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I'm tutoring someone, and I'm stuck on one of her problems. The equation is $\sqrt{x+14}\le x-16$. She hasn't been taught the quadratic formula or how to factor these problems yet. Is there a way to solve this problem without using either of these methods? It's possible that the teacher wrote the problem without checking to make sure that it is solvable with what they know, but I want to make sure.

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  • $\begingroup$ First of all, state the question formally. Second of all, your going to need to say which methods she is allowed to use $\endgroup$ – Quality Mar 4 '15 at 1:18
  • $\begingroup$ How she can know about $\sqrt{..}$ without knowing how to square? $\endgroup$ – Elaqqad Mar 4 '15 at 1:21
  • $\begingroup$ What do you mean factor? Does she not know about $(x+2)^2=x^2+4x+4$? $\endgroup$ – Gonate Mar 4 '15 at 1:28
  • $\begingroup$ @Gonate She does know how to square things, but she hasn't been taught how to go in reverse. She can take $(x+2)^2$ and get $x^2+4x+4$, but she doesn't know how to find $(x+2)^2$ from $x^2+4x+4$ yet. $\endgroup$ – Jacob Jones Mar 4 '15 at 4:38
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it may be easier to make a change of variable $u = x + 14, x = u - 14.$ now we have $$\sqrt u \le u - 30$$ the equality is satisfied by $$u = 36$$ and the inequality for $$ u > 36$$ this can be argued by picking test points on either side of $u = 36.$ what we have now is $$\sqrt u \le u - 30 \text{ for } u \ge 36.$$ we can now translate it back to the original $x$ variable $$ \sqrt{x+14} \le x - 16 \text{ for } x \ge 22.$$

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  • $\begingroup$ $\sqrt{6} \neq 6-30.$ Maybe equality when $u=36.$ $\endgroup$ – coffeemath Mar 4 '15 at 2:11
  • $\begingroup$ @coffeemath, thanks for spotting the error. $\endgroup$ – abel Mar 4 '15 at 2:16
  • $\begingroup$ abel: If you're like me, at some point you had $6=36-30$ and just used the wrong notation for what $u$ denotes. (+1 on answer) $\endgroup$ – coffeemath Mar 4 '15 at 2:24
  • $\begingroup$ @coffeemath, thanks. i seem to make a lot o errors like this. $\endgroup$ – abel Mar 4 '15 at 3:13

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