Question

Question

Answer 1

If sinA and cos A are roots of the equation:\[x^2 + \sqrt{3}x +k=0\]
then what is k

Answer 2

@jim_thompson5910

Answer 3

If k and m are roots of a polynomial, then we know

x = k, x = m
x-k = 0, x-m = 0
(x-k)(x-m) = 0
x^2 - mx - kx + km = 0
x^2 - (m+k)x + km = 0

Answer 4

notice how we have x^2 - (m+k)x + km = 0

k and m are the roots. We have m+k and km

km is the product of the roots
m+k is the sum of the roots

Answer 5

If sinA and cos A are roots of the equation

\[\Large x^2 + \sqrt{3}x +k=0\]

then k has to be equal to the product of the roots (as shown above)

Answer 6

so k = sinAxcosA

Answer 7

yes, and you can optionally use a trig identity to get

k = sin(A)*cos(A) = (1/2)*sin(2A)

I used the second one on this page

http://www.sosmath.com/trig/douangl/douangl.html

Answer 8

I understand but how will I find the numerical value of k

Answer 9

\[
\Large x^2 + \sqrt{3}x \ \ \ \ \ \ \ +k=0\\
\Large x^2 - (p+q)x+pq = 0
\]

\[
\Large x^2 + \sqrt{3}x \ \ \ \ \ \ \ +k=0\\
\Large x^2 - (\sin(A)+\cos(A))x+\sin(A)*\cos(A) = 0
\]
-------------------------------------------------------
so we can see that
\[\Large - (\sin(A)+\cos(A))=\sqrt{3}\]
use this new equation to find the value of A

Answer 10

sorry I cant solve it, can u walk me through it

Answer 11

I don't know if there is a way to solve this by hand

Answer 12

so I think the only method is to use technology like a graphing calculator

Answer 13

you have ended up with two equations:\[\sin(A)+\cos(A)=-\sqrt{3}\tag{1}\]\[\sin(A)\cos(A)=k\tag{2}\]Now, if we square equation (1) we get:\[\sin^2(A)+\cos^2(A)+2\sin(A)\cos(A)=3\]\[\therefore1+2\sin(A)\cos(A)=3\tag{3}\]Now just substitute (2) into (3) and you are done

Answer 14

@asnaseer can u type it again I cant read it