Question

Twice the product of the zeroes of the polynomial 1
23x2 - 26x + 161 is 14p. Find p.​

Answer 1

Given : polynomial 23x² - 26x + 161

Twice the product of the zeroes of the polynomial is 14p

To Find : value of p

Solution:

polynomial ax² + bx + c

Sum of zeroes = - b/a

Product of zeroes = c/a

23x² - 26x + 161

a = 23

b = -26

c = 161

Product of zeroes =  161/23  =  7

Twice the product of the zeroes = 2 (7) = 14

Twice the product of the zeroes of the polynomial is 14p

=> 14p = 14

=> p = 1

Hence value of the p is 1

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