Question

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

(a) $\frac{1}{3}$
(b) $\frac{-1}{3}$
(c) $\frac{2}{3}$
(d) $\frac{-2}{3}$

Answer 1

(c) $\frac{2}{3}$
Here, $\mathrm{p}\left(x\right)=x^2-2x+3k$
Comparing the given polynomial with $a{x}^2+bx+c$, we get:
$a=1,b=-2\mathrm{and}c=3k$
It is given that $\alpha \mathrm{and}\beta$ are the roots of the polynomial.
$$\begin{gathered} \therefore \mathrm{\alpha }+\mathrm{}\mathrm{\beta }=-\frac{b}{a} \\ =>\mathrm{\alpha }+\mathrm{\beta }=-\left(\frac{-2}{1}\right) \\ =>\mathrm{\alpha }+\mathrm{\beta }=2...\left(\mathrm{i}\right) \end{gathered}$$

Also, $\alpha \beta$=$\frac{c}{a}$
$$\begin{gathered} =>\mathrm{\alpha \beta }=\frac{3k}{1} \\ =>\mathrm{\alpha \beta }=3k...\left(\mathrm{ii}\right) \\ \mathrm{Now},\mathrm{\alpha }+\mathrm{}\mathrm{\beta }=\mathrm{\alpha \beta } \\ =>2=3k[\mathrm{Using}(\mathrm{i})\mathrm{and}(\mathrm{ii}\left)\right] \\ =>k=\frac{2}{3} \end{gathered}$$