anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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experimentX
  • experimentX
6a-(36a^2 - 4 (2-2a+9a^2))^(1/2) > 6
experimentX
  • experimentX
or 6a+(36a^2 - 4 (2-2a+9a^2))^(1/2) > 6
anonymous
  • anonymous
above ans is correct 6a-(36a^2 - 4 (2-2a+9a^2))^(1/2) > 6

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experimentX
  • experimentX
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
experimentX
  • experimentX
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+>
experimentX
  • experimentX
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
anonymous
  • anonymous
preparing for IIT-JEE a>11/9
anonymous
  • anonymous
http://www.goiit.com/posts/show/390349.htm go to this

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