Question

SOLVE







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

Answer 1

anything else given for alpha and beta ? or you need k in terms of alpha and beta ?

Answer 2

Hint:
\((x-\alpha)(x-\beta)=x^2+7x+K\)

Answer 3

question has not sufficient data

Answer 4

it has

Answer 5

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

Answer 6

\(\alpha + \beta = -7\) and \(\alpha \times \beta = k\)

Answer 7

\((x-\alpha)(x-\beta)=x^2-(\alpha+\beta)x+\alpha\beta=x^2+7x+K\)\)
can you finish this

Answer 8

still, you'll not get a value for k, you'll be able to express k in terms of either alpha or beta.

Answer 9

yes.

Answer 10

good enough

Answer 11

@hartnn
is absolutely correct

Answer 12

so is @myko

Answer 13

I am trying to say something but checking my method first. It seems somewhat different .

Answer 14

\((\alpha + \beta)^2 = 49\)
\(\alpha^2 + \beta^2 + 2(k) = 49\)
\(\alpha^2 + \beta^2 = 49-2k \)

Now find : \((\alpha - \beta)^2 \) = \(\alpha^2 + \beta^2 - 2k\) = \(49 - 4k\) .

So \(\alpha + \beta = 7\) and \(\alpha - \beta = \sqrt{49-4k}\)

Answer 15

2 equations 3 unknowns.....

Answer 16

\(2\alpha = -7 + \sqrt{49-4k}\)
\(2\alpha + 7 = \sqrt{49-4k}\)
\((2\alpha + 7)^2 = 49-4k\)
\(4\alpha^2 + 49 + 28 \alpha = 49 -4k\)
\(4\alpha^2 + \cancel{49} + 28\alpha = \cancel{49} - 4k\)
\(4\alpha (\alpha + 7) = -4k\)
\(k = \cfrac{\cancel{4}^{-1}\alpha(\alpha+7)}{\cancel{4}^1} \)
\(k = -\alpha^2 -7\alpha \)

Answer 17

@myko yes but after simplifying we get much better answer for k in terms of only one root \(\alpha\) .

Answer 18

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.

Answer 19

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.

Answer 20

I don't see why it is better or worse... It is exactly same amount of info

Answer 21

agreed @mathslover

Answer 22

Thank you Hartnn .