+0  
 
+2
91
1
avatar+147 

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 24, 2021
 #1
avatar+11297 
+1

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

 

Hello powclaire!

 

\(\dfrac{\sqrt{61}}{3} =\dfrac{-5+ \sqrt{25+12a}}{2a} -\dfrac{-5- \sqrt{25+12a}}{2a}\\ =\dfrac{-5+ \sqrt{25+12a}+5+\sqrt{25+12a}}{2a} \)

\(\dfrac{\sqrt{25+12a}}{a}=\dfrac{\sqrt{61}}{3}\\ \dfrac{25+12a}{a^2}=\dfrac{61}{9}\)

\(61a^2-108a+225=0 \)

 

\(a=3\)

laugh  !

 Feb 24, 2021
edited by asinus  Feb 24, 2021

16 Online Users

avatar