MHB Range of Values For Inequalities

AI Thread Summary
The discussion revolves around solving the inequality -1 < a - b < 10 with the constraints -3 ≤ b ≤ 1. The initial attempt to find the range of a resulted in -4 < a < 11, leading to the conclusion that 16 < a² < 121. However, the correct interpretation shows that squaring the inequality gives 0 ≤ a² < 121, which is a more accurate representation of the range for a². The key takeaway is that squaring the boundaries of a must consider the non-negativity of a², resulting in the final inequality. This highlights the importance of correctly applying mathematical operations to inequalities.
kc1895
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Here is a basic inequality question for which I cannot understand the answer:

If $$-1<a-b<10 ,and -3\le b\le1$$ then what inequality represents the range of values of a2?

I plug-in -3 and 1 for b for boundaries and get -4<a<11.
Since the boundary is for a2, the range would be 16<a2<121.

But why would this be incorrect?

Thanks for your help!
 
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Given that:

$$-4<a<11$$

this implies:

$$0\le|a|<11$$

Now, using:

$$|x|\equiv\sqrt{x^2}$$

we may write:

$$0\le\sqrt{a^2}<11$$

And so squaring through the compound inequality, we obtain:

$$0\le a^2<121$$
 
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Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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