Prove it guys........
Answers
SOLUTION:-
Given:
(x-b)(x-c)+ (x-c)(x-a) +(x-a)(x-b)
=) x²-bx-cx+bx+x²-cx-ax+ca+x²-ax-bx+ab
=) x²+x²+x²-ax-ax-bx-bx-cx-cx+ab+bc+ca
=) 3x² -2ax -2bx -2cx +(ab+bc+ca)
=) 3x² -2ax (a+b+c) +(ab+bc +ca)
=) 3x² - 2(a+b+c)x + (ab+bc +ca)
Now,
Discriminate for above equation;
D=b² - 4ac
So,
D= {-2(a+b+c)}² - 4×3(ab+bc+ca)
=) 4(a+b+c)² - 12(ab+bc+ca)
=) 4[(a+b+c)² - 3(ab+bc+ca)]
=) 2×2×[a²+b²+c²+2ab+2bc+2ca-3ab-3bc-3ac
=) 2×2 ×[a² +b² +c² -ab-bc-ca]
=) 2[2a² +2b²+2c² -2ab-2bc -2ca]
=) 2[(a²+b²-2ab)+(b²+c²-2bc)+(c²+a²-2ca]
=) 2[(a-b)² + (b-c)² +(c-a)²]
Now,
Since, square of any quantity is either 0 or greater than 0 that is it's always +ve.
=) (a-b)² , (b-c)² ,(c-a)² are all +ve. no.
So,
2[(a-b)² + (b-c)² + (c-a)²
Discriminate is positive that is D=0.
It's Real roots.
For roots be equal & discriminate D=0
=) 2[(a-b)² + (b-c)² +(c-a)²]=0
=) (a-b)² + (b-c)² + (c-a)² =0
Since square number are always positive thus the above sum can be 0 only;
=) a-b = 0
=) b-c = 0
=) c-a = 0
So,
⚫a= b
⚫b= c
⚫c= a
Therefore,
=) a= b = c [proved]