Question

If cosecα, cotα are the roots of $ax^{2} + bx + c = 0$ then,

1) $a^{4} + b^{4} + 4ab^{2}c = 0$

2) $a^{4} - b^4 + 4ab^2c = 0$

3) $a^2 - b^2 = 2ac$

4) $a^2 + b^2 - 2abc = 0$


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Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Answer:

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}^

By above identity we get,

{bc+ab-2ac}^2=0

bc+ab-2ac=0

b(a+c)=2ac

l

b=2ac/(a+c)

l

Hence if zeroes of given quadratic

equation are equal then,

b=2ac/(a+c)

1/b = (a+c)/2ac

2/b = a/(ac) + c/ac

2/b = 1/c + 1/a

"1/a + 1/c = 2/b"__________proved

I think my answer is capable to clear your confusion..

Step-by-step explanation:

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