MHBSolving for Real Solutions: Systems of Equations with Powers
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The discussion revolves around solving a system of equations involving variables a, b, c, and d. Two solutions are identified: (-2, -1, -1, 1) and (1, 2, -1, 1). Participants confirm the validity of these solutions through various algebraic manipulations and substitutions. The approach includes rewriting equations for clarity and deriving relationships between the variables. Ultimately, the consensus is that these are the only solutions to the problem.
which leads us to $a = 1$ and $a = -2$ from which we can find the remaining variables $b, c$ and $d$ leading to the solutions that Opalg found.
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anemone
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Opalg said:
The only way I can do this is by guesswork.
[sp]$a-b-c+d=1 = (-2) +3$
$a^2+b^2-c^2-d^2=3 = (-2)^2 - 1$
$a^3-b^3-c^3+d^3=-5 = (-2)^3 + 3$
$a^4+b^4-c^4-d^4=15 = (-2)^4 - 1$.
When you write them like that it's easy to see that $(a,b,c,d) = (-2,-1,-1,1)$ is a solution. But is it unique?
Edit (in haste). It looks as though $(a,b,c,d) = (1,2,-1,1)$ is another solution.[/sp]
Thanks for participating, Opalg!:)
I like the way you have the given values of the 4 equations rewritten as $(-2) +3$, $(-2)^2 - 1$, $(-2)^3 + 3$, $(-2)^4 - 1$. And yes, these two are the only solutions to the problem. Good observation, Opalg!
Jester said:
My solution
I'll solve the first equation for $c$ so
$c =a-b+d-1$
This give the second equation
$2\,ab-2\,ad+2\,a+2\,bd-2\,b-2\,{d}^{2}+2\,d-4=0$
or solving for $b$
$b = {\dfrac {ad-a+{d}^{2}-d+2}{a+d-1}}$.
Note that $a+d-1 \ne 0$ since if this was true, there's no solution to the system. With these two assignments, the remaining equations become
And squaring both sides of the equation $(a-b)^2=(1+c-d)^2$ yields $a^2+b^2-2ab=1+c^2+d^2-2cd+2c-2d\;(2)$.[/TD]
[TD]$a^2+b^2-c^2-d^2=3$ gives $a^2+b^2=3+c^2+d^2\;\;(3)$ .
Replacing it into (2) gives $3+c^2+d^2-2ab=1+c^2+d^2-2cd+2c-2d$ or
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Just chatting with my son about Maths and he casually mentioned that 0 would be the midpoint of the number line from -inf to +inf. I wondered whether it wouldn’t be more accurate to say there is no single midpoint. Couldn’t you make an argument that any real number is exactly halfway between -inf and +inf?
A power has two parts. Base and Exponent.
A number 423 in base 10 can be written in other bases as well:
1. 4* 10^2 + 2*10^1 + 3*10^0 = 423
2. 1*7^3 + 1*7^2 + 4*7^1 + 3*7^0 = 1143
3. 7*60^1 + 3*60^0 = 73
All three expressions are equal in quantity. But I have written the multiplier of powers to form numbers in different bases. Is this what place value system is in essence ?