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anonymous
 3 years ago
SOLVE
if alpha and beta are zeroes of x^2+7x+K. find value of k
anonymous
 3 years ago
SOLVE if alpha and beta are zeroes of x^2+7x+K. find value of k

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hartnn
 3 years 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 ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hint: \((x\alpha)(x\beta)=x^2+7x+K\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0question has not sufficient data

hartnn
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.1\(\alpha + \beta = 7\) and \(\alpha \times \beta = k\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\((x\alpha)(x\beta)=x^2(\alpha+\beta)x+\alpha\beta=x^2+7x+K\)\) can you finish this

hartnn
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@hartnn is absolutely correct

mathslover
 3 years 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
 3 years 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 \(\alpha  \beta = \sqrt{494k}\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.02 equations 3 unknowns.....

mathslover
 3 years 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
 3 years 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
 3 years 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
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I don't see why it is better or worse... It is exactly same amount of info
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