If one zero of a polynomial (a2+ 9)x2 + 13x +6a is the reciprocal of the other then find the value of a.
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Answered by
48
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
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
Answered by
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
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