Question

if alpha +beeta are zeros of x^2
-p(x+1)+c such that (alpha+1)(beeta +1)=0,find the value of c​

Answer 1

Answer:

GIVEN EQUATION :–

${x}^{2} - p(x + 1) + c = 0$

IT'S ROOTS ARE –

$\alpha \: \: \: and \: \: \beta$

=》

$( \alpha + 1)( \beta + 1) = 0$

$\alpha \beta + ( \alpha + \beta ) + 1 = 0 \\ \\ \frac{c - p}{1} + \frac{p}{1} + 1 = 0 \\ \\ c - p + p + 1 = 0 \\ \\ c = - 1$

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