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. Share with your friends Share 10 Lovina Kansal answered this Dear student 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=0⇒m=3 or m=-3(which is not possible)Putting m=3 in eq(A), we get9+3a+12=0⇒3a=-21⇒a=-7Putting m=3 in eq(B), we get9+3b+15=0⇒3b=-24⇒b=-8 Regards 57 View Full Answer Jegannathan Anandaraman answered this 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) 7 Shivam Pandey answered this question not clear -24
