anonymous
  • anonymous
SOLVE if alpha and beta are zeroes of x^2+7x+K. find value of k
Mathematics
jamiebookeater
  • jamiebookeater
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

hartnn
  • hartnn
anything else given for alpha and beta ? or you need k in terms of alpha and beta ?
anonymous
  • anonymous
Hint: \((x-\alpha)(x-\beta)=x^2+7x+K\)
anonymous
  • anonymous
question has not sufficient data

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
it has
hartnn
  • hartnn
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
mathslover
  • mathslover
\(\alpha + \beta = -7\) and \(\alpha \times \beta = k\)
anonymous
  • anonymous
\((x-\alpha)(x-\beta)=x^2-(\alpha+\beta)x+\alpha\beta=x^2+7x+K\)\) can you finish this
hartnn
  • hartnn
still, you'll not get a value for k, you'll be able to express k in terms of either alpha or beta.
anonymous
  • anonymous
yes.
anonymous
  • anonymous
good enough
anonymous
  • anonymous
@hartnn is absolutely correct
hartnn
  • hartnn
so is @myko
mathslover
  • mathslover
I am trying to say something but checking my method first. It seems somewhat different .
mathslover
  • mathslover
\((\alpha + \beta)^2 = 49\) \(\alpha^2 + \beta^2 + 2(k) = 49\) \(\alpha^2 + \beta^2 = 49-2k \) Now find : \((\alpha - \beta)^2 \) = \(\alpha^2 + \beta^2 - 2k\) = \(49 - 4k\) . So \(\alpha + \beta = 7\) and \(\alpha - \beta = \sqrt{49-4k}\)
anonymous
  • anonymous
2 equations 3 unknowns.....
mathslover
  • mathslover
\(2\alpha = -7 + \sqrt{49-4k}\) \(2\alpha + 7 = \sqrt{49-4k}\) \((2\alpha + 7)^2 = 49-4k\) \(4\alpha^2 + 49 + 28 \alpha = 49 -4k\) \(4\alpha^2 + \cancel{49} + 28\alpha = \cancel{49} - 4k\) \(4\alpha (\alpha + 7) = -4k\) \(k = \cfrac{\cancel{4}^{-1}\alpha(\alpha+7)}{\cancel{4}^1} \) \(k = -\alpha^2 -7\alpha \)
mathslover
  • mathslover
@myko yes but after simplifying we get much better answer for k in terms of only one root \(\alpha\) .
hartnn
  • hartnn
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.
mathslover
  • mathslover
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.
anonymous
  • anonymous
I don't see why it is better or worse... It is exactly same amount of info
hartnn
  • hartnn
mathslover
  • mathslover
Thank you Hartnn .

Looking for something else?

Not the answer you are looking for? Search for more explanations.