VCAA 2006 questin

[Elnino_Gerrard]

Question 3 part e. Find the values of K, where k is a positive real number  for which the equation has one or more solutions for x.
3-ke^x-e^-x
Now i know how to get the answer using the discrimant formula, i subbed in a =e^x and got a quadratic. But the thing is how do we know whether for that k value we wont get a=  a negative number therefore no solution for x?

[Elnino_Gerrard]

?

[luffy]

That a value will not be determined by the discriminant. It is changed by the + or - part of the quadratic formula. For example, in the case of e^x, only the '+' component of the quadratic formula would be included and the negative part would be rejected.

I'm kinda a noob with these forums and am not too clear on how to put in fractions yet. So this may not have made much sense.

[Elnino_Gerrard]

That a value will not be determined by the discriminant. It is changed by the + or - part of the quadratic formula. For example, in the case of e^x, only the '+' component of the quadratic formula would be included and the negative part would be rejected.

I'm kinda a noob with these forums and am not too clear on how to put in fractions yet. So this may not have made much sense.

what im trying to say is i get how to use the discriminant as in saying dis>0 for 2 solutions but what of one those solutions is negative?

[davidle_10]

it says one or more solutions which means that it is greater than or equal to zero. and then it is stated in the question that k positive.