If one zero of the polynomial (a^2+9)x^2+13x+6a, is reciprocal of the other, find the value of a.
Answers
Consider a second degree polynomial p(x) = ax² + bx + c whose zeroes are reciprocals to each other.
We just call the zeroes α and 1/α.
Hence the factorized form of p(x) will be (x - α)(x - 1/α) if a = 1.
We may take the factorized form of p(x) as (mx - mα)(nx - n/α).
(mx - mα)(nx - n/α) = ax² + bx + c
⇒ mnx² - mnx/α - mnαx + mn = ax² + bx + c
⇒ mnx² - (mn/α + mnα)x + mn = ax² + bx + c
From this, we get,
→ a = mn
→ b = - (mn/α + mnα)
→ c = mn
Seems that a = c. Hence we can say that,
"If the zeroes of a second degree polynomial are reciprocals to each other, then the coefficients of x² and x⁰ will be equal."
Consider the polynomial (a² + 9)x² + 13x + 6a.
Since zeroes are reciprocals to each other, the coefficients of x² and x⁰ are equal. So,
a² + 9 = 6a
⇒ a² - 6a + 9 = 0
⇒ (a - 3)² = 0
⇒ a = 3
Hence, 3 is the answer!
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