Question

If alpha and beta are the zeroes of px2 - 2x+3p and alpha+beta=alpha.beta,then find the value of p.
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pls answer it fast !!!!!!!!!!!!!!

Answer 1

Answer:

$Value \: of \: p = \frac{2}{3}$

Step-by-step explanation:

Compare given Quadratic expression px²-2x+3p with ax²+bx+c=0 we get

a = p, b = -2 , c = 3p

and

$\alpha \: and \: \beta \: are \\ two \: zeroes .$

$i) Sum\: of \: the \: zeroes = \frac{-b}{a}$

$\implies \alpha+\beta= \frac{-(-2)}{p}=\frac{2}{p}$

$ii)Product\:of\:the\: zeroes =\frac{c}{a}$

$\alpha \beta = \frac{3p}{p}=3$

Now ,

According to the problem given,

$\alpha + \beta = \alpha \beta$

$\implies \frac{2}{p}=3$

$\implies \frac{p}{2}=\frac{1}{3}$

$\implies p = \frac{2}{3}$

Therefore,

$Value \: of \: p = \frac{2}{3}$

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Answer 2

Answer:

Step-by-step explanation:

ompare given Quadratic expression px²-2x+3p with ax²+bx+c=0 we get

a = p, b = -2 , c = 3p

and  $\alpha +\beta =-b/a\\ =-(-2)/p\\\ =2/p\\\alpha \beta =c/a \\ =3p/p\\ =3\\\alpha +\beta =\alpha \beta (given)\\2/p=3\\p=2/3$

∴p=2/3