Q4 of 30 Page 36

Find a quadratic polynomial whose zeros are 2 + √3 and 2 – √3.

(x2 – 2x + k) the remainder comes out to be (x + a). Find k and a.


As the remainder is x + a


⇒ x + a = (2k – 9)x + (k2 – 8k + 10)


Comparing constant and coefficient of x of LHS and RHS


1 = 2k – 9 and a = k2 – 8k + 10


⇒ 2k – 9 = 1


⇒ 2k = 10


⇒ k = 5


Put k = 5 in a = k2 – 8k + 10


⇒ a = 52 – 8(5) + 10


⇒ a = 25 – 40 + 10


⇒ a = 25 – 30


⇒ a = –5


Hence k = 5 and a = –5


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