Question

If one zero of the polynomial (a²-9) x² + 13x + 6a is reciprocal of the other, find the value of a

Answer 1

Answer:

Answer :-

→ a = 3 .

Step-by-step explanation :-

Let one zero of the following polynomial be α .

Then, the other zero is 1/α .

therefore∴ product of zeros = ( α × 1/α ) = 1 .

But, product of zeros = ( constant term )( coefficient of x² ) = 6a(a² + 9 )

therefore∴ 6a/(a² + 9) = 1 .

⇒ a² + 9 = 6a .

⇒ a² + 9 - 6a = 0 .

⇒ a² + 3² - 2(a)(3) = 0 .

⇒ ( a - 3 )² = 0 .

⇒ a - 3 = 0 .

therefore∴ a = 3 .

Hence, a = 3 .

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Answer 2

Step-by-step explanation:

let the zeroes be p and 1/p

product of zeroes= p×1/p = 1 =c/a

1=6a/a²-9

a²-9= 6a

a²-6a -9a =0

here D = b²-4ac = 36 + 36

D= 72

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