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If one zero of quadratic polynomial f(x)= kt^2 +2t +3k is equation to their product, find the value of k.

Posted by Amit Patle 6 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 7 months ago

Let the two zeroes of the polynomial f(t) = kt² + 2t + 3k be α and β.we know that
Sum of the roots = {tex}-\frac ba{/tex}
Product of the roots = {tex}\frac ca{/tex}
And we have a = k, b = 2 and c = 3k.
Sum of the zeroes = α + β = {tex}-\frac {{ 2}}{k}{/tex}
Product of the zeroes = α {tex}\times{/tex} β = 3
Now,
{tex}\frac{{ - 2}}{k} ={/tex} 3 {tex}\Rightarrow{/tex} 3k = - 2 {tex}\Rightarrow k = \frac{{ - 2}}{3}{/tex}
Hence, required value of k is {tex} \frac{{ - 2}}{3}{/tex}.

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