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 February 26th, 2011, 09:14 PM #1 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Tangent Let 0° < ? < ? < 90°. If the quadratic equation x² - 3ax + 1 - 4a = 0 has to have two roots tan ? and tan ?, determine the value of a. Also determine the value of $\tan{\frac{\alpha\,+\,\beta}{2}}.$
 February 27th, 2011, 08:44 PM #2 Global Moderator   Joined: Dec 2006 Posts: 19,058 Thanks: 1618 What are the answers supposed to be?
 February 27th, 2011, 10:32 PM #3 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 $\frac{2}{9}\,=<\,a\,=<\,\frac{1}{4},\,\tan{\frac{\al pha\,+\,\beta}{2}}\,=\,\frac{1}{3}$
 February 28th, 2011, 01:09 AM #4 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Tangent The discriminant is 9a² - 4(1 - 4a) = 9a² + 16a - 4. For this to be positive, a must be outside the interval (-8±10)/9. So a < -2 or a > 2/9. But the roots of the equation are both positive, so a > 0 and 3a > sqrt(9a² + 16a - 4) Solving, 9a² > 9a² + 16a - 4 so a < 1/4. Finally, tan ? + tan ? = 3a, and tan ? tan ? = 1 - 4a. Therefore tan (?+?) = 3a/4a = 3/4. Applying the double-angle formula for tan, and solving 3/4 = 2t/(1-t²) for t, you find that t is either -3 or 1/3. But t is positive, so t = 1/3.
 March 1st, 2011, 07:31 AM #5 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Thanks aswoods!

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