Question

if the sum of the zeros of the quadratic polynomial k y square + 2 Y - 3 K is equal to twice their product find the value of k​

Answer 1

According to the question,

(α+β) = 2(αβ)

This implies

$\frac{ - b}{a} = 2( \frac{c}{a} )$

$= > \frac{ - 2}{k} = 2( \frac{ - 3k}{k} )$

$= > \frac{ - 2}{k} = 2( - 3) \: \: \: \: (k \: \: and \: k \: get \: cancelled)$

$= > \frac{ - 2}{k} = - 6$

$= > - 2 = - 6k$

$= > k = \frac{ - 2}{ - 6}$

$= > k = \frac{1}{3}$

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Hence,

$\blue{\fbox{\huge{\star{k={\frac{1}{3}}}}}}$

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