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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f(x) = (a²+9)x²+3x+6a
if one zero is the reciprocal of the other.
product of the zeroes = 1.
6a/a²+9 = 1
6a = a² + 9
0 = a² - 6a + 9
a² - 3a - 3a + 9
a (a - 3) -3(a - 3)
(a - 3)(a - 3)
(a - 3)²=0
thus a-3 =0 i.e. a =3
thus the value of a is 3
if one zero is the reciprocal of the other.
product of the zeroes = 1.
6a/a²+9 = 1
6a = a² + 9
0 = a² - 6a + 9
a² - 3a - 3a + 9
a (a - 3) -3(a - 3)
(a - 3)(a - 3)
(a - 3)²=0
thus a-3 =0 i.e. a =3
thus the value of a is 3
hummi1:
In equation there is 13x but you put 3x
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◆Ekansh Nimbalkar◆
hello friend here is your required answer
hello friend here is your required answer
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