If the roots of the quadratic equation a(b-c)x2 +b(c-a)x+c(a-b)=0 then show that 1/a,1/b,1/c are in ap
Answers
Given quadratic equation,
a(b-c)x^2+b(c-a)^2+c(a-b)=0
also zeroes of given equation are equal.
Therefore,
discriminant=0
b^2-4ac=0
{b(c-a)}^2-4{a(b-c)}{c(a-b)}=0
b^2(c^2+a^2-2ac)-4{ab-ac}{ac-bc}=0
b^2c^2+a^2b^2-2ab^2c-4{a^2bc-ab^2c-a^2c^2+abc^2}=0
b^2c^2+a^2b^2-2ab^2c-4a^2bc+4ab^2c+4a^2c^2-4abc^2=0
b^2c^2+a^2b^2+2ab^2c-4a^2bc +4a^2c^2-4abc^2=0
b^2c^2+a^2b^2+4a^2c^2+2ab^2c -4a^2bc-4abc^2=0
We know that,
a^2+b^2+c^2+2ab+2bc+2ac={a+b+c}^2
By above identity we get,
{bc+ab-2ac}^2=0
bc+ab-2ac=0
b(a+c)=2ac
b=2ac/(a+c)
Hence if zeroes of given quadratic equation are equal then,
b=2ac/(a+c)
Answer:
Step-by-step explanation:
d=b^2 -4ac
b(c-a)^2 - 4a(b-c)c(a-b) = 0
(bc - ab)^2 - 4(ab-ac)(ac-cb) = 0
b^2c^2 + a^2b^2 -2ab^2c -4(a^bc - ab^2c -a^2c^2 + ac^2b) = 0
b^2c^2 + a^2b^2 -2ab^2c - 4a^2bc + 4ab^2c +4a^2c^2 - 4ac^ = 0
b^2c^2 + a^2b^2 +2ab^2c - 4a^2bc +4a^2c^2 - 4ac^2b = 0
(bc + cb)^2 - 4ac(ab- ac + cb) = 0
(bc + cb)^2 - 4ac(ab -c(a-b))=0