if roots of the equation (b-c) x²+(c-a) x+(a-b) =0 are equal, then show that a, b, c are in AP
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(b-c)x²+(c-a)x+(a-b)=0
Comparing with quadratic equation
Ax²+Bx+C=0
A=(b-c),B=c-a,C=a-b
Discriminant is eqwual to zero when roots are equal
D=B²-4AC=0
D=(c-a)²−4(b-c)(a-b)=0
D=(c²+a²−2ac)-4(ba-ac-b²+bc)=0
D=c²+a²−2ac-4ab+4ac+4b²-4bc=0
c²+a²+2ac-4b(a+c)+4b²=0
(a+c)²-4b(a+c)+4b²=0
[(a+c)-2b]²=0
a+c=2b,
Thus, a , b, c are in A.P
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Bhriti182
Bhriti182
17.08.2017
Math
Secondary School
answered • expert verified
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.
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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