ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Elnino_Gerrard on November 07, 2010, 05:28:09 pm
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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?
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?
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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.
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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?
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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.