Answers

2014-09-12T16:54:19+05:30
Lets suppose the zero factor is x = b and its reciprocal is x = 1/b 

x = b & x = 1/b <---- as they are zeros. 
sum of zeros = -b/a 
b + 1/b = -13/(a^2 + 9) 
(b^2 + 1)/b = -13/(a^2 + 9) <--- two variables 


product of zeros = c/a 
b * 1/b = 6a/(a^2 + 9) 
1 = 6a/(a^2 + 9) 
(a^2 + 9) = 6a 
a^2 - 6a + 9 = 0 
(a - 3)^2 = 0 ----> a = 3 
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2014-09-12T17:02:55+05:30
Given expression is 9x² + 13x +(a² + 6a) = 0

Let two reciprocal zeroes be k and 1/k.
For the general quadratic equation ax² +bx + c = 0
 Product of zeroes = c/a.

Hence for given equation, product kx(1/k) = 1 = (a² + 6a) / 9
                                                      or a² + 6a - 9 = 0
                  from here a = [-6 + 3√13]/2, [-6 - 3√13]/2

Calculate the values again as per this method if co-efficients of a² and a are other than I have assumed
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