Question

if one zero of polynomial p(x) = (a²+9) x²+45a+6a is reciprocal of the other,find the value of 'a'

Answer 1

P(x)=(a²+9)x²+45a+6a

Given condition,

On zero is reciprocal to the other

$\sf{let \: one \: zero \: be \: \alpha \: }$

$\sf{other \: zero \: would \: be \: \frac{1}{ \alpha } }$

•Product of zeros of a polynomial: constant term/x²coefficient

Now,

$\sf{ \alpha . \frac{1}{ \alpha } = \frac{6a}{a {}^{2} + 9 } } \\ \\ \implies \: \sf{ \frac{6a}{a {}^{2} + 9 } = 1 } \\ \\ \implies \: \sf{a {}^{2} + 9 = 6a } \\ \\ \implies \: \sf{a {}^{2} - 6a + 9 = 0 } \\ \\ \implies \: \sf{a {}^{2} - 3a - 3a + 9 = 0 } \\ \\ \implies \: \sf{(a - 3)(a - 3) = 0} \\ \\ \implies \: \sf{(a - 3) {}^{2} = 0 \implies \: a - 3 = 0 } \\ \\ \: \implies \underline{\boxed{\sf{a = 3}}}$

The value of a is 3