if one zero of polynomial (a2 +9)x2 +13x +6a is reciprocal of other find value of a

Asked by achalgoel | 9th Jan, 2010, 09:17: PM

Expert Answer:

x = {-13 ±(169 - 4(6a)(a2+9))}/(2(a2+9))

x = {-13 ±(169 - 24a3 - 216a)}/(2(a2+9))

Since the zeros i.e. x1 and x2 are reciprocals,

{-13 +(169 - 24a3 - 216a)}/(2(a2+9)) = (2(a2+9))/{-13 -(169 - 24a3 - 216a)}

(-13)2 - (169 - 24a3 - 216a) = (2(a2+9))2

24a(a2 + 9) = 4(a2+9)2

24a = 4(a2+9)

4a2 -24a+36 = 0

a2 - 6a + 9 = 0

a = 3

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Answered by  | 9th Jan, 2010, 11:23: PM

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