Given that the absolute value of the difference of the two roots ax^2 + 5x - 3 = 0 is \(\frac{\sqrt{61}}{3}\), and a is positive, what is the value of a?
+36916 Roots are 5+- sqrt(25+4(a)(3) / 2a
[5 + sqrt(25+12a) - 5 + sqrt(25+12a) ] /2a = sqrt61/3
2 sqrt (25+12a)/2a = sqrt(61)/3
sqrt(25+12a)/a = sqrt(61)/3 results in a = 3
+128475 1/ (2a) * abs [( -5 + sqrt [ 25 + 12a] ) - ( - 5 - sqrt [ 25 - 12a] ) ] =
1/ (2a) * abs ( 2 sqrt [ 25 + 12a ) =
1/a * abs (sqrt [ 25 + 12a ] ) = sqrt (61) / 3
sqrt ( 25 + 12a) = a* sqrt(61) / 3
25 + 12a = 61a^2 / 9
225 + 108a = 61a^2
61a^2 - 108a - 225 = 0
(61a + 75) (a - 3) = 0
The second factor provides a positive solution for a ....a = 3
