The equation x^2 + (k-5)x + 1 = 0 has equal roots. Determine the possible values of k.

So the first part of this question is regarding your knowledge and understanding on the subject matter. You should know that for real and equal roots, (b2 - 4ac)=0. (For real and unequal roots, it is >0 and for non-real roots, it is <0.)
Once we know this, we can find the values of a, b and c using the equation. Where a =1, b=(k-5) and c=1.Next, we need to solve for (b2 - 4ac)=0.
(k-5)2- (411) = 0 (k-5)2- 4 = 0
The next step is to expand the brackets:
k2 - 10k +25 - 4 = 0k2 - 10k +21 = 0
Finally, we are left with an equation which can be factored to find the values of k, as follows:
(k-7) (k-3) = 0
Therefore, k = 7 or k =3.



Related Maths Scottish Highers answers

All answers ▸

The line, L, makes an angle of 30 degrees with the positive direction of the x-axis. Find the equation of the line perpendicular to L, passing through (0,-4).


y=x^3-3x^2+2x+5 a)Write down the coordinates of P the point where the curve crosses the x-axis. b)Determine the equation of the tangent to the curve at P. c)Find the coordinates of Q, the point where this tangent meets the curve again.


Express '2x^2 + 8x + 30' in the form 'a(x+b)^2 + c'


Given that dy/dx = 6x*2 - 3x + 4 And y =14 when x=2. Express y in terms of x