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anonymous

  • 5 years ago

how do i factor: 14x^2-47x+7=0

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  1. anonymous
    • 5 years ago
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    You don't.

  2. anonymous
    • 5 years ago
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    oh.. ok well my original problem i needed to solve was -42x^2+141x=-21 and i thought i was on the right track... maybe not?

  3. anonymous
    • 5 years ago
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    find delta and factor using the solutions

  4. anonymous
    • 5 years ago
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    what is delta?

  5. anonymous
    • 5 years ago
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    The only thing that will make 14x^2 is (7x )*(2x ). Now, the only factors that give the 7 is 7 and 1. With this knowledge you get (7x-1)*(2x-7)

  6. anonymous
    • 5 years ago
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    Wrong; is it possible that taking out 3 was all you were meant to do? Because it can;t go any further.

  7. anonymous
    • 5 years ago
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    hmm.. maybe but the example i followed came up with two terms which then i put to the zero factor property... but maybe this is as far as this one goes.

  8. anonymous
    • 5 years ago
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    what about the p and q thing?

  9. anonymous
    • 5 years ago
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    I put it through the answer check and it said that 14x^2-47x+7 was wrong

  10. anonymous
    • 5 years ago
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    it said the methods to solving were either the quadratic formula or by completing the square.

  11. anonymous
    • 5 years ago
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    \[\sqrt{47^2-4 \times 7 \times 14} \notin \mathbb{Z^+}\] therefore it won't factorise Completing the square doesn't look too friendly on this one; just go with the formula

  12. anonymous
    • 5 years ago
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    \[x = \frac{47 \pm \sqrt{1817}}{28}\] ...dunoo how that will go in your 'answer check'

  13. anonymous
    • 5 years ago
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    haha yeah.. probably not that well.. but thanks for the help @INewton

  14. anonymous
    • 5 years ago
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    just an update... apparently 14x^2-47x+7=) factors into (7x+1)(2x-7)... which in turn would make the real number solutions: -1/7 and 7/2

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