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There is only one real value of a for which the quadratic equation `ax^2+(a+3)x+a-3=0` has two positive integral solutions. The product of these two solutions is

A

0.09

B

0.08

C

0.06

D

12

Text Solution

Verified by Experts

The correct Answer is:
B

Sum of the roots `=-((a+3))/a=-I^(+)`[let]
`:.a=(-3/(I^(+)+1))`..i
Product of the roots `=alpha beta=(a-3)/a=I^(+)+2`….ii
and `D=(a+3)^(2)-4a(a+3)`
`=9/((I^(+)+1)^(2)){(I^(+)-2)^(2)-12}` [from Eq. (i)]
D must be perfect square then `I^(+)=6`
From Eq. (ii)
Product of the roots `=I^(+)+2=6+2-8`
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