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1.Find the minimum value of3x^2+5y^2+4z^2-4xy-4xz-4x-2y+25 for real numbers x,y and z.2.suppose a,b,c are real numbers such that ('a' is a negative real number) ab-a=b+119, bc-b=c+59, ca-c=a+71. Find a+b+c

Gaurav Saini
9 Points
2 years ago
1. We would try to make whole squares to get minimum value
Rewrite as
$x^2-4xy+4y^{2}+x^{2}-4xz+4z^{2}+x^{2}-4x+4+y^{2}-2y+1+20$
(x-2y)+(x-2z)2+(x-2)2+(y-1)2+20
squares of any number >=0
So minimum value of expression =20

2.  ab-a=b+119 , bc-b=c+59 , ca-c =a+71
ab-a-b=119 , bc-b-c=59, ca-c-a=71
Adding 1 on both sides we get
ab-a-b+1=120
a(b-1)-(b-1)=120
(a-1)(b-1)=120 ----(1)

similarly adding 1 to both side in other two equation we will get
(b-1)(c-1)=60 -----(2)

(a-1)(c-1)=72 -----(3)

divide 1 by 2
a-1/c-1=2
and from 3  putting value of (c-1)
(a-1)2/72=2
(a-1)2=144
a-1=-12 (a is negative )
a=-11
Similarly
b=-9
c=-5