Check my working? (1 Viewer)

[Azreil]

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2006exams/pdf_doc/maths_06.pdf

Question 7c)

(c) (i) Write down the discriminant of 2x<SUP>2 +(k –2)x +8, where k is a constant.
dis = b^2-4ac
= (k-2)^2 - 4(2)(8)
= k^2- 4k - 60
(ii) Hence, or otherwise, find the values of k for which the parabola y =2x^2 +kx +9 does not intersect the line y =2x +1.
2x+1 = 2x^2+kx+9
2x^2 + (k-2)x + 8 = 0
occurs when
k^2 - 4k - 60 >= 0
(k+6)(k-10) >= 0
k =< -6 or k >= 10
therefore does not occur when -6 =< k =< 10
</SUP><SUP></SUP>
<SUP>Can anyone tell me if this is right please?</SUP>

 

[Js^-1]

Yep 100%, well done.

[Edit]: Except for the = .

 

[eskimoh]

if it doesnt intersect the line, shouldnt the discriminate just be > 0 rather than >= 0?

 

[Js^-1]

Yeh actually good point, if it was = 0 then there would be a single point of intersection.

 

[eskimoh]

yep
so everything is right except the discrim is > 0 and so i think the answer would be

k<-6 and k>10

 

[Azreil]

Thanks :]