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If the zero of polynomial ax square+bc+c are in ratio 4:5 prove that 20 b square=81ac

Posted by Ayushi Maurya 4 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

Yogita Ingle 4 years, 9 months ago

Let roots 4k and 5k
4k+5k =-b/a
9k =-b/a
K=-b/9a..... (1)
4k*5k=c/a
20k²=c/a
Sub ... (1)
20 (-b/9a)²=c/a
20 (b²/81a²)=c/a
20 b²= 81 ac

0Thank You

Ranbeer Boparai 4 years, 9 months ago

Let the zeroes be 4p, 5p. We know that the sum of the zeroes=-b/a. Then, 4p+5p=9p=-b/a. p=-b/9a, then, p^2=b^2/81a^2...(1) And the product of zeroes is c/a. Then 4p×5p=c/a. 20p^2=c/a then, p^2=c/20a...(2) Using (1) and (2). b^2/81a^2=c/20a. Then, 20b^2=81ac.

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CBSE > Class 10 > Mathematics