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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0k =(5√59)/6=(5i√ 59 )/6= 0.83331.2802i

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Raise k to the 2nd power

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0check the discriminant \(\huge\color{reD}{\rm b^24ac}\) `Discriminant` you can use this to find if the equation is factorable or not if ` b^24ac > 0` then there are 2 real zeros if ` b^24ac = 0` then there is one real root if ` b^24ac < 0` then you will get two complex roots (no xintercept)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0((3 • k2)  5k) + 7 = 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Simplify 3k25k + 7 Trying to factor by splitting the middle term 2.1 Factoring 3k25k+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 Step1 : Multiply the coefficient of the first term by the constant 3 • 7 = 21 Step2 : 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 3k25k+7 = 0

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1if u got a quadratic equation of the form\[ak^2+bk+c=0\]then the roots are given by\[k=\frac{ b \pm \sqrt{b^24\times a \times c} }{ 2a }\] 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 :)

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0so since i get 109 after finding b^2  4ac i got 109

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0no that's not correct b^2 4ac isn't equal to 109

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0looks right so it's negative b^24ac < 0 so no xintercepts you will get complex roots

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0so since 59 is less than 0 it't not factorable?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0right if the discriminant is less than 0 then no

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0unless you are able to use complex numbers, if not then no.
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