Rational roots - standard form of equation

Janinever
Messages
13
Reaction score
0
Hi everybody!

I've hit a blank with regards to this 1 equation on a old exam paper - think I've overloaded myself a bit and just feel a bit like a airhead at the moment!

I understand the actual method and getting to the answer but it starts off with a equation which you then need to get to a standard form to use the formula to check if the roots are rational.

The initial equation is this :

4x^2 + 2kx - k = 2x

Then on the memorandum this is taken to standard form so the info can be substituted into the formula Δ = b^2 - 4ac

the standard form they then write that initial equation in is :

4x^2 + 2(k-1)x-k = 0

How do they get from 4x^2 + 2kx - k = 2x to this 4x^2 + 2(k-1)x-k = 0

The rest I understand entirely. Literally just the above has be scratching my head!

Please help :)

Thank you!
 
Mathematics news on Phys.org
Janinever said:
How do they get from 4x^2 + 2kx - k = 2x to this 4x^2 + 2(k-1)x-k = 0

Sometimes you overlook the obvious answers. Just subtract 2x from both sides.
 
Thank you!
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top