Question

HELP ME FRIENDS PLZZZ ​

Answer 1

SOLUTION ☺️

=) Polynomial is (a^2+9)x^2+13x+6a

Let one zero be B then other zero will be reciprocal of it 1/B

Therefore,

Product of zeroes= constant term/coefficient of x^2

=) 1 (as B×1/b= 1)

=) 6a/(a^2+9)= 1

=) 6a= a^2+9

=) a^2-6a+9= 0

=) (a-3)^2=0

=) a-3= 0

=) a= 3

Polynomial will be (3^2+9)x^2+13x+6×3

=) 18x^2+13x+18

This polynomial will be have seems roots because,

=) b^2- 4ac<0

hope it helps ✔️