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February 25th, 2019, 10:46 AM  #1 
Newbie Joined: Jan 2018 From: Belgium Posts: 7 Thanks: 0  Inequality with square roots
Hello, When practising, I was trying to solve following problem: sqrt (x  3) + sqrt (x + 5) > sqrt (x + 4) The solution is apparently x > 3, but I get x > ((12 + sqrt (912))/6 (which is 3.03322).Could anyone help me and give me some pointers as to how to tackle this problem please? I start by actually squaring the numbers (condition: x > 3, otherwise the numbers in the square root will get negative), so I get: x  3 + 2(sqrt((x3)(x+5))) + x + 5 > x + 4 Then I arrange the numbers: 2 * sqrt (x^2 + 2x  15) > x + 2 I now want to square again (no negative number under the square root under condition x > 3): 4(x^2 + 2x  15) > x^2 4x + 4 I work with this until I get: 3x^2 + 12x  64 > 0. Now, trying to factor the final one, I get D=144+768=912; with roots (12+sqrt(912))/6). Taking the condition x>3 into account, my solutions seems to be: x>(12+sqrt(912))/6. Thanks in advance. 
February 25th, 2019, 11:56 AM  #2  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 550  Quote:
And if we are dealing with complex numbers, order is not relevant. So, for the problem to make any sense, we start with a condition that $x \ge 3 \implies \sqrt{x  3} \ge 0.$ Furthermore, $5 > 4 \implies x + 5 > x + 4 \text { for any value of } x.$ $\therefore x \ge 3 \implies \sqrt{x + 5} > \sqrt{x + 4} \implies \sqrt{x  3} + \sqrt{x + 5}> 0 + \sqrt{x + 4} = \sqrt{x + 4}.$ When you start squaring things, you can get spurious results.  
February 25th, 2019, 12:30 PM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,122 Thanks: 1003  
February 25th, 2019, 04:49 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,373 Thanks: 2010 
If x = 3 when squaring the second time, one effectively has 0 > 1, and that shouldn't be squared to give 0 > 1.


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inequality, roots, square 
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