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The solutions of  may be expressed in the form (a-b^1/2)/c and , (a+ b^1/2)/c where , , and  are prime numbers. Find .The solutions of $x(x-3)=1$ may be expressed in the form $\frac{a+\sqrt{b}}{c}$ and $\frac{a-\sqrt{b}}{c}$, where $a$, $b$, and $c$ are prime numbers. Find $abc$.

Guest Sep 27, 2018
edited by Guest  Sep 27, 2018
 #1
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\(x(x-3)=1\)

 

\($\frac{a+\sqrt{b}}{c}$ and $\frac{a-\sqrt{b}}{c}$\)

 

 

x( x - 3)  = 1

 

x^2 - 3x   = 1     subtract 1 from  both sides

 

x^2  - 3x  + 1    = 0  

 

Using the quadratic formula...   we have

 

x  =  [3 ±√ [9 - 4(1)(1) ]  / 2   =  [ 3 ±√ 5 ] / 2

 

So  abc  =  3 * 5 * 2    =  30

 

 

cool cool cool

CPhill  Sep 27, 2018

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