
Kindly correct my 10a and why 10b answer k shows different sign?
Kindly correct my 10a and why 10b answer k shows different sign?
The concept of D is incorrect. To check for positive or negative definite quadratic function, always check D < 0.
Quadratic function is always negative if:
Quadratic function is always positive if:
10 (b) is using the concept in green.
So there is a sign difference in your working only.
From this step:
-3k2 + 12 < 0 (multiply by -1/3)
k2 – 4 > 0
(k + 2) (k – 2) > 0 ———–|———–|———–
-2 2
For the expression to be > 0, the solution should be found in the red part on the number line, meaning k < -2 or k > 2
Since a > 0 => k+ 2 > 0 => k > -2, so reject k < -2
The solution for k is k > 2
So is my answer for 10a is ok?
To check for real roots, only need to determine if D > 0 so condition 2 to delete because not necessary for checking real roots(that is only required for checking positive definite)
Based on your working of D,
k2 – 8k + 16 = (k-4)2
This means that for all values of k, (k-4)2 > 0
Hence (k-2)x2 + kx + 2 = 0 has real roots for all values of k.
yes, i am really confused sometime, but thanks again for your this explanation. 🙂 cheers.