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$$2^{y}+2(3^{y}) > 3(4^{y})$$ and $$y=3x^2+2x-2$$

Which of the following is a possible value of $x$?

$A -1.5 $

$B-2.5 $

$C -0.5 $

$D+0.7 $

$E+1.2$

$********$

I could just conclude from the $1st$ inequality that $$y∊(-∞, 0)$$ Should I put values of $y$ e.g. $-1, -2$ and then check or is there some other way?

I couldn't approach further. Please help.

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$2^y-4^y>2(4^y-3^y)$ => RHS>0 and LHS <0, for y>0. Clearly, y=0 is not a solution –  lab bhattacharjee Jul 23 '12 at 6:39
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2 Answers

up vote 1 down vote accepted

The roots of the quadratic are $\frac{-2\pm\sqrt{28}}{6}$. So $y$ is negative only for $\frac{-2-\sqrt{28}}{6}\lt x \lt \frac{-2+\sqrt{28}}{6}$. The smaller root is about $-1.2$, and the larger is about $0.5$.

Combine this with your observation that $y \le 0$, and note that only one of the given values of $x$ lies in the right range.

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Option $C$ sir if I get you right. –  TheApe Jul 23 '12 at 7:06
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@TheApe: Yes, exactly. –  André Nicolas Jul 23 '12 at 7:08
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I'd say there's no clever thought required. You've got five possible values for $x$. Each of these five values gives you a unique value for $y$. Try each of these five values in the original inequality, and see which values make it true and which make it false. If you have a decent calculator, you should be able to check each of the five possibilities in about 30 seconds apiece; so you'll have the answer in under 3 minutes.

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