Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

See more answers at
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.

Join Brainly to access

this expert answer


To see the expert answer you'll need to create a free account at Brainly

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

it has
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.
good enough
@hartnn is absolutely correct
so is @myko
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
Thank you Hartnn .

Not the answer you are looking for?

Search for more explanations.

Ask your own question