SOLUTION: Find the range of k for which the quadratic equation x^2-kx+3k-2=0 has two different solutions, one solution is within -1< x <0 an the other solution is within 1< x <2. Let f(x

Algebra ->  Functions -> SOLUTION: Find the range of k for which the quadratic equation x^2-kx+3k-2=0 has two different solutions, one solution is within -1< x <0 an the other solution is within 1< x <2. Let f(x      Log On


   

Question 1018321: Find the range of k for which the quadratic equation x^2-kx+3k-2=0 has two different solutions, one solution is within -1< x <0 an the other solution is within 1< x <2.
Let f(x)= x^2-kx+3k-2
Please show all working out. Thank you
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
There may be a better way to get to the same answer.
Here is a cumbersome one.

Using the quadratic formula, we find that the solutions to x%5E2-kx%2B3k-2=0 ,
if any, are given by
.
For the solutions to exist, we need to have
k%5E2-12x%2B8%3E=0 , and only if k%5E2-12x%2B8%3E0 there are two different solutions.
Since the solutions to k%5E2-12x%2B8%3E=0 are
,
two different solutions to x%5E2-kx%2B3k-2=0 exist only if
k%3C6-2sqrt%287%29+ or k%3E6%2B2sqrt%287%29+ .

If k%3E6%2B2sqrt%287%29 , .
So, if k%3E6%2B2sqrt%287%29 , there is one solution (the one above) that is
neither within -1%3C+x+%3C0+ nor within 1%3C+x+%3C2 .

If k%3C6-2sqrt%287%29=about0.7 ,
and x%5E2-kx%2B3k-2=0 is to have
one solution within -1%3C+x+%3C0+ , and
another solution within 1%3C+x+%3C2 , it must be
-1%3Ck%2F2-sqrt%28k%5E2-12k%2B8%29%2F2%3C0+ , and 1%3Ck%2F2%2Bsqrt%28k%5E2-12k%2B8%29%2F2%3C2 .

1%3Ck%2F2%2Bsqrt%28k%5E2-12k%2B8%29%2F2%3C2-->2%3Ck%2Bsqrt%28k%5E2-12k%2B8%29%3C4-->2-k%3Csqrt%28k%5E2-12x%2B8%29%3C4-k
Since 2-k%3E0 ,
2-k%3Csqrt%28k%5E2-12k%2B8%29%3C4-k-->%282-k%29%5E2%3Ck%5E2-12k%2B8%3C%284-k%29%5E2-->4-4k%2Bk%5E2%3Ck%5E2-12k%2B8%3C16-8k%2Bk%5E2-->4-4k%3C-12k%2B8%3C16-8k-->4%2B8k%3C8%3C16%2B4k-->system%288k%3C4%2C%22and%22%2C-8%3C4k%29-->system%28k%3C1%2F2%2C%22and%22%2C-2%3Ck%29-->-2%3Ck%3C1%2F2

-1%3Ck%2F2-sqrt%28k%5E2-12x%2B8%29%2F2%3C0-->-2%3Ck-sqrt%28k%5E2-12x%2B8%29%3C0+-->-2-k%3C-sqrt%28k%5E2-12x%2B8%29%3C-k-->k%2B2%3Esqrt%28k%5E2-12x%2B8%29%3Ek .
As we already found out (above) that we need k%3E-2 , k%2B2%3E0 , so
k%2B2%3Esqrt%28k%5E2-12x%2B8%29<-->%28k%2B2%29%5E2%3Ek%5E2-12x%2B8<-->k%5E2%2B4k%2B4%3Ek%5E2-12x%2B8<-->4k%2B4%3E-12x%2B8<-->16k%3E4<-->k%3E1%2F4 .

In sum, the answer is highlight%281%2F4%3Ck%3C1%2F2%29 .