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 February 25th, 2019, 10:46 AM #1 Newbie   Joined: Jan 2018 From: Belgium Posts: 8 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((x-3)(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
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Quote:
 Originally Posted by Atrend 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((x-3)(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.
$x < 3 \implies \sqrt{x - 3} \not \in \mathbb R.$

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
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Quote:
 Originally Posted by Atrend sqrt (x - 3) + sqrt (x + 5) > sqrt (x + 4) The solution is apparently x > 3 ....
If x=3 then:
sqrt(0) + sqrt(8) > sqrt(7) February 25th, 2019, 04:49 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,965 Thanks: 2214 If x = 3 when squaring the second time, one effectively has 0 > -1, and that shouldn't be squared to give 0 > 1. Tags inequality, roots, square Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post CRen1234 Algebra 2 May 22nd, 2014 02:04 PM fe phi fo Elementary Math 4 May 13th, 2013 07:21 PM drewm Algebra 3 July 16th, 2011 07:09 PM micheal2345 Algebra 11 November 8th, 2009 07:41 AM romeroom Algebra 7 March 17th, 2009 01:53 PM

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