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

#1**+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