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calculusxy

  • one year ago

would 3k^2 + 5k + 7 = 0 have no solutions? if so, why?

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  1. calculusxy
    • one year ago
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    @Lily2913

  2. anonymous
    • one year ago
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    k =(5-√-59)/6=(5-i√ 59 )/6= 0.8333-1.2802i

  3. calculusxy
    • one year ago
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    i don't understand

  4. anonymous
    • one year ago
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    Raise k to the 2nd power

  5. Nnesha
    • one year ago
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    check the discriminant \(\huge\color{reD}{\rm b^2-4ac}\) `Discriminant` you can use this to find if the equation is factorable or not if ` b^2-4ac > 0` then there are 2 real zeros if ` b^2-4ac = 0` then there is one real root if ` b^2-4ac < 0` then you will get two complex roots (no -x-intercept)

  6. anonymous
    • one year ago
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    ((3 • k2) - 5k) + 7 = 0

  7. anonymous
    • one year ago
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    Simplify 3k2-5k + 7 Trying to factor by splitting the middle term 2.1 Factoring 3k2-5k+7 The first term is, 3k2 its coefficient is 3 . The middle term is, -5k its coefficient is -5 . The last term, "the constant", is +7 Step-1 : Multiply the coefficient of the first term by the constant 3 • 7 = 21 Step-2 : Find two factors of 21 whose sum equals the coefficient of the middle term, which is -5 . -21 + -1 = -22 -7 + -3 = -10 -3 + -7 = -10 -1 + -21 = -22 1 + 21 = 22 3 + 7 = 10 7 + 3 = 10 21 + 1 = 22 Observation : No two such factors can be found !! Conclusion : Trinomial can not be factored Equation at the end of step 2 : 3k2 - 5k + 7 = 0 Step 3 : Solve 3k2-5k+7 = 0

  8. imqwerty
    • one year ago
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    if u got a quadratic equation of the form-\[ak^2+bk+c=0\]then the roots are given by-\[k=\frac{ -b \pm \sqrt{b^2-4\times a \times c} }{ 2a }\] if the square root part is less than 0 i.e., it is negative will mean that the square root cannot be operated and hence there are no real roots try to calculate the square root part :)

  9. calculusxy
    • one year ago
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    so since i get 109 after finding b^2 - 4ac i got 109

  10. Nnesha
    • one year ago
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    no that's not correct b^2 -4ac isn't equal to 109

  11. calculusxy
    • one year ago
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    -59?

  12. imqwerty
    • one year ago
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    yes :)

  13. Nnesha
    • one year ago
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    looks right so it's negative b^2-4ac < 0 so no x-intercepts you will get complex roots

  14. calculusxy
    • one year ago
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    so since -59 is less than 0 it't not factorable?

  15. Nnesha
    • one year ago
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    right if the discriminant is less than 0 then no

  16. anonymous
    • one year ago
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    unless you are able to use complex numbers, if not then no.

  17. calculusxy
    • one year ago
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    thx to all :)

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