Question

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Q. If the roots of the equation $(b - c)x^2 + (c - a)x + (a - b) = 0$ are equal, then show that $a, b, c$ are in A.P.

( Related to Arithmetic Progression of 10th standard )

Answer 1

Answer:

                       $\huge\boxed{SOLUTION;}$

                             

we know that when roots of quadratic equations are equal then

=) Discriminant (D) = 0

Given equation is ,

=) (b-c)x2 + (c-a)x +(a-b) = 0

D = (c-a)^2 - 4(b-c)(a-b) = 0

=)(c^2 + a^2 -2ac) - 4(ab -b^2 -ac + bc )=0

=) c^2 + a^2 -2ac - 4ab + 4b^2 +4ac -4bc =0

=) a^2 + 4b^2 +c^2 -4ab - 4bc + 2ac =0

=) ( a -2b + c)^2 = 0

=) a -2b + c = 0

=) a +c = 2b ---------*(1)

we can write this expression as

=) (b - a )= (c -b )

so this is required condition for a , b ,c therefore a, b ,c are in A.P

                $\huge\boxed{proved}$

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Answer 2

Refer to the attatchment...