We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
108
2
avatar

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+18460 
+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+101870 
+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

5 Online Users

avatar
avatar