Question

If one zero of a polynomial (a2+ 9)x2 + 13x +6a is the reciprocal of the other then find the value of a.

Answer 1

let alpha & 1/alpha be the 2 zeroes
a=a^2+9, b=13, c=6a
pr. of zeroes= c/a
alpha×1/alpha =6a/a^2+9
1=6a/a^2+9
a^2+9=6a
a^2-6a+9=0
(a-3)^2 =0 ( (a-b)^2= a^2+b^2-2ab)
a=3

Answer 2

Answer:

Step-by-step explanation:

Let the other zero be α

Therefore, the other zero is   1/a

​  

 

Now, α×  1/a=1

α

1

​  

=  

a  

2

−9

6a

​  

 

=>1=  

a  

2

−9

6a

​  

 

=>a  

2

+9−6a=0

=>a  

2

−6a+9=0

=>a  

2

−3a−3a+9=0

=>a(a−3)−3(a−3)=0

=>(a−3)(a−3)=0

=>a=3 and a=3