Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

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

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

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

hartnnBest ResponseYou've already chosen the best response.3
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
 one year ago

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

mykoBest 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
 one year ago

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

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

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

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

mykoBest ResponseYou've already chosen the best response.1
2 equations 3 unknowns.....
 one year ago

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

mathsloverBest 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 \(\alpha\) .
 one year ago

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

mathsloverBest ResponseYou've already chosen the best response.1
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.
 one year ago

mykoBest ResponseYou've already chosen the best response.1
I don't see why it is better or worse... It is exactly same amount of info
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.