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Q3 of 113 Page 71

If α, β are the zeros of kx2 – 2x + 3k such that α + β = αβ then k = ?

We have,

p (x) = x2 – 2x + 3k


Now by comparing the given polynomial with ax2 + bx + c, we get:


a = 1, b = - 2 and c = 3k


In the question it is given that, are the roots of the given polynomial


∴ =


= -( )


= 2 (i)


:


=


=


= 3k (ii)


Hence, by using (i) and (ii), we have


=


2 = 3k


k =

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Questions · 113
2. Polynomials
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