plz ans this super easy question
Answers
Answer:
D=0 ( because roots are equal)
a=( b-c ) , b= c-a , c= a-b
D= b^2 - 4ac
0 = (c-a)^2 -4×(b-c) × (a-b)
0 = c^2 +a^2 -2ca - 4× ab-b^2 -ac +bc
0 = c^2 +a^2 - 2ca - 4ab -4b^2 -4ac +4bc
a^2 -4b^2 +c^2 -4ab-4bc-4ac =0
according to identity this( a+b+c)^2 above equation is equal to
(a-2b+c)^2 = 0
a-2b+c=√0
a+c-2b =0
a+c= 2b
hence proved
Answer:
Plz makes me brainliest
Step-by-step explanation:
D=0 ( because roots are equal)
a=( b-c ) , b= c-a , c= a-b
D= b^2 - 4ac
0 = (c-a)^2 -4×(b-c) × (a-b)
0 = c^2 +a^2 -2ca - 4× ab-b^2 -ac +bc
0 = c^2 +a^2 - 2ca - 4ab -4b^2 -4ac +4bc
a^2 -4b^2 +c^2 -4ab-4bc-4ac =0
according to identity this( a+b+c)^2 above equation is equal to
(a-2b+c)^2 = 0
a-2b+c=√0
a+c-2b =0
a+c= 2b
hence proved
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