If \(554_b\) is the base \(b\) representation of the square of the number whose base \(b\) representation is \(24_b,\) then find \(b\).

- h0pe

- jamiebookeater

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- myininaya

wait so it kind of looks like it is saying this:
\[554_b=(24_b) \cdot (24_b)\]

- myininaya

and of course we are to find the b there

- h0pe

I think that looks about right

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## More answers

- myininaya

\[5 \cdot b^2+ 5 \cdot b+4=(2 \cdot b+4)(2 \cdot b +4)\]

- myininaya

this looks like a quadratic equation

- h0pe

so then \((2b+4)(2b+4)\) is \(4b^2+8b+8\) right?

- myininaya

well not exactly 4(4)=16
and 2b(4)+4(2b)=16b

- myininaya

check that middle term and last term you have

- h0pe

They are 16s whoops \(4b2+16b+16\)

- h0pe

\(4b^2\)

- myininaya

right so you have
\[5b^2+5b+4=4b^2+16b+16\]

- myininaya

you need to some subtraction on both sides

- h0pe

okay

- h0pe

so \(b^2=11b+12\)

- myininaya

\[\text{ or } b^2-11b-12=0\]

- myininaya

which you can factor the left hand side

- myininaya

or you can use the quadratic formula if you really want to
but factoring the left hand expression is not too bad

- myininaya

you will get one b that makes sense
and the other b that makes no sense

- h0pe

what do you mean by factoring? sorry it's been a while since I've used that. 0.0

- myininaya

like
to factor something like x^2-5x+6
you can look for two numbers that multiply to be 6 and add up to be -5
so -2(-3)=-6 and -2+(-3)=-5
so x^2-5x+6 can be written as (x-2)(x-3)
and if you wanted to solve x^2-5x+6=0
you can write this first as (x-2)(x-3)=0
then set both factors equal to zero
x-2=0 or x-3=0
so x=2 or x=3
you can find two numbers that multiply to be -12 and add up to be -11

- h0pe

Right, I remember now

- h0pe

For the case of \(b^2âˆ’11bâˆ’12=0\) it would be -12 and 1, wouldn't it be?
So the equation would be \((b-12)(b+1)\) right?

- myininaya

bingo so you have to solve (b-12)(b+1)=0
which means you have either b-12=0 or b+1=0 (or both <--both will not work in this case and I hope you will see why )

- h0pe

The answer has to be \(b=12\), because there aren't negative bases.

- myininaya

cool stuff :)

- h0pe

Thanks :)

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