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msingh
Group Title
SOLVE
if alpha and beta are zeroes of x^2+7x+K. find value of k
 one year ago
 one year ago
msingh Group Title
SOLVE if alpha and beta are zeroes of x^2+7x+K. find value of k
 one year ago
 one year ago

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hartnn Group TitleBest ResponseYou've already chosen the best response.3
anything else given for alpha and beta ? or you need k in terms of alpha and beta ?
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.1
Hint: \((x\alpha)(x\beta)=x^2+7x+K\)
 one year ago

msingh Group TitleBest ResponseYou've already chosen the best response.0
question has not sufficient data
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
well, product of roots(zeros) for ax^2+bx+c =0 is just c/a so here, k/1 = alpha*beta , gives k=alpha*beta
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.1
\(\alpha + \beta = 7\) and \(\alpha \times \beta = k\)
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.1
\((x\alpha)(x\beta)=x^2(\alpha+\beta)x+\alpha\beta=x^2+7x+K\)\) can you finish this
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
still, you'll not get a value for k, you'll be able to express k in terms of either alpha or beta.
 one year ago

msingh Group TitleBest ResponseYou've already chosen the best response.0
@hartnn is absolutely correct
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
so is @myko
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.1
I am trying to say something but checking my method first. It seems somewhat different .
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.1
\((\alpha + \beta)^2 = 49\) \(\alpha^2 + \beta^2 + 2(k) = 49\) \(\alpha^2 + \beta^2 = 492k \) Now find : \((\alpha  \beta)^2 \) = \(\alpha^2 + \beta^2  2k\) = \(49  4k\) . So \(\alpha + \beta = 7\) and \(\alpha  \beta = \sqrt{494k}\)
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.1
2 equations 3 unknowns.....
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.1
\(2\alpha = 7 + \sqrt{494k}\) \(2\alpha + 7 = \sqrt{494k}\) \((2\alpha + 7)^2 = 494k\) \(4\alpha^2 + 49 + 28 \alpha = 49 4k\) \(4\alpha^2 + \cancel{49} + 28\alpha = \cancel{49}  4k\) \(4\alpha (\alpha + 7) = 4k\) \(k = \cfrac{\cancel{4}^{1}\alpha(\alpha+7)}{\cancel{4}^1} \) \(k = \alpha^2 7\alpha \)
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.1
@myko yes but after simplifying we get much better answer for k in terms of only one root \(\alpha\) .
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
you could have got that by using : k= x^27x since, alpha (or beta) is a root, x= alpha k= alpha^27*alpha.
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.1
I just presented my method . Though, there may be some arguments like its longer method or something else . But I just wanted to let the asker know about this method also. There may be many other methods to arrive the same answer that I given above, but if all the methods are known to a person, then it is better than knowing only 1 method , as per my opinion. Oh yeah, right Hartnn.
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.1
I don't see why it is better or worse... It is exactly same amount of info
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
agreed @mathslover
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.1
Thank you Hartnn .
 one year ago
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