The number of all common roots of the equation x4−10x3+37x2−60x+36=0 and the transformed equation of it obtained by increasing any two distinct roots of it by 1 , keeping the other two roots fixed, is
Let ‌f(x)=x4−10x3+37x2−60x+36=0 ‌‌ At ‌x=1,f(1)=1−10+37−60+36≠0 ‌‌ At ‌x=2,f(2)=(24−10(23+37(22... ‌−60(2+36=0 ‌‌‌=16−80+148−120+36 ‌=200−200=0 So, x=2 is a factor of f(x) Similarly x=3 is a factor of f(x) ∴f(x)=(x−2)2(x−3)2 ∴ Original roots are 2,2,3,3 The distinct roots are 2 and 3 There are three possible pairs of distinct roots (i) Increase the two 2s by 1 (ii) Increase the two 3s by 1 (iii) Increase one 2 and one 3 by 1 Let us assume we choose one root 2 and one root 3 to increase by 1 Since, these distinct roots.
Increase one ' 2 ' by 1:2+1=3 Increase one ' 3 ' by 1:3+1=4 Other two roots ( 2 and 3 ) are kept fixed ∴ New sets of roots of he transformed equation is 3,4,2,3 Let the transformed equation be g(x). its roots are 2,3,3,4 ∴g(x)‌=(x−2)(x−3)(x−3)(x−4) ‌=(x−2)(x−3)2(x−4) The common roots are value of x for which f(x)=0 and g(x)=0 ∴ Common roots are 2,3,3 ∴ Number of common roots =3