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

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!

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

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

$61a^2-108a+225=0$

$a=3$

  !