Question

if both the roots of the equation
x^2-6ax+2-2a+9a^2=0 exceed 3 ,then
a)a<1 b)a>11/9 c) a>3/2 d) a<5/2

Answer 1

6a-(36a^2 - 4 (2-2a+9a^2))^(1/2) > 6

Answer 2

or

6a+(36a^2 - 4 (2-2a+9a^2))^(1/2) > 6

Answer 3

above ans is correct
6a-(36a^2 - 4 (2-2a+9a^2))^(1/2) > 6

Answer 4

from wolfram calculator we have, http://www.wolframalpha.com/input/?i=6a-%2836a%5E2+-+4+%282-2a%2B9a%5E2%29%29%5E%281%2F2%29+%3E+6
a > 11/9

Answer 5

for other equation we have
a > 1
http://www.wolframalpha.com/input/?i=6a%2B%2836a%5E2+-+4+%282-2a%2B9a%5E2%29%29%5E%281%2F2%29+ >

Answer 6

so it seems that ... a is positive, and the least value of root is the first one ... so it seems your answer is a > 11/9

Answer 7

preparing for IIT-JEE
a>11/9

Answer 8

http://www.goiit.com/posts/show/390349.htm
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