If equation x2+ax+12=0, x2+bx+15=0 and x2+(a+b)x+36=0 have a common positive root. Find the value of a and b.

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We have,x2+ax+12=0     ...(1)x2+bx+15=0     ...(2)x2+a+bx+36=0      ...(3)Let the common root be m.So, m will satisfy all three equations i.em2+am+12=0    ...(A)m2+bm+15=0     ....(B)m2+(a+b)m+36=0     ....(C)Adding (A) and (B), we get2m2+a+bm+27=0    ..(D)Subtracting (C) form (D), wem2-9=0m=3 or m=-3(which is not possible)Putting m=3 in eq(A), we get9+3a+12=03a=-21a=-7Putting m=3 in eq(B), we get9+3b+15=03b=-24b=-8
Regards

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Hi Ipsita Dash, let k be the common root
So k^2 + a k + 12 = 0 --(1)
k^2 + b k + 15 = 0 --(2)
And k^2 + (a+b) k + 36 = 0 ---(3)
(3) - (2) ==> a k + 21 = 0 ---(4)
(1) - (4) ==> k^2 - 9 = 0 ===> k = +3 or -3
If k = 3, then a = - 7 
And if k = - 3 then a = +7
From 2, if k = +3, b = 8
If k = - 3 then b = -8
Set of values of a, b are (-7. -8) or (7, 8)
 
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question not clear
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