Question

If one zero of the polynomial (a 2 +9)x 2 + 13x + 6a is reciprocal of the other. find the value of a.

Answer 1

given,α=α,β=1/α
we know that , αβ=c/a
                     ⇒α×1/α=6a/a²+9
                     ⇒1=6a/a²+9
                      ⇒a²+9=6a
                     ⇒a²-6a+9=0
                     ⇒a²-3a-3a+9=0
                     ⇒a(a-3)-3(a-3)=0
                     ⇒(a-3)(a-3)=0
                  ⇒a-3=0
                   ⇒a=3

Answer 2

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