SOLUTION: calculate for which value(s) of k the equetion (x^2+2x-11)÷2(x-3)=k will have no real roots. (8)

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: calculate for which value(s) of k the equetion (x^2+2x-11)÷2(x-3)=k will have no real roots. (8)       Log On


   

Question 847204: calculate for which value(s) of k the equetion (x^2+2x-11)÷2(x-3)=k will have no real roots. (8)
Answer by swincher4391(1107) About Me  (Show Source):
You can put this solution on YOUR website!
%28x%5E2%2B2x-11%29%2F%282%2A%28x-3%29%29+=+k
%28x%5E2%2B2x-11%29%2F%282x-6%29+=+k
Multiply both sides by 2x-6.
x%5E2+%2B2x+-11+=++%282x-6%29%2Ak
x%5E2+%2B+2x+-+11+=+2k%2Ax+-+6k
x%5E2+%2B+2x+-+2kx+-+11+%2B6k+=+0
x%5E2+%2B+%282-2k%29x+%2B+%286k+-11%29+=+0
From here you can use the discriminant to figure out where you will have no real roots. Remember that b^2-4ac is our discriminant and when it is negative is when we'll have no real roots.
So set b^2 - 4ac < 0.
%282-2k%29%5E2+-+4%2A1%2A%286k-11%29+%3C+0
4-4k%2B4k%5E2+-+24k+%2B+44+%3C+0
4k%5E2+-+32k+%2B+48+%3C0
4%28k%5E2+-+8k+%2B+12%29+%3C0
4%28k-6%29%28k-2%29%3C0
Using a number line to find where (k-6)(k-2) < 0
<--------2--------------6---------->
We find that between 2 and 6 is where we get a negative value.
So for all 2%3Ck%3C6 we have no real roots.