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
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
Regards,
Team,
TopperLearning.
Answered by
| 9th Jan, 2010,
11:23: PM
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