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

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

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

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

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

myko
 one year ago
Best 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

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

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

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

mathslover
 one year ago
Best 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 \(\alpha  \beta = \sqrt{494k}\)

myko
 one year ago
Best ResponseYou've already chosen the best response.12 equations 3 unknowns.....

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

mathslover
 one year ago
Best 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 \(\alpha\) .

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

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1I 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.

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