Question 1177575: The quadratic equation x² + kx + 36=0 has
two different real roots. Find the set of possible values of k.
Found 3 solutions by greenestamps, Edwin McCravy, ikleyn:
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!
The product of a and b has to be 36.
Choose any two numbers a and b whose product is 36; the sum (a+b) will be one of the possible values for k.
One example: a=9, b=4; ab=36; a+b=13. One possible value of k is 13.
One more: a=-3, b=-12; ab=36; a+b=-15. Another possible value of k is -15.
Note the statement of the problem is not complete; with no restriction on the values of a and b, there are an infinite number of possible values of k. For example...
a=8, b=4.5; ab=36; a+b=12.5; another possible value of k is 12.5.
From that example, you can see that you could choose ANY value for a; then b would be 36/a, and you would get another value for a+b.
So undoubtedly the question was supposed to ask for the set of possible INTEGER values of k.
I'll let you finish the problem; you should find 18 different possible integer values for k.
Note that two of those 18 values of k will be 12 and -12, where the a and b are both 6 or both -6. Since the problem says the roots are different, you should find 16 different integer values of k in which the two roots are different.
Answer by Edwin McCravy(20054)  (Show Source):
You can put this solution on YOUR website!
Let the roots be p and q.
Then p+q = -k and pq=36
Solve the 2nd equation for p and substitute in the first and you'll get:
 and 
Pick any nonzero number for one root q, then the other root p will be 36/q,
and k will be . Here are some possible values.
if q = -10, then p = -18/5 and k = 68/5
if q = -9, then p = -4 and k = 13
if q = -8, then p = -9/2 and k = 25/2
if q = -7, then p = -36/7 and k = 85/7
if q = -6, then p = -6 and k = 12
if q = -5, then p = -36/5 and k = 25/2
if q = -4, then p = -9 and k = 13
if q = -3, then p = -12 and k = 15
if q = -2, then p = -18 and k = 20
if q = -1, then p = -36 and k = 37
if q = 1, then p = 36 and k = -37
if q = 2, then p = 18 and k = -20
if q = 3, then p = 12 and k = -15
if q = 4, then p = 9 and k = -13
if q = 5, then p = 7.2 and k = 25/2
if q = 6, then p = 6 and k = -12
if q = 7, then p = 36/7 and k = -85/7
if q = 8, then p = 4.5 and k = -25/2
if q = 9, then p = 4 and k = -13
if q = 10, then p = 18.5 and k = -137/2
I lined through 2 of them because the roots are not different in those.
Edwin
Answer by ikleyn(52776)  (Show Source):
You can put this solution on YOUR website! .
I will show you a STANDARD and very SIMPLE WAY,
on how this problem (and many other similar problems) should be treated.
Since the problem talks about different real roots of a quadratic equation, it means that its discriminant is positive.
The discriminant is b^2 - 4ac = k^2 - 4*36 = k^2 - 144.
So, the discriminant must be positive
k^2 - 144 > 0.
It implies
k^2 > 144,
which means that EITHER k > 12 OR k < - 12.
ANSWER. The set of all possible values of k is { k | k > 12 or k < - 12 }.
It is the union of two infinite semi-intervals (-oo,12) U (12,oo).
Solved.
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