h0pe
  • h0pe
If \(554_b\) is the base \(b\) representation of the square of the number whose base \(b\) representation is \(24_b,\) then find \(b\).
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
myininaya
  • myininaya
wait so it kind of looks like it is saying this: \[554_b=(24_b) \cdot (24_b)\]
myininaya
  • myininaya
and of course we are to find the b there
h0pe
  • h0pe
I think that looks about right

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

myininaya
  • myininaya
\[5 \cdot b^2+ 5 \cdot b+4=(2 \cdot b+4)(2 \cdot b +4)\]
myininaya
  • myininaya
this looks like a quadratic equation
h0pe
  • h0pe
so then \((2b+4)(2b+4)\) is \(4b^2+8b+8\) right?
myininaya
  • myininaya
well not exactly 4(4)=16 and 2b(4)+4(2b)=16b
myininaya
  • myininaya
check that middle term and last term you have
h0pe
  • h0pe
They are 16s whoops \(4b2+16b+16\)
h0pe
  • h0pe
\(4b^2\)
myininaya
  • myininaya
right so you have \[5b^2+5b+4=4b^2+16b+16\]
myininaya
  • myininaya
you need to some subtraction on both sides
h0pe
  • h0pe
okay
h0pe
  • h0pe
so \(b^2=11b+12\)
myininaya
  • myininaya
\[\text{ or } b^2-11b-12=0\]
myininaya
  • myininaya
which you can factor the left hand side
myininaya
  • myininaya
or you can use the quadratic formula if you really want to but factoring the left hand expression is not too bad
myininaya
  • myininaya
you will get one b that makes sense and the other b that makes no sense
h0pe
  • h0pe
what do you mean by factoring? sorry it's been a while since I've used that. 0.0
myininaya
  • 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
  • h0pe
Right, I remember now
h0pe
  • h0pe
For the case of \(b^2−11b−12=0\) it would be -12 and 1, wouldn't it be? So the equation would be \((b-12)(b+1)\) right?
myininaya
  • 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
  • h0pe
The answer has to be \(b=12\), because there aren't negative bases.
myininaya
  • myininaya
cool stuff :)
h0pe
  • h0pe
Thanks :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.