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ANT101 changed this question to a completely different one while i was working on it.

I have changed it back to the original question that I have answered.

 

 

Find the absolute value of the difference of the solutions of     \(x^2-5x+5\)

ant101  Aug 22, 2018
edited by ant101  Aug 22, 2018
edited by Melody  Aug 22, 2018
edited by Melody  Aug 22, 2018
 #1
avatar+93866 
+1

Find the absolute value of the difference of the solutions of \(x^2-5x+5=0\)

 

The solutions of any quadratic equation are

 

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\x=\frac{-b}{2a}\pm \frac{ \sqrt{b^2-4ac} }{2a} \)

 

So the ablosute value difference of the roots is

 

\(\text{abs value of diff of roots} \\ =\left|\left[ \frac{-b}{2a}+ \frac{ \sqrt{b^2-4ac} }{2a}\right] - \left[ \frac{-b}{2a}- \frac{ \sqrt{b^2-4ac} }{2a}\right] \right|\\ =\left|\left[ \frac{-b}{2a}+ \frac{ \sqrt{b^2-4ac} }{2a}\right] + \left[ \frac{b}{2a}+\frac{ \sqrt{b^2-4ac} }{2a}\right] \right|\\ =\left| \frac{ \sqrt{b^2-4ac} }{2a}+ \frac{ \sqrt{b^2-4ac} }{2a} \right|\\ =\left| \frac{ \sqrt{b^2-4ac} }{a} \right|\\ \)

 

for your question

\(x^2-5x+5=0\)

 

a=1    b=-5        c=5

 

\(\text{abs value of diff of roots }= \frac{ \sqrt{25-20}}{1}=\sqrt5\)

Melody  Aug 22, 2018
 #3
avatar+591 
+1

Thanks! So, my internet connection was working very slow, and I quickly realized I posted the same question twice! I clicked on my latest post, and changed it to another question, leaving the question before the latest one as it is.

ant101  Aug 22, 2018
 #2
avatar+1198 
+1

Ants can be such pains-in-the-ass! Especially at picnics … and on math forums. smileyLOLlaugh

GingerAle  Aug 22, 2018
 #4
avatar+93866 
0

Hi Ginger,

LOL  This ant just made an understandable mistake.  No big deal.  :)

Melody  Aug 22, 2018

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