$$213_x=139_{10}$$
$$21_x=1021_4$$
How would I solve this
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What do quadratic forms have to do with this? – Marc van Leeuwen Mar 11 '13 at 15:16
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1You might want to consider accepting an answer to your questions: do accept an answer (one per question), you simply click on the $\checkmark$ to the left of the answer you'd like to accept. Plus you get 2 reputation points for each answer you accept. When you get 30+ rep, you can upvote as many answers as you'd like. – amWhy Mar 11 '13 at 17:10
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I'm not sure wether i should reply or not since this smells like homework; anyway, just think what base x means:
$213_x = 139_{10} \Rightarrow 2x^2+x+3 = 139$
$21_x = 1021_{4} \Rightarrow 2x + 1 = 4^3 + 2\cdot 4 + 1$
Now solve the 2 equations and you're done
PS : on the right side of the $\Rightarrow$ , every number is assumed to be $\text{base 10}$
kaharas
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$$213_x=139_{10}$$ $$2\cdot x^2+1\cdot x^1+3\cdot x^0=1\cdot 10^2+3\cdot 10^1+9\cdot10^0$$ $$2x^2+x+3=139$$ $$2x^2+x-136=0\Rightarrow x_{1,2}=\frac{-1\pm\sqrt{1+8\cdot 136}}{4}=\frac{-1\pm33}{4}$$ $$x_1=8$$ the base $$x_2=-17/2$$ is not a correct base see comments below.
similarly $$21_x=1021_4$$ $$2x+1=4^3+2\cdot 4 +1=73\Rightarrow x=36$$
Adi Dani
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Bases are normally positive integers only, sometimes negative integers make sense. Rationals I've never seen. – vonbrand Mar 11 '13 at 18:03
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you never seen, me too but that does not mean that they do not exist, the example above shows that they exist. Can you give the reason that such bases are wrong to use. – Adi Dani Mar 11 '13 at 18:13
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