Question

If the roots of a quadratic equation (b-c) x2 + (c-a)x + (a-b) = 0 are real and equal,then prove that 2b = a+c

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

(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...