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anonymous

  • 5 years ago

how do i get my quadratic answer in decimal form

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  1. amistre64
    • 5 years ago
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    divide......

  2. amistre64
    • 5 years ago
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    or do you mean that sqrt part?

  3. anonymous
    • 5 years ago
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    \[33\pm \sqrt{65}\over 32\]

  4. amistre64
    • 5 years ago
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    split it up int0 its parts: 33/32 + sqrt(65)/32 33/32 - sqrt(65)/32 .969 ----------- 33 | 32.00000 - 297 ---- 230 - 198 ---- 320 - 297 ----- 230 <- this is repeating so we use this symbol -- .96 that is .96 with a bar above it to indicate that it is a repeating decimal

  5. amistre64
    • 5 years ago
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    the trick to doing the sqrt part is really to know that x^2 + 2xy + y^2 is the key to your results. We we seperate the number up into groups of 2 starting at the decimal and going in both directions, for example: 12345.654 becomes 1 23 45. 65 40 00 00 00

  6. amistre64
    • 5 years ago
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    your sqert(65) is already in a group of "2" so we have: ------------------ \/65. 00 00 00 00 00

  7. amistre64
    • 5 years ago
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    now we need to find a number for "x^2" that is close to but less than the first grouping; in this case close to but less than 65. 8^2 = 64; so lets use that: 8. ------------------ \/65. 00 00 00 00 00 -64 -------- 1 00 we have our "x^2, now we need to "2x" and find "y" to add to this. 8. ------------------ \/65. 00 00 00 00 00 -64 -------- 2x+y| 1 00 16 __ | 0 is the only thing we can use for y because 160 is greater than 100 and we cant use anything else.

  8. amistre64
    • 5 years ago
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    8. 0 y ------------------ \/65. 00 00 00 00 00 -64 -------- 160 | 1 00 ------- 160+ y | 1 00 00 and continue

  9. anonymous
    • 5 years ago
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    sweet, thanks!

  10. amistre64
    • 5 years ago
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    its either that or use a calculator :)

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