Question

if zero of the polynomial (a2+9)x2+13x+6a is reciproca of the other , find the value of a

Answer 1

i think the answer is a= -3
do alpha multiplied by 1/alpha=c/a
alpha multiplied by 1/alpha=1,then
c/a=6a/a2+9
equate it
1=6a/a2+9
a= -3

Answer 2

Let one One zero = a, the other be = 1/a
Product = c/a = 6a/ a2 + 9
a* 1/a = 6a / a2 +9
a2 + 9 = 6a
a2 + 9 - 6a = 0 [This is in the form of (a - b) 2]
a2 + 32 – 2*a*3 = 0
[a – 3]2 = 0
a – 3 = 0
a = 3
     (or you may factorise)
a2 + 9 - 6a = 0
a2 - 6a + 9 = 0
a2 - 3a -3a + 9
a (a - 3) -3 (a - 3) = 0
(a - 3) (a - 3) = 0
a = 3 , a= 3
The answer is a = 3