So this one question has been pissing me off, and I have tried passing it and going back but it is really annoying.

Sqrt(x+4)=x^2-5

Now after working it out, I square each side to get:

X+4=x^4-10x^2+25

Then move the LHS to the RHS

0=x^4-10x^2-x+21

But that is where I get stuck. I can't seem to find a way to solve that...
Help please? Lol

X=42

it always equals 42

Through derivation you can find out how many solutions there are and a general area in which to look for them.

Not sure if you have some set way you need to solve it, but if you have a graphing calculator/program, then starting at the first equality and plotting both sides to find the intersect is probably the easiest way.

0=x^4-10x^2-x+21
0 = (x^4 -10x^2 + 21) - x
0 = (x^2 - 3)(x^2 - 7) - x

that's as far as I'm able to get without looking up stuff from old math classes :p

6x9

Duh

try to substitute x^2 with u or something

Ferrari's method should work fine, you already even have a depressed quartic so the first step is done for you. Duh...

or ya just plug it into a calculator:P

Ya I have tried substituting... have gotten nowhere heh thanks though :P

-2.495325943, -1.882801176, 1.621432679, 2.756694440

Keep in mind that two of those will not be solutions of the original equation. But it's easy enough to know that x^2-5 should be positive and so |x| > sqrt(5)

Thing is, this is math 12, should I know Ferrari's Method in math 12 or do you learn it a lot later in schooling?

PS. I suck at math :)

twiztid, I dont' think they expect you to know Ferrari's Method. I don't think anyone is ever taught it, in high school/undergrad or grad school. Sorry my "duh" was meant to be completely sarcastic... (and to be clear the only reason I knew about ferrari's method is 2 minutes on google and then wikipedia;))

I think the point of the question is to plug it into calculator and then determine which solutions are actually solutions? Not exactly sure tbh...either that or your prof messed up and one of the constants is wrong or w/ever making the solutions not "nice."

Ferrari's method = fast method (as in fast cars) = using a calculator?

there is no solution...

http://www.myalgebra.com/algebra_solver.aspx
copy and paste your equation in(have to use lowercase S in Sqrt for it to take) and it tells no solution. very useful tool though, hasn't been wrong yet(as long as you punch things in correctly with proper parenthesis' for it)

Dukey: it's a 4th order polynomial there's always 4 solutions per fundamental law of algebra.


http://www.wolframalpha.com/...?i=x%2B4%3D+%28x^2-5%29^2

not true toma. Try x^4+100=0.

you are a nerd.

Bobby: why would that one not still have four solutions?

Isn't it just +-sqrt(i)sqrt(10) and +-sqrt(-i)sqrt(10)?

2nd semester of abstract algebra in undergrad....my professor was kinda nuts tho

oh, also don't forget your imaginary roots =]

lol that sucks Ozzite, the method doesn't look like much fun to learn:P. And ya you guys are right of course about 4 complex roots, but I thought from the context of problem we were looking for real solution. My apologies if I misunderstood you toma!

There's always 4 complex roots, unless roots are repeated.




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If f(x) = x^4-10x^2-x+21
Then f'(x) = x^3 - 20x - 1

Where f'(x) = 0, f(x) has either a minimum or maximum (or inflection point I think in special cases). Since f'(x) is cubic, there will be as many as 3 solutions. Plugging these solutions into f(x) will tell you how many times the graph of f(x) crosses the axis, i.e. how many solutions there are.

Unfortunately, f'(x) doesn't appear to factor easily, so finding the roots of f'(x) is also a rather difficult task.

f''(x) = 3x^2-20
If we let u=sqrt(20)/sqrt(3) (because I'm lazy) Then
f''(x) = 3(x+u)(x-u)
f''(x) crosses the axis twice, thus f'(x) has 0 slope twice. Since f'(x) has a positive leading coefficient, this means it has a maximum at -u and a minimum at u

f'(x) = x^3 - 20x - 1
if we approximate u=2.582 then
f'(-u)= -17.213 + 51.64 - 1 = positive
f'(u) = 17.213-51.64 -1 = negative


Thus, f'(x) crosses the axis 3 times
Since f'(x) crosses the axis 3 times and f(x) has a positive leading coefficient, f(x) has 2 minima and 1 maximum

Find these minima and the maximum and you'll know how many real roots f(x) has.

True, but also X>=-4 so its

(X>=sqrt(5) or X<=-sqrt(5))
and
X>=-4


as for the actual solution, im not sure if this works but:

Sqrt(x+4)=x^2-5
log(Sqrt(x+4))= log(x^2-5)
10^log(Sqrt(x+4))= 10^log(x^2-5)
10^(1/2*log(x+4))= 10^log(x^2-5)
10^1/2 * 10^log(x+4))= 10^log(x^2-5)
10^1/2 * (x+4)= x^2-5
x^2 - sqrt(10)*x - 4*sqrt(10) - 5 = 0
which is
x^2 - b*x - c = 0
where
b= sqrt(10)
c=4*sqrt(10)-5

etc

using the fact that 10^log(x)=x and that log(x^2)=2*log(x)

I dont have time to check it out though.

we didn't really need to learn it per se, but there was an afternoon where the prof just went off on the board and went through the whole solution method and stuff.

I have a bellybutton

hahaha chevs thanks for contributing.