Question

If one zero of the polynomial (a² + 9) x² + 13x + 6a is reciprocal of the other , Find the value of a ?

Answer 1

Answer:

let g,h be the zeroes of (a^2+9)x^2+13x+6a=0

g=1/h

g+h=-13/a^2+9

gh=6a/a^2+9

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

6a/a^2+9=1

6a=a^2+9

a^2-6a+9=0

a^2-3a-3a+9=0

a(a-3)-3(a-3)=0

(a-3)^2=0

a=3

Answer 2

Answer:

a=3

Step-by-step explanation:

Let the zeroes of p(x) be y and 1/y

P(x)= (a^2+9)x^2+ 13x + 6a

a=a^2 + 9, b=13x, c=6a

Product of zeroes = c/a

y x 1/y = 6a/a^2+9

1= 6a/a^2+9

a^2+9=6a

a^2 - 6a+9

On factorising

a^2 - 3a-3a+9

a(a-3)-3(a-3)

(a-3)(a-3)

a=3