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I have the following inequation: $\sqrt{t^{2}-t-12}<7-t$. Can I just set both sides of the inequation to the power of two, or is there any condition under which exponentiating is an equivalent operation?

Thanks

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    $\begingroup$ Do you mean possibly $(\sqrt{t^{2}-t-12})^2<(7-t)^2$ or $2^{\sqrt{t^{2}-t-12}}<2^{7-t}$? If the former, you need to consider what happens when squaring negative numbers $\endgroup$ – Henry Sep 22 at 14:46
  • $\begingroup$ @Henry the solution to a square root is always a positive number, so it's not an issue in this problem. $\endgroup$ – Matthew Daly Sep 22 at 14:47
  • $\begingroup$ @MatthewDaly what is $\sqrt{t^{2}-t-12}$ when $t=1$ ? Or $7-t$ when $t=100$ ? $\endgroup$ – Henry Sep 22 at 14:49
  • $\begingroup$ @Henry I see your point now. The inequality isn't more difficult in this case, but you need to keep the appropriate domain in mind. $\endgroup$ – Matthew Daly Sep 22 at 14:52
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This is not an equation; it's an inequality.

However, you have that $\exp\text{LHS}<\exp\text{RHS}$ because the exponential function is monotonic.

Oh, it seems the operation you want to perform is not exponentiation, but empowerment -- raising to a power. You may always do this if the power you want to raise both sides to is odd or a fraction with odd denominator since such functions are monotonic. However, when they are even, or have even denominator, you have to make sure that both sides are nonnegative. Otherwise, you need to simultaneously reverse the inequality, too.

Thus, you may square both sides without changing the $<$ to $>$ only whenever $7-t\ge 0.$ Of course, $\text{LHS},$ being a square root, is always nonnegative whenever it is real.

You have used many terms wrongly here, so let me correct the last: equivalent operation. Well, it seems like what you mean is order-preserving operation. What we have are equivalence relations, which are quite different things from operations that respect linear orders.

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    $\begingroup$ More exactly, it is an inequation. $\endgroup$ – Bernard Sep 22 at 14:50
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    $\begingroup$ @Bernard That may be a non-idiomatic synonym for inequality. $\endgroup$ – Allawonder Sep 22 at 14:51
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    $\begingroup$ For me (and others) an inequality is a relation. An inequation is a problem – an inequality with unknown(s). $\endgroup$ – Bernard Sep 22 at 15:00
  • $\begingroup$ @Bernard There's no difference between an inequality involving variables or one containing only numbers. They are both relations between numbers. $\endgroup$ – Allawonder Sep 23 at 16:23
  • $\begingroup$ One is usually a statement universally valid (possibly under some condition) e.g. $|x+y|\le |x|+|y|$ is kown as the triangle inequality, and the other in a problem with unknowns., e.g. ‘solve for $(x,y)$ in $|x-2|+|y-3| <1$’. You wouldn't speak of the equality $x^2+2x-5=0$, do you? $\endgroup$ – Bernard Sep 23 at 16:32
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Hints:

  1. You have to specify the domain of validity of the inequation. Here the condition is $t^2-t-12\ge 0$. As the roots of this quadratic polynomial are $-3$ and $4$, you obtain $\;(-\infty,-3]\cup[4,+\infty)$.
  2. In the domain of validity, you have the equivalence $$\sqrt{A}<B\iff A<B^2 \; \color{red}{\text{ and }B\ge 0}.$$
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It is valid to exponentiate if you preserve the inequality, that is, if $$a<b\iff a^c<b^c$$

We would have a problem if we were to do this as follows: $$-2<1\iff (-2)^2<1^2$$

In your case there is no problem, since the squareroot is defined to be positive, we only have to make sure that it stays that way. We have to look at $t:t^2-t-12\geq 0$, that is, $t\in (-\infty,-3]\cup [4,\infty)$, and also $t< 7$ (otherwise we would have $\sqrt{t^2-t-12}<0$, which is a problem). This forces us to have $t\in (-\infty,-3]\cup [4,7)$.

Given these conditions we know that $$\sqrt{t^2-t-12}<7-t\iff t^2-t-12<(7-t)^2$$ This simplifies to $$t<\frac{61}{13}$$

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It must be $$t^2-t-12\geq 0$$ and $$7-t>0$$, then we get by raising to the power two: $$t^2-t-12<49+t^2-14t$$ nand we get $$t<\frac{61}{13}$$ Finally we get $$t\le -2\sqrt{3}$$

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