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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     $$x^2-5x+5$$

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

#1
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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

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

$$x^2-5x+5=0$$

a=1    b=-5        c=5

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

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Aug 22, 2018
#3
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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
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Ants can be such pains-in-the-ass! Especially at picnics … and on math forums. LOL Aug 22, 2018
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Hi Ginger,

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

Melody  Aug 22, 2018