If $\alpha,\ \beta$ are zeroes of $x^2-6x+k$, what is the value of $k$ if $3\alpha+2\beta=20$?
Given: $\alpha,\ \beta$ are zeroes of $x^2-6x+k$ and $3\alpha+2\beta=20$.
To do: To find the value of $k$.
Solution:
$3\alpha+2\beta=20\ ...............( i)$ [given]
$\alpha+\beta=-\frac{b}{c}$
$=-( -\frac{6}{1})$
$\alpha+\beta=6\ ..............( ii)$
$\alpha\times \beta=\frac{c}{a}$
$=\frac{k}{1}$
$\alpha\times \beta=k\ ..........( iii)$
By elimination method in equation $( i)$ and $( ii)$
$3\alpha+2\beta=20$
$2\times (\alpha+\beta=6)$
now,
$3\alpha+2\beta=20$
$2\alpha+2\beta=12$
we get, $\alpha=8$
Substituting $\alpha=8$ in equation $( ii)$
$\alpha+\beta=6$
$\Rightarrow 8+\beta=6$
$\Rightarrow \beta=6-8$
$\Rightarrow \beta=-2$
if $\alpha=8$ and $\beta=-2$
By equation $( iii)$ that is
$\alpha\times\beta=k$
$8\times(-2)=-16$
Therefore, $k=-16$
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