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     \(x^2-5x+5\)

 Aug 22, 2018
edited by ant101  Aug 22, 2018
edited by Melody  Aug 22, 2018
edited by Melody  Aug 22, 2018

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



a=1    b=-5        c=5


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

 Aug 22, 2018

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

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

 Aug 22, 2018

Hi Ginger,

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

Melody  Aug 22, 2018

7 Online Users


New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.