Multiplied top and bottom by √ 3-√2+1 giving us 2√2 below and 4(√6-1) above which simplifies to 2√3 -√2
@XJWill1
Күн бұрын
While it is not the most efficient for this problem, there is a general method for rationalizing a denominator that will work efficiently even for much more complicated denominators (eg., those with higher degree roots). It just needs the minimal polynomial for the denominator. Once you have that, the rest is just simple algebra. To find the minimal polynomial, choose a variable x and set it equal to the expression: x = sqrt(3) + sqrt(2) - 1 Now manipulate both sides of the equation until there is a polynomial with rational coefficients. x^4 + 4*x^3 - 4*x^2 - 16*x - 8 = 0 This is the minimal polynomial, call it m(x) . To find the rationalizing factor for the denominator, compute m(x) / x and drop the 1/x term: x^3 + 4*x^2 - 4*x - 16 Now that polynomial, with x replaced by the value of the denominator, is the rationalizing factor. In this problem, the factor is sqrt(8) * sqrt(3) + sqrt(8) - 4 And multiplying the denominator by that factor results in (sqrt(3) + sqrt(2) - 1 ) * (sqrt(8) * sqrt(3) + sqrt(8) - 4 ) = 8 So the desired result is the numerator multiplied by the rationalizing factor and divided by 8: 1/8 * (sqrt(6) - 2*sqrt(3) + sqrt(2) + 4) * (sqrt(8) * sqrt(3) + sqrt(8) - 4 ) = 2*sqrt(3) - sqrt(2)
@XJWill1
Күн бұрын
Here is a derivation showing why the above method works. If the denominator is d, then multiply d by a factor that produces a rational number c d*f(d) = c Write this as a function of x x*f(x) - c = 0 where d is a root of that equation. To find the function f(x), assume there is a polynomial function m(x) with rational coefficients such that m(x) = x*f(x) - c which must satisfy m(d) = 0 which is the definition of the minimal polynomial for the denominator. Solve for f(x) f(x) = m(x)/x + c/x and choose c to eliminate the k/x term from m(x)/x
@SyberMath
6 сағат бұрын
Wow! Mind-blowing. How do you know these things?
@XJWill1
6 сағат бұрын
@@SyberMath Lots of reading. If you type a radical expression into WA, one of the things it will output is the minimal polynomial for the expression. Then I started reading about minimal polynomials.
@Barteqw90
Күн бұрын
3a-c+d=1, you lost something
@itsphoenixingtime
12 сағат бұрын
Not going to lie, after seeing the answer, I just felt like if I went with the initial idea of rationalising twice, I would have solved it quicker than trying to find a "slick elegant" method by trying to break up the top to fit the bottom. I tried but I ended up messing up, so I guess this is definitely a lesson to be taken away from.
@SyberMath
6 сағат бұрын
Really hard to without going backwards with how I came up with the problem
@neuralwarp
21 сағат бұрын
Let u=root2, v=root3, then simplify (4+uv-u²v+u)÷(v+u-1)
@SyberMath
21 сағат бұрын
I thought about it but was not sure if it would work
I think I will try to force out common factor the denominator and see what happens. If it looks disgusting maybe I will multiply top and bottom by √3-√2+1 (almost conjugate but not quite a^2-b^2 and will cancel √3 terms). 4+√2(√3+√2-1-√2+1)-2(√3+√2-1-√2+1)+√2=> √2-2 plus a fraction: 4-2+√2+2√2-2+√2= 4√2/... (√3+√2-1)*(√3-√2+1)=3-2+2√2-1=2√2 so then 4√2/...=2(√3-√2+1) So the whole thing reduces to √2-2+2(√3-√2+1)= 2√3-√2 To verify (2√3-√2)(√3+√2-1)=6+2√6-2√3-√6-2+√2= 4+√6-2√3+√2
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