Quadratic equation blues.

I was asked last night about performing maths with quadratic equations and I could show how to solve most problems but there was one that has left me completely baffled.  After about 45 minutes last night I had to give up.  (Probably missed something obvious and was making life way to complex for myself.)

Given the following quadratic equation:
4x.x - 35x + q = 0
what is the value of q.  One is also told the raltionship between two values of x which solve the equation.  Call those solution n and m then n = m squared (n = m.m).

I can use formulae from vieta to obtain things like m + n = 35/4 and m.n = q/4 and the formula for solving the quadratic but I then just get a horrendous mess with quadratic or higher equations when trying to simplify things.

Can someone point me in the right direction please.
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AndyAinscowFreelance programmer / ConsultantAsked:
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n=0, m=0, q=0 is one solution
AndyAinscowFreelance programmer / ConsultantAuthor Commented:
Okaay, but how would one approach that calculating that formally.

(Also note that does not fit the vieta equations because the square of zero is zero so one ends up with 0+0 = 35/4)
AndyAinscowFreelance programmer / ConsultantAuthor Commented:
There are some formulae developed by Vieta concerning polynomials in general and quadratics in specific.  


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4n.n - 35n + q = 0
4n.n.n.n - 35n.n + q = 0
4n.n.n.n - 31n.n + 35n = 0
n(4n.n.n - 31n + 35)=0
n=0 or (4n.n.n - 31n + 35)=0
given n.n + n - 35/4 = 0
n = (-1+- sqrt(1.1 + 35)/2.1
AndyAinscowFreelance programmer / ConsultantAuthor Commented:
Thanks.  I'll have a think about that.

I'm getting a horrible feeling this was a very nasty question in the chapter concerning quadratics.  (The ones prior to this were done very rapidly with the help of the vieta equations).  If I am correct then I think it is out of order to have that in a book for someone to be learning the maths - it could really wreck the pupils confidence.
AndyAinscowFreelance programmer / ConsultantAuthor Commented:
Still having problems.  One of the solutions for x is to be the square of the other.  That is a constraint given in the question.

from the part
n = (-1+- sqrt(1.1 + 35)/2.1

that I get two values 5/2 and -7/2 unfortunately one isn't the square of the other (and neither related to zero).
4n.n.n.n - 39n.n + 35n = 0
n(4n.n.n - 39n + 35)=0
n=0 or (4n.n.n - 39n + 35)=0
5/2 and -7/2 are two values for n
the corresponding values for m would be 25/4 and 49/4
Sorry, I had n and m switched from your original statement
AndyAinscowFreelance programmer / ConsultantAuthor Commented:
S**t, I've got to think about it, that might have been my mistake.
Just take m=n.n and n+m=35/4
combine them into a quadratic in n
solve for n, then use n to find m and q

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AndyAinscowFreelance programmer / ConsultantAuthor Commented:
Done that and got two pairs of roots (hopefully correctly).
5/2 and 25/4
-7/2 and 49/4

Now I put these back into the original eqation to cross check.  The first pair give a different value for q with each substition.  The second pair both give q = -171.5

So in the original eqation the value of q is -171.5 and the pair values of x are -7/2 and 49/4

Many thanks.  I'd done part of this procedure then made a mistake in thinking that the 5/2 and -7/2 should be the pair of solutions of x and one should the the square of the other NOT each independant which I would than have to work with further.
AndyAinscowFreelance programmer / ConsultantAuthor Commented:
:-)  Thanks.
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