Question

The set of values of p for which the roots of the equation


$3x {}^{2} + 2x + (p - 1)p = 0$
are of opposite sign is _______?​

Answer 1

This is a question on quadratic equations funda.

For ax2+bx+c=0,

if the two roots are of different signs, their product must be –ve => c/a < 0

=> p(p-1)/3 < 0 => p(p-1) < 0 => 0 < p < 1

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Answer 2

Heya....

Products of the roots is negative

$\frac{(p - 1)p}{3} < 0$

$(p - 0)(p - 1) < 0$

$i.e \: 0 < p < 1$

Hence set of values is (0,1)

Hope it helps