If one zero of the polynomial (a² + 9) x² + 13x + 6a is reciprocal of the other , Find the value of a ?
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Answered by
15
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
Answered by
8
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
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