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Math Problem: Irrational Inequation
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<blockquote data-quote="Deboog" data-source="post: 9597888" data-attributes="member: 330941"><p>Alright. So after looking at it more, I realized that Sabykos is actually right. When square both sides the FIRST time it is fine because you can prove that both sides are positive. When you square both sides the SECOND time it is not fine because you cannot prove that both sides are positive. In fact, the right side can very well be negative. A workaround is to, as a first step multiply both sides by root(4-3x) + root(x+1) which will move the root mess over to the right side. Then, after you square both sides and simplify you will get 16x^2 - 4x -11 > 8root[(4-3x)(x+1)] Thankfully, we can prove both sides are positive, the short version is square roots are positive, and so a square root times 8 is positive, and the left is greater than a positive so it's also positive.</p></blockquote><p></p>
[QUOTE="Deboog, post: 9597888, member: 330941"] Alright. So after looking at it more, I realized that Sabykos is actually right. When square both sides the FIRST time it is fine because you can prove that both sides are positive. When you square both sides the SECOND time it is not fine because you cannot prove that both sides are positive. In fact, the right side can very well be negative. A workaround is to, as a first step multiply both sides by root(4-3x) + root(x+1) which will move the root mess over to the right side. Then, after you square both sides and simplify you will get 16x^2 - 4x -11 > 8root[(4-3x)(x+1)] Thankfully, we can prove both sides are positive, the short version is square roots are positive, and so a square root times 8 is positive, and the left is greater than a positive so it's also positive. [/QUOTE]
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