Question

51. Assertion (A) If one zero of the polynomial p(x) = (k+18)x² + 11x + 4k is the reciprocal of the other zero, then k=6 Reason (R) If (x-a) and (x - B) are the factor of the polynomial p(x), then a and ß are the zeroes of the p(x)

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Answer 1

Assertion and reason both are correct but reason is not the correct explanation of assertion.

Solution: For the polynomial $f(x) = (k + 18) {x}^{2} + 11x + 4k$ one of the root is the reciprocal of the other root.

Let the root be a.

Then, other root= 1/a

Product of roots

= a × 1/a

= 1

But for a quadratic polynomial $a {x}^{2} + bx + c$ the product of the roots is given by c/a.

Here, a = k+18, b= 11 , c = 4k

Product of roots

= c/a

= 4k/k+18

Therefore,

4k / k+18 = 1

=> 4k = k+18

=> 4k-k = 18

=> 3k = 18

=> k= 18/3

=> k = 6

Therefore, the value of k is 6.

The remainder theorem states that when f(x) is divided by x-a, the remainder is given by f(a).

If (x-a) and (x-b) are the factors of p(x) that means that they divide f(x) completely leaving remainder 0. The remainder is given by f(a) and f(b) which is 0. Therefore, a and b are the zeroes of p(x).