My Math Forum Irrational Inequation, how does one solve this?

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 April 15th, 2017, 01:20 PM #1 Newbie   Joined: Apr 2017 From: brazil Posts: 4 Thanks: 0 Irrational Inequation, how does one solve this? x + (x2-10x+9)^(1/3) > [x+2(x2-10x+9)^(1/2)]^(1/2) Solution: -45/4 < x < 1 or x >= 9 Last edited by skipjack; April 17th, 2017 at 02:54 AM.
 April 16th, 2017, 01:09 AM #2 Global Moderator   Joined: Dec 2006 Posts: 18,954 Thanks: 1601 Is there a typing error? The solution given is incorrect, as can be verified by substituting x = 0.
 April 16th, 2017, 09:58 AM #3 Newbie   Joined: Apr 2017 From: brazil Posts: 4 Thanks: 0 not sure, I don't even know what to do with it. It must be an error, but still, how do I solve the question then?
 April 16th, 2017, 03:04 PM #4 Global Moderator   Joined: Dec 2006 Posts: 18,954 Thanks: 1601 Where did you get the question from? It appears that x + (x² - 10x + 9)^(1/2) > [x + 2(x² - 10x + 9)^(1/2)]^(1/2) was intended. Thanks from germaneagle
 April 16th, 2017, 04:21 PM #5 Newbie   Joined: Apr 2017 From: brazil Posts: 4 Thanks: 0 can no one solve this question ?
 April 16th, 2017, 05:33 PM #6 Newbie   Joined: Apr 2017 From: brazil Posts: 4 Thanks: 0 thank you guys!
 April 17th, 2017, 03:01 AM #7 Global Moderator   Joined: Dec 2006 Posts: 18,954 Thanks: 1601 Consider x + √(x² - 10x + 9) = √(x + 2√(x² - 10x + 9)). Squaring both sides gives 2x² - 10x + 9 + 2x√(x² - 10x + 9) = x + 2√(x² - 10x + 9). Rearranging that gives 2x² - 11x + 9 = 2√(x² - 10x + 9) - 2x√(x² - 10x + 9), i.e., (x - 1)(2x - 9) = -2(x - 1)√(x² - 10x + 9). Hence x = 1 or 2x - 9 = -2√(x² - 10x + 9). Squaring that gives (2x - 9)² = 4(x² - 10x + 9), i.e. 4x = -45, so x = -45/4. Solving the intended original inequation is now straightforward.

 Tags inequation, irracional, irrational, solve

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