if one zero of the polynomial ( a^2+9)x^2 +13x + 6a is reciprocal of the other, then find the value of a
Answers
Answer:Given ,
One of the zero of the polynomial is reciprocal of the other.
And the given expression is quadriatic in X with the zeroes of the polynomial being it's roots.
We know that , product of roots of a quariatic , ax^2 + bx + c = 0 is given by c/a.
SO , here ,
AS the zeroes are reciprocal to each other , their product = 1
So, 6a/(a^2 + 9) = 1
a^2-6a + 9 = 0
(a-3)^2 = 0
a = 3
Answer:
Step-by-step explanation:
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polynomial is (a2+9)x2 + 13x + 6a
Let one zero be b then other zero will be reciprocal of it i.e.1/b.
∴ product of the zeroes = constant term/cofficient of x2 = 1 (as b*1/b = 1)
6a/(a2+9) = 1
⇒ 6a = a2+9
⇒ a2 -6a + 9 = 0
⇒ (a-3)2 = 0
⇒ a - 3 = 0
⇒ a = 3
polynomial will be (32+9)x2 + 13x + 6*3
= 18x2 + 13x + 18
This polynomial will be have imaginary roots because b2-4ac<0