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h0pe

  • one year ago

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

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  1. myininaya
    • one year ago
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    wait so it kind of looks like it is saying this: \[554_b=(24_b) \cdot (24_b)\]

  2. myininaya
    • one year ago
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    and of course we are to find the b there

  3. h0pe
    • one year ago
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    I think that looks about right

  4. myininaya
    • one year ago
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    \[5 \cdot b^2+ 5 \cdot b+4=(2 \cdot b+4)(2 \cdot b +4)\]

  5. myininaya
    • one year ago
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    this looks like a quadratic equation

  6. h0pe
    • one year ago
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    so then \((2b+4)(2b+4)\) is \(4b^2+8b+8\) right?

  7. myininaya
    • one year ago
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    well not exactly 4(4)=16 and 2b(4)+4(2b)=16b

  8. myininaya
    • one year ago
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    check that middle term and last term you have

  9. h0pe
    • one year ago
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    They are 16s whoops \(4b2+16b+16\)

  10. h0pe
    • one year ago
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    \(4b^2\)

  11. myininaya
    • one year ago
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    right so you have \[5b^2+5b+4=4b^2+16b+16\]

  12. myininaya
    • one year ago
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    you need to some subtraction on both sides

  13. h0pe
    • one year ago
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    okay

  14. h0pe
    • one year ago
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    so \(b^2=11b+12\)

  15. myininaya
    • one year ago
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    \[\text{ or } b^2-11b-12=0\]

  16. myininaya
    • one year ago
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    which you can factor the left hand side

  17. myininaya
    • one year ago
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    or you can use the quadratic formula if you really want to but factoring the left hand expression is not too bad

  18. myininaya
    • one year ago
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    you will get one b that makes sense and the other b that makes no sense

  19. h0pe
    • one year ago
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    what do you mean by factoring? sorry it's been a while since I've used that. 0.0

  20. myininaya
    • one year ago
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    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

  21. h0pe
    • one year ago
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    Right, I remember now

  22. h0pe
    • one year ago
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    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?

  23. myininaya
    • one year ago
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    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 )

  24. h0pe
    • one year ago
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    The answer has to be \(b=12\), because there aren't negative bases.

  25. myininaya
    • one year ago
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    cool stuff :)

  26. h0pe
    • one year ago
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    Thanks :)

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