β)+αβ]
=a2[(α2)(β2)+1]>0 [α<2 and β<2]
a and 5a+2b+c have same sign.

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