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Given that the absolute value of the difference of the two roots of \(ax^2 + 5x - 3 = 0\) is \(\frac{\sqrt{61}}{3}\), and a is positive, what is the value of a?

 Feb 28, 2021
edited by jxc516  Feb 28, 2021
 #1
avatar+113133 
+2

\(\left|difference\;\;of\;\;roots\right|=\left| [{-b + \sqrt{b^2-4ac} \over 2a}]-[{-b- \sqrt{b^2-4ac} \over 2a}]\right|\)

 

Can you take it from there jxc?

 Feb 28, 2021
edited by Melody  Feb 28, 2021
edited by Melody  Feb 28, 2021
 #2
avatar+201 
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yes I solved it

 

Thank you! :)

jxc516  Mar 1, 2021
 #3
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You responded only after I complained on other posts, \

 

You posted this question afterwards.

https://web2.0calc.com/questions/help-help-help_11

 

How about you show us what you have already tried for yourself instead of just posting bare questions.

 

In fact please go to the link above.  And explain to us how Textot got his answer.  

 

If you do not know then ask Textot questions about it.

Encourage or even push him/her to guide you through it.

Melody  Mar 1, 2021
edited by Melody  Mar 1, 2021
 #4
avatar+201 
+2

ok lemme get my paper...

 

\(\frac{-b+\sqrt{b^2-4ac}}{2a}-\frac{-b-\sqrt{b^2-4ac}}{2a}\Rightarrow \frac{-b+b+\sqrt{b^2-4ac}+\sqrt{b^2-4ac}}{2a}\Rightarrow\frac{2\sqrt{b^2-4ac}}{2a}\Rightarrow \frac{\sqrt{b^2-4ac}}{a}\Rightarrow \frac{\sqrt{61}}{3} \implies \boxed{a=3}\)

 

I might have explained it a little thinly on accident

jxc516  Mar 1, 2021
 #5
avatar+113133 
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thanks for your response jxc.     smiley

Melody  Mar 1, 2021

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