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anonymous

  • 5 years ago

Does anyone understand the zero product principle for quadric equations?

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  1. anonymous
    • 5 years ago
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    \[2m ^{2} + 10m=-12\]

  2. anonymous
    • 5 years ago
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    factorize the eq. then equate both to zero. you will get the two roots

  3. anonymous
    • 5 years ago
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    Well it goes like this First rearrange the eqn as follows: \[2m ^{2} + 10m + 12 = 0\]

  4. anonymous
    • 5 years ago
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    or the formula for fractions verse the equations without fractions would help. I have the information mixed up--the (x+a)(x+b)=0 is all that is stated

  5. anonymous
    • 5 years ago
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    Now the product of the coefficient of m^2 and the constant terms +12 is 24 We have to find factors p and q of 24 so that pq = 24 and p + q = 10

  6. anonymous
    • 5 years ago
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    Can you give me the value of p and q ??

  7. anonymous
    • 5 years ago
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    ???????

  8. anonymous
    • 5 years ago
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    6 and 4

  9. anonymous
    • 5 years ago
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    would you happen to know the formulas?

  10. anonymous
    • 5 years ago
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    Absolutely correct !!!! Ok If we take p as 6 and q as 4 pq = 6 x 4 = 24 and 6+ 4 = 10 So, now we can rewrite the equation as \[2m ^{2} + 6m + 4m + 24 = 0\]

  11. anonymous
    • 5 years ago
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    I'm solving it step by step for u

  12. anonymous
    • 5 years ago
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    Sorry, I made a mistake with the last eqn !!! It will be \[2m ^{2} + 6m + 4m + 12 = 0\]

  13. anonymous
    • 5 years ago
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    Now we take out common factors as follows: \[2m(m + 3) +4(m + 3) = 0\]

  14. anonymous
    • 5 years ago
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    Now (m + 3) is common and so we can rewrite as follows: \[(m + 3) (2m + 4) = 0\]

  15. anonymous
    • 5 years ago
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    Got it till here????

  16. anonymous
    • 5 years ago
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    Now according to the zero product principal, a x b = 0 if either a=0 or b=0 So (m+3)(2m+4) = 0 means that either m+3 = 0 ie. m = -3 or 2m +4 = 0 ie 2m = -4 ie m = -4/2 ie m = -2

  17. anonymous
    • 5 years ago
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    Hope it is clear now???

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