# If one zeroes of polynomial 1a2-9x2+13x+6a is reciprocal of other, find the value of a

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is the question correct?

i mean should it be (a2+9)x2+13x+6a??

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is the question correct?

i mean should it be (a2+9)x2+13x+6a??

Log in to add a comment

x = b & x = 1/b <---- as they are zeros.

sum of zeros = -b/a

b + 1/b = -13/(a^2 + 9)

(b^2 + 1)/b = -13/(a^2 + 9) <--- two variables

product of zeros = c/a

b * 1/b = 6a/(a^2 + 9)

1 = 6a/(a^2 + 9)

(a^2 + 9) = 6a

a^2 - 6a + 9 = 0

(a - 3)^2 = 0 ----> a = 3

Let two reciprocal zeroes be k and 1/k.

For the general quadratic equation ax² +bx + c = 0

Product of zeroes = c/a.

Hence for given equation, product kx(1/k) = 1 = (a² + 6a) / 9

or a² + 6a - 9 = 0

from here a = [-6 + 3√13]/2, [-6 - 3√13]/2

Calculate the values again as per this method if co-efficients of a² and a are other than I have assumed