Find Quadratic Equation Given 1 Root: (2^(1/2)+1)

AI Thread Summary
To find the quadratic equation given one root as (2^(1/2)+1), the other root must be (1-(2^(1/2))) to ensure both roots yield rational coefficients. The quadratic can be expressed in the form a(x - x_0)(x - x_1), leading to the equation x^2 - 2x - 1 = 0. The discriminant must be a perfect square for the coefficients to remain rational. The calculations confirm that the derived quadratic equation is correct. Thus, the final quadratic equation is x^2 - 2x - 1 = 0.
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Homework Statement



If one root of a quadratic equation with rational co-efficient is (2^(1/2)+1) , then find the quadratic equation.

Homework Equations



x=(-b+d^1/2)/2a
d=b^2-4ac


The Attempt at a Solution



Well, I can't quite understand this question. Please help me to understand what is given in the following statemnt.
 
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If the only square root in the solution is \sqrt{2}, what must d be? Since the only difference between roots of a quadratic equation is that \pm before the square root, what must the other root be?

Another, more "sophisticated" method:
Any quadratic, ax^2+ bx+ c can be written a(x- x_0)(x- x_1)= ax^2- a(x_0+x_1)+ ax_0x_1 where x_0 and x_1 are roots of the equation. If one root is 1+ \sqrt{2} what must the other be so that both x_0+ x_1 and x_0x_1 are rational?
 
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Well, Please check this. I think the othr root is 1-(2^1/2). thre fore the quadratic equation is x^2-2x-1=0
Am I right?
 
Yes, that is correct.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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