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

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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

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