Math, asked by IASaspirant, 11 months ago

plz ans this super easy question​

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Answered by taniyatariyal3
1

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

Answered by Lilmezzi1115
0

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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