If the root of quadratic equation (a-b)x^+(b-c)x+(c-a)=0
are equal prove that 2a=b+c
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2
Answer:
solved
Step-by-step explanation:
(a-b)x²+(b-c)x+(c-a)=0
(b-c)² = 4(a-b)(c-a)
b²-2bc+c² = 4{ac+ba-bc-a²)
Answered by
107
(b-c)x² + x(c-a) + (a-b) =0
To have real and equal roots, Discriminant, D =0
(c-a)² - 4(a-b)(b-c) =0
⇒c² +a² - 2ac - 4(ab -ac -b²+ bc) =0
⇒c² + a² -2ac - 4ab +4ac +4b² -4bc =0
⇒a² +4b² +c² - 4ab - 4bc +2ac =0
⇒(a -2b +c)² =0
∴(a -2b +c) =0
⇒(a+c) =2b (proved )
a,b and c are in A.P , where b is A.M
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