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February 16th, 2019, 08:07 AM  #1 
Newbie Joined: Jan 2018 From: Belgium Posts: 7 Thanks: 0  Factor polynomial
Hello, When making some exercices to improve my general understanding of basic mathematics, I stumbled upon following problem: to factorise X^2 + ((sqrt 2)+(sqrt 3)) x + (sqrt 6) I do not recognise the form (a+b)^2, so I tried going for the discriminant, which would be ((sqrt 2)+(sqrt 3))^2  4*(sqrt 6) = 2 + 3 + 2((sqrt 2)+(sqrt 3))  4 (sqrt 6) = 5  2(sqrt6) I thus there are two roots to this polynomial, but I do not know how to calculate them. When trying to use the formula (b + (sqrt D))/2a, the numbers (especially the square roots) get to hard for me to work with. The book lists the solutions (x+ (sqrt 2)) * (x+ (sqrt 3)), but not how to get there. Could anyone give me a hint here please? 
February 16th, 2019, 08:13 AM  #2  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,226 Thanks: 908 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  
February 16th, 2019, 08:26 AM  #3 
Newbie Joined: Jan 2018 From: Belgium Posts: 7 Thanks: 0 
Oh, wow, I didn't know that equation. Thank you!

February 16th, 2019, 09:21 AM  #4 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,226 Thanks: 908 Math Focus: Wibbly wobbly timeywimey stuff.  
February 16th, 2019, 12:21 PM  #5  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,226 Thanks: 908 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle x^2 + ( \sqrt{3} + \sqrt{2} )x + \sqrt{6} = 0$ $\displaystyle x = \dfrac{(\sqrt{3} + \sqrt{2} ) \pm \sqrt{ ( \sqrt{3} + \sqrt{2} )^2  4 (1) \sqrt{6} }}{2 (1)}$ $\displaystyle x = \dfrac{  ( \sqrt{3} + \sqrt{2} ) \pm \sqrt{3 + 2 + 2\sqrt{6}  4 \sqrt{6} } }{2}$ $\displaystyle x = \dfrac{ ( \sqrt{3} + \sqrt{2} ) \pm \sqrt{3 + 2  2 \sqrt{6} } }{2}$ $\displaystyle x = \dfrac{ (\sqrt{3} + \sqrt{2} ) \pm \sqrt{(\sqrt{3}  \sqrt{2})^2}}{2}$ $\displaystyle x = \dfrac{ (\sqrt{3} + \sqrt{2} ) \pm (\sqrt{3}  \sqrt{2} )}{2}$ I leave it to you to finish this. Dan  

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