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msingh
SOLVE if alpha and beta are zeroes of x^2+7x+K. find value of k
anything else given for alpha and beta ? or you need k in terms of alpha and beta ?
Hint: \((x-\alpha)(x-\beta)=x^2+7x+K\)
question has not sufficient data
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
\(\alpha + \beta = -7\) and \(\alpha \times \beta = k\)
\((x-\alpha)(x-\beta)=x^2-(\alpha+\beta)x+\alpha\beta=x^2+7x+K\)\) can you finish this
still, you'll not get a value for k, you'll be able to express k in terms of either alpha or beta.
@hartnn is absolutely correct
I am trying to say something but checking my method first. It seems somewhat different .
\((\alpha + \beta)^2 = 49\) \(\alpha^2 + \beta^2 + 2(k) = 49\) \(\alpha^2 + \beta^2 = 49-2k \) Now find : \((\alpha - \beta)^2 \) = \(\alpha^2 + \beta^2 - 2k\) = \(49 - 4k\) . So \(\alpha + \beta = 7\) and \(\alpha - \beta = \sqrt{49-4k}\)
2 equations 3 unknowns.....
\(2\alpha = -7 + \sqrt{49-4k}\) \(2\alpha + 7 = \sqrt{49-4k}\) \((2\alpha + 7)^2 = 49-4k\) \(4\alpha^2 + 49 + 28 \alpha = 49 -4k\) \(4\alpha^2 + \cancel{49} + 28\alpha = \cancel{49} - 4k\) \(4\alpha (\alpha + 7) = -4k\) \(k = \cfrac{\cancel{4}^{-1}\alpha(\alpha+7)}{\cancel{4}^1} \) \(k = -\alpha^2 -7\alpha \)
@myko yes but after simplifying we get much better answer for k in terms of only one root \(\alpha\) .
you could have got that by using : k= -x^2-7x since, alpha (or beta) is a root, x= alpha k= -alpha^2-7*alpha.
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.
I don't see why it is better or worse... It is exactly same amount of info