If the zeores of the polynomial x²-px+q, be in the ratio of 2:3, prove that 6p² = 25q
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Answered by
30
Here,f(x) = x² - px + q
let the zeros be 2y and 3y
so,6y² = q/1 = q
=> y = √q/√6
and 2y + 3y = p/1 = p
=> 2√q/√6 + 3√q/√6 = p
=> √2q/√3 + √3q/√2 = p
=> (2√q+3√q)/√6 = p
=> 5√q = √6p
=> 25q = 6p² [proved]
let the zeros be 2y and 3y
so,6y² = q/1 = q
=> y = √q/√6
and 2y + 3y = p/1 = p
=> 2√q/√6 + 3√q/√6 = p
=> √2q/√3 + √3q/√2 = p
=> (2√q+3√q)/√6 = p
=> 5√q = √6p
=> 25q = 6p² [proved]
Answered by
10
Hello mate,
Here is your answer,
Let the zeroes be 2a and 3a.
Sum of 0's
p = 5a
a = p/5
Product of 0's
q = 6a2
q = 6(p/5)2
q = 6p2/25
6p2=25q
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