another challenging problem (1 Viewer)

[Affinity]

a>0

find the sum of all solutions for x in the close interval [-2,6] for:

a^( (x^2 - 5)*|cos(x)| ) = (sqrt(a))^( |x|cos(x) )

 

[freaking_out]

Originally posted by Affinity
a>0

find the sum of all solutions for x in the close interval [-2,6] for:

a^( (x^2 - 5)*|cos(x)| ) = (sqrt(a))^( |x|cos(x) )

give us a hint on where to start?

 

[ND]

3pi/2? Or am i missing solutions?

edit: oh wait, between -2 and 6: 3pi/2 + 2.5?

 

[phenol]

3pi/2?

 

[MyLuv]

Originally posted by Affinity
a>0

find the sum of all solutions for x in the close interval [-2,6] for:

a^( (x^2 - 5)*|cos(x)| ) = (sqrt(a))^( |x|cos(x) )

Is sqrt mean square root?
If so then ,by dividing with RHS ,the equation go to:
(x^2-5)lcos(x)l = lxlcosx/2
square--> [cos(x)]^2*(x^2-5)^2=x^2*[cosx]^2*1/4
---> cos x =0 is solution ie x=+-(pi/2) or 3pi/2 is solution(between -2 and 6)
then divide for cosx^2 the equation is:
x^4 - 10x^2+25=x^2/4--> sum of the root is 0
---> SUm of the root for the original = 3pi/2

 

[phenol]

nice, i case bashed

 

[ND]

Re: Re: another challenging problem

Originally posted by MyLuv
Is sqrt mean square root?
If so then ,by dividing with RHS ,the equation go to:
(x^2-5)lcos(x)l = lxlcosx/2
square--> [cos(x)]^2*(x^2-5)^2=x^2*[cosx]^2*1/4
---> cos x =0 is solution ie x=+-(pi/2) or 3pi/2 is solution(between -2 and 6)
then divide for cosx^2 the equation is:
x^4 - 10x^2+25=x^2/4--> sum of the root is 0
---> SUm of the root for the original = 3pi/2

I did something kinda similar, but then i realised one of teh roots in there is -2.5, and hence can't be counted, so i got 3pi/2+2.5.

 

[MyLuv]

Re: Re: Re: another challenging problem

Originally posted by ND
I did something kinda similar, but then i realised one of teh roots in there is -2.5, and hence can't be counted, so i got 3pi/2+2.5.

Er,I think the root -2.5 can be from the 2nd equation but it also has other roots that sum to 0

 

[ND]

Yeh, the roots are symmetrical but because the interval isn't, 2.5 can be counted in the sum but -2.5 can't. Same as the way 3pi/2 was counted but -3pi/2 wasn't.

 

[underthesun]

??? 2 and -2? and pi/2 and 3pi/2 and -pi/2? am I lost?

 

[MyLuv]

Originally posted by ND
Yeh, the roots are symmetrical but because the interval isn't, 2.5 can be counted in the sum but -2.5 can't. Same as the way 3pi/2 was counted but -3pi/2 wasn't.

Err..ye ,I see.That means we need to solve the other equation too

 

[ND]

Well here's what i did:

a^( (x^2 - 5)*|cos(x)| ) = (sqrt(a))^( |x|cos(x) )
square both sides:
a^[(2x^2-10)*|cosx|]=a^[|x|cosx]
(2x^2-10)*|cosx|=|x|cosx
2x^2-10=|x| (noting solutions to cosx=0)
so then 2x^2+x-10=0 or 2x^2-x-10=0
solving these gives x=+-2, +-2.5

 

[underthesun]

I don't think +2.5 is solution, is it? because the (x^2-5) will equal the |x| part, but cos(2.5) won't equal |cos(2.5)|. Im a bit confused on the wording

Anyways, is -2 in the interval?

 

[ND]

Yep you're right, no solution greater than sqrt5 would work...

 

[Affinity]

3Pi/2 is the official answer

x = 2, -2, -pi/2 pi/2 3pi/2

basically what ND did.
first to find possible solutions

squaring both sides, equating indices

(2x^2 - 10)|cos(x)| = |x| cos(x)

cos(x) = 0 solutions,
-pi/2, pi/2 3pi/2

else if cos(x) != 0

(2x^2 -10) = |x| cos(x)/|cos(x)|
(2|x|^2 - 10) = |x| * sgn[cos(x)] (sign of cos(x))

2|x|^2 (+/-) |x| - 10 = 0

solving gives |x| = 2.5, 2 are possible solutions
so x= -2.5, -2, 2, 2.5 are possible,
then you test evey one in teh original
etc.