1.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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We know that equation ax
2
+bx+c=0
Then sum of roots =
a
−b
and product of roots=
a
c
Let the other zero be α
Therefore, the other zero is
α
1
Now, α×
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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