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Could someone please outline their steps in solving: $$2^{2x+1} -5\cdot2^x +2 = 0$$ The answer is $x = \pm1$, which I have found through trial and error, but I would like to find algebraically.

Cheese Cake
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TB2
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    Hint: Try letting $y = 2^x$. – Moo Mar 05 '17 at 03:05
  • Using Moo's hint, what is the natural thing to do with $2^{2x+1}$? – quasi Mar 05 '17 at 03:11
  • See http://math.stackexchange.com/questions/2169968/sum-of-reciprocal-of-the-positive-divisors-of-1800/2169975#2169975 and http://math.stackexchange.com/questions/384090/find-all-real-numbers-x-for-which-frac8x27x12x18x-frac76 – lab bhattacharjee Mar 05 '17 at 03:49

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$$2^{2x+1} -5\cdot2^x +2 = 0$$

$$ 2^{2x}2^{1} - 5 \cdot 2^x + 2=0$$

Let $t=2^x$

$$ 2t^2-5t+2=0$$

$$ (2t-1)(t-2) =0$$

$$ t=\frac{1}{2} ,~~t=2$$

$$ 2^x=\frac{1}{2}=2^{-1} . ~~ 2^x=2 $$

$$ x=? , ~~ x=?$$

bigfocalchord
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