Solve this equation please

[soccerboi]

Solve 2^x=x²-1
Thanks

[pi]

Solve 2^x=x²-1
Thanks

Haha, you can't solve those at VCE level (I have no idea to anyway). The best you can do is find x=3 via inspection, and find the other solutions (not sure if there are others, but there probably are) by "solving" on a CAS (or similar)


edit: using a CAS, you'll find that x=-1.98 and x=3.407 too

[soccerboi]

Here's the actual ques:
the number of solutions to the equation 2^x=x²-1 is:
A 0
B 1
C 2
D 3
E 4

Is there a way to answer this without using CAS?

[kensan]

Wouldn't it be two solutions because it's a quadratic?

[pi]

Well, the answer would be D. I don't think there is, but I guess you'llhave to wait for an answer from kamil, TT, etc. to be sure of that

As its a Multi-Choice question, the CAS is allowed, so this must be a CAS-only question imo

Wouldn't it be two solutions because it's a quadratic?

No, because it has a 2^x and NOT a x²

[TrueTears]

Here's the actual ques:
the number of solutions to the equation 2^x=x²-1 is:
A 0
B 1
C 2
D 3
E 4

Is there a way to answer this without using CAS?

There are a few ways, one method is to expand (maybe expand up to the third or fourth polynomial), I expanded it to the third



So our question essentially becomes





Now the question is reduced down into how many zeros does this cubic polynomial have, which is doable even if it doesn't have rational roots, one could go on to approximate the roots using Newton's method (http://en.wikipedia.org/wiki/Newton%27s_method) or in general the Householder method (http://en.wikipedia.org/wiki/Householder%27s_method), turns out there are 3.


Note the above is prone to error and depends on how well you approximate, and if you did attempt the above on an exam, then you better have a fricken fast brain

I leave it to kamil to think of a more elegant solution ;]

[yawho]

Use the gradients at x=3 to understand why 3 intersections

[Bhootnike]

mm since its mc and you're given the options, why wouldnt you just sub in each option and see which one is valid ?

2^3  = 3^2 -1
8 = 8

[TrueTears]

mm since its mc and you're given the options, why wouldnt you just sub in each option and see which one is valid ?

2^3  = 3^2 -1
8 = 8

think you misinterpreted the question the options are the number of zeros not the actual zeros themselves

just by coincidence x = 3 is a zero and there are 3 zeros lol

[Bhootnike]

ohhh right haha i didnt look at the question written with the mc answers, i just assumed it was the same as

Solve 2^x=x²-1
Thanks

[1i1ii1i]

answer is option D =3

[1i1ii1i]

it doesn't ask to solve it asks for number of solutions!

[1i1ii1i]

is it on that chapter 1 algebraic techniques test because that was pretty hard

[kensan]

triple post ftw