Question

If the roots of the equation (b-c)x2 +(c-a) x +(a-b) = 0 are equal show that a, b, c are in AP.

Need full explanation!
Step by step!

Answer 1

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

proved
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Hope it will help u ^_^

Answer 2

Answer:

Step-by-step explanation:

2= -(c-a)/(b-c)

=> 2=(a-c)/(b-c)

=>2b-2c=a-c

=>2b=a+c

=>b=(a+c)/2

Hence proved.

Hope it helps.