Q6 of 113 Page 71

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

We have,

(a + 9) x2 – 13x + 6a = 0


Here, we have:


A = (a2 + 9), B = 13 and C = 6a


Let us assume be the zeros of the given polynomial


∴ Product =


=


1 =


a2 + 9 = 6a


a2 – 6a + 9 = 0


a2 – 2 × a × 3 + 32 = 0


(a – 3)2 = 0


a – 3 = 0


a = 3


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