Maths Methods 3/4 Help Thread 2011

[luken93]

e^2x +be^x +1 = 0 where b is a real constant
for what values of b will x have no real solutions?
thx in advance

Let e^x = A
A^2 + bA + 1 = 0
No Solutions exist when b^2 - 4ac < 0
b^2 - 4(1)(1) < 0
b^2 < 4
-2 < b < 2

See below.

[thushan]

e^2x +be^x +1 = 0 where b is a real constant
for what values of b will x have no real solutions?
thx in advance

solve for e^x:

e^x = (-b +- root(b^2 - 4))/2.

Two conditions: b^2 - 4 >=0 AND -b + root(b^2 - 4) >0  --> we take the greater root to be >0 as we only need at least 1 solution for x (given e^x > 0 for all x)

condition 1: IbI >= 2.

condition 2: b < root(b^2 - 4) => this is always true where IbI >=2.

Hence, where IbI < 2, x will have no real solutions.

[thushan]

luke, its actually -2<b<2

and i think you have to show that for b>2 and b<-2 the higher root is > 0.

[duquesne9995]

e^2x +be^x +1 = 0 where b is a real constant
for what values of b will x have no real solutions?
thx in advance

solve for e^x:

e^x = (-b +- root(b^2 - 4))/2.

Two conditions: b^2 - 4 >=0 AND -b + root(b^2 - 4) >0  --> we take the greater root to be >0 as we only need at least 1 solution for x (given e^x > 0 for all x)

condition 1: IbI >= 2.

condition 2: b < root(b^2 - 4) => this is always true where IbI >=2.

Hence, where IbI < 2, x will have no real solutions.

but when I solve in my calc when b=3 or b=2, it comes out with no solutions as well

[thushan]

Crud, you're right, i made a mistake. Lemme fix:

 b < root(b^2 - 4) => this is only true for b<or =-2.

Hence, there are only solutions when b< or = -2.

Hence x has no real solutions when b > -2.

Sorry about that!

[duquesne9995]

Crud, you're right, i made a mistake. Lemme fix:

 b < root(b^2 - 4) => this is only true for b<or =-2.

Hence, there are only solutions when b< or = -2.

Hence x has no real solutions when b > -2.

Sorry about that!

ok thanks for that
so, using your condition 1 and condition 2, i take the union of both intervals?

[thushan]

Intersection. Condition 1 and 2 were for values of b when b is defined.

[duquesne9995]

Intersection. Condition 1 and 2 were for values of b when b is defined.

okay thanks so much!
was so confused before. but now I get it.

[thushan]

no worries!

[BlueSky_3]

@Thushan: what do you mean by the higher root? And why was Luken93's solution wrong?

[BlueSky_3]

Anyone?

[paulsterio]

@Thushan: what do you mean by the higher root? And why was Luken93's solution wrong?

I think it was because he didn't consider that e^(x) = +ve number
So you could solve it as a quadratic say, but if you end up with 2 negative solutions for e^(x) - there would still be no solutions

Hence, like Thushan said, you have to consider both the cases, there to be 1 or 2 solutions for the quadratic itself, i.e. discriminant>0 and then you will have to ensure that one of the solutions are positive, i.e. that one of "-b+sqrt(discriminant)" or "-b-sqrt(discriminant)" are positive as well

[abeybaby]

thats funny, we did that question today in MM lecture!

[duquesne9995]

thats funny, we did that question today in MM lecture!

which MM lecture is that? at your school or at a company/tutor?

[abeybaby]

Company, Connect Education. One of my tutoring students gave me that question and it was a good question so I just mentioned it in passing