SOLUTION: Find the constant k such that the quadratic 2x^2 + 3x - 20x + 3x^2 + k has a double root.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Find the constant k such that the quadratic 2x^2 + 3x - 20x + 3x^2 + k has a double root.      Log On


   



Question 1209248: Find the constant k such that the quadratic 2x^2 + 3x - 20x + 3x^2 + k has a double root.
Found 2 solutions by Edwin McCravy, mccravyedwin:
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!

2x%5E2+%2B+3x+-+20x+%2B+3x%5E2+%2B+k

5x%5E2+-+17x+%2B+k

It will have a double root if its discriminant b2-4ac = 0

discriminant+=+%28-17%29%5E2-4%285%29%28k%29+=+0
289-20k+=+0
-20k=-289
k=%28-289%29%2F%28-20%29
k=289%2F20    <-- answer

Checking:
2x%5E2+%2B+3x+-+20x+%2B+3x%5E2+%2B+289%2F20=0
5x%5E2+-+17x+%2B+289%2F20=0
100x%5E2-340x%2B289=0
%2810x-17%29%2810x-17%29=0
10x-17=0;  10x-17=0
   10x=17;    10x=17
     x=17%2F10;     x=17%2F10

So both roots are the same, so it has a double root if k=289%2F20

So k=289%2F20 is correct.

Edwin

Answer by mccravyedwin(407) About Me  (Show Source):
You can put this solution on YOUR website!
You could also do it this way:

2x%5E2+%2B+3x+-+20x+%2B+3x%5E2+%2B+k

5x%5E2+-+17x+%2B+k

That would have a double root if it would factor this way,
as the sum of the square roots of the first and last terms,
with the same sign between them as the middle term has:

%28sqrt%285%29x-sqrt%28k%29%29%5E2

5kx%5E2-2sqrt%285k%29x%2Bk%5E2

Then the middle term's coefficient would equal -17

-2sqrt%285k%29=-17

4%2A5k=289

20k=289

k=289%2F20 

Edwin