Question

Find all values of p for which the roots of the equation (p-3)x^2 -2px + 5p = 0 are real and positive

Answer 1

apply the concepts of location of roots
for real and positive roots
two condition need to be filled
1. D>=0
2. -b/2a >0

$for \: first \: d \geqslant 0 \\ p^{2} - (5p(p - 3)) \geqslant 0\\ 4p ^{2} - 15p \leqslant 0 \\ p(4p - 15) \leqslant 0 \\ 0 \leqslant p \leqslant \frac{15}{4}$
$for \: second \: \frac{ - b}{2a} \geqslant 0 \\ \frac{p}{p - 3} \geqslant 0 \\ 0 < p < 3$
intersection of above two will give us
$0 < p < 3$

Answer 2

Answer:

You have to use the method of location of roots to get the answer correctly...checknout my solution and mark me brainliest if you like the soln...thank you.

Step-by-step explanation:

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