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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?

 Apr 25, 2019
 #1
avatar+19792 
+2

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

 Apr 25, 2019
 #2
avatar+106539 
+1

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

 

cool cool cool

 Apr 25, 2019

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