If α, β are the zeroes of
Question:

If $\alpha, \beta$ are the zeroes of $k x^{2}-2 x+3 k$ such that $\alpha+\beta=\alpha \beta$, then $k=?$

(a) $\frac{1}{3}$

(b) $\frac{-1}{3}$

(c) $\frac{2}{3}$

(d) $\frac{-2}{3}$

 

Solution:

(c) $\frac{2}{3}$

Here, $\mathrm{p}(x)=x^{2}-2 x+3 k$

Comparing the given polynomial with $a x^{2}+b x+c$, we get:

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

It is given that $\alpha$ and $\beta$ are the roots of the polynomial.

$\therefore \alpha+\beta=-\frac{b}{a}$

$=>\alpha+\beta=-\left(\frac{-2}{1}\right)$

$=>\alpha+\beta=2 \quad \ldots(\mathrm{i})$

Also, $\alpha \beta=\frac{c}{a}$

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

$=>\alpha \beta=3 k \quad \ldots($ ii $)$

Now, $\alpha+\beta=\alpha \beta$

$=>2=3 k \quad[\mathrm{Using}(\mathrm{i})$ and $(\mathrm{ii})]$

$=>k=\frac{2}{3}$

Administrator

Leave a comment

Please enter comment.
Please enter your name.