The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Interesting Misprint Question

The quadratic x2+4x-3=0 was misprinted as x2+4x-3=0 on a worksheet of mine. I have now been set it for homework after expressing interest in it.
Can anyone offer me advice on solving it algebraically, graphically by plotting y=4x and y=3-x2 I can see there are two solutions.
Thanks in advance! :smile:
Original post by ThreePoint14159
The quadratic x2+4x-3=0 was misprinted as x2+4x-3=0 on a worksheet of mine. I have now been set it for homework after expressing interest in it.
Can anyone offer me advice on solving it algebraically, graphically by plotting y=4x and y=3-x2 I can see there are two solutions.
Thanks in advance! :smile:


You are going to need to use the graphical method you used or a numerical method.
The iterative formula:

xn+1=ln⁡(3−xn2)ln⁡4\displaystyle x_{n+1}=\frac{\ln(3-x_n^2)}{\ln 4} converges to the positive root fairly quickly.

The iterative formula:

xn+1=3−4xnxn\displaystyle x_{n+1}=\frac{3-4^{x_n}}{x_n} converges to the negative root incredibly slowly.

Newton Raphson finds either root so long as you start near enough:

xn+1=xn−xn2+4xn−32xn+4xnln⁡4\displaystyle x_{n+1}=x_n - \frac{{x_n}^2 + 4^{x_n}-3}{2x_n + 4^{x_n} \ln 4}
(edited 10 years ago)
Original post by Mr M
The iterative formula:

xn+1=ln⁡(3−xn2)ln⁡4\displaystyle x_{n+1}=\frac{\ln(3-x_n^2)}{\ln 4} converges to the positive root fairly quickly.

The iterative formula:

xn+1=3−4xnxn\displaystyle x_{n+1}=\frac{3-4^{x_n}}{x_n} converges to the negative root incredibly slowly.

Great! Thanks very much. :smile:
Original post by ThreePoint14159
Great! Thanks very much. :smile:


You are welcome. Just checking you know how to use an iterative formula?
Original post by Mr M
You are welcome. Just checking you know how to use an iterative formula?

I do, incidentally just been doing Newton-Raphson in FP1 :smile:
I had considered doing numerical methods, just took me a while to differentiate 4x for Newton Raphson but I'm sorted now!
Original post by ThreePoint14159
I do, incidentally just been doing Newton-Raphson in FP1 :smile:
I had considered doing numerical methods, just took me a while to differentiate 4x for Newton Raphson but I'm sorted now!


Good.
Reply 7
Avatar for arkanm
arkanm
5
Here's a nice way to prove there are no integral solutions:

(*) 3−x2=4x3-x^2=4^x.

Case 1: If x<-1 then LHS<0 but RHS is always positive, hence no solutions.

Case 2: If x=-1 then 2=1/2, contradiction.

Case 3: If x=0 then 3=1, contradiction.

Case 4: If x is at least 1 then the RHS has at least one factor of 4. Therefore taking mod 4 of both sides yields 3−x2≡0  ⟺  x2≡3(mod4)3-x^2\equiv 0\iff x^2\equiv 3\pmod 4, which is absurd since 3 isn't a quadratic residue modulo 4.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you