Question

3x^2 +kx-s=0
if s>0 for what values of k does the equation have two real solutions?

Answer 1

The quadratic equation will give you real solutions when the discriminant (the stuff under the square root sign) is non-negative. If the discriminant is equal to zero you have one real solution. If it is greater than zero, you have two real solutions. If it is negative you have no solutions. Given:\[ax^2+bx+c=0\]The quadratic formula is \[x=[-b \pm \sqrt{b^2-4ac}]/2a\]So in your case we have:\[a=3, b=k , c=-s\]All we care about is the stuff under the root sign must be either zero or a positive real number for their to be real solutions:\[\sqrt{b^2-4ac}\ >0\]For your example:\[\sqrt{k^2-4(3)(-s)}>0\]Or,\[k^2+12s >0\]Now take a look at the expression above. We know that s is a positive number because it is given in the problem. Also, the square of ANY number (positive or negative) is also a positive number (or zero). Therefore, the expression is true for ALL values of k (even zero, since s cannot equal zero). k can take on any value whatsoever and the equation will have two real roots/solutions.

Answer 2

thank you very much!!! Did you have a source or did you just know it?

Answer 3

I have quite a bit of experience with Math so no source needed. I re-read it this morning and it still looks correct :) Thanks for the medal!