SOLUTION: find the range of values which the line y=kx+3 meets the curve xy+20=5y

Algebra ->  Graphs -> SOLUTION: find the range of values which the line y=kx+3 meets the curve xy+20=5y      Log On


   



Question 712086: find the range of values which the line y=kx+3 meets the curve xy+20=5y
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
You should be able to substitute y=kx+3 into xy+20=5y and solve for k. That should help to find kx%5E2%2B3%281-5k%29x%2B5=0. Knowledge of the general solution to a quadratic equation suggests k must not be zero.

Carry out the steps you find x=%28-3%281-5k%29-sqrt%28225k%5E2-110k%2B9%29%29%2F%282k%29
or x=%28-3%281-5k%29%2Bsqrt%28225k%5E2-110k%2B9%29%29%2F%282k%29

From that you may want to be sure to know how the discriminant should be greater than or equal to zero. Solve 225k%5E2-110k%2B9%3E=0. Again use solution to quadratic equation:

k=%2811-2%2Asqrt%2810%29%29%2F45 or k=%2811%2B2%2Asqrt%2810%29%29%2F45 But those values should be first checked as critical points. The k quadratic should be greater than OR EQUAL TO ZERO. There are three intervals of k to check the discriminant expression as 225k%5E2-110k%2B9%3E=0.

Pushing ahead according to testing k in those intervals, the critical points in decimalized form would be close to k at 0.10389 and at 0.385. I picked to check values 0, 0.2, and 0.4. The results of the expression>=0 went like this:

at 0: 9%3E=0, yes.
at 0.2: -4%3E=0, no.
at 0.4: 1%3E=0, yes.

Based on that, it seems k should be this:
k%3C=%2811-2%2Asqrt%2810%29%29%2F45
OR
k%3E=%2811%2B2%2Asqrt%2810%29%29%2F45

This was based on expecting complex values of k to cause lack of meaningful solutions to the original set of equations.