If x∈R,  then the maximum and minimum values of  x2+14x+9x2+2x+3  are

Let the given condition x2+14x+9x2+2x+3  =m,  then   ⇒(1−m)x2+(14−2m)x+  (9−3m)=0But x∈R,  therefore, (14−2m)2−4.(1−m)(9−3m)  ≥0               (∵ if x  is real ⇒b2−4ac≥0 )⇒    4m2−56m+196−12m2+48m−36≥0       ⇒    −8m2−8m+160  ≥0       ⇒    m2+m−20  ≤  0       ⇒    (m+5)  (m−4)  ≤0       ⇒-5⩽m≤4 ∴  maximum value = 4and minimum value = −5