Question

find the value of 'p' for which -4 is a zero of x^{2} -2x-(7p+3)

Answer 1

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$Find \ the \ value \ of \ 'p ' \ for \ which \ -4 \ is \ a \\ zero \ of \ x^{2} -2x-(7p+3)$

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$\pink{\tt{Let \ other \ zero \beta}}$

$\boxed{\red{\bold{Sum \ of \ zeroes:}}}$

$\red{\implies}-4 + \beta = \displaystyle \frac{-b}{a}$

$\red{\implies}-4 + \beta = \displaystyle \frac{-(-2)}{1}$

$\red{\implies}-4 + \beta = 2$

$\red{\implies}\beta = 2+4 = 6$

$\boxed{\red{\bold{Product\ of \ zeroes:}}}$

$\green{\implies}-4 × \beta = \displaystyle \frac{c}{a}$

$\green{\implies}-4 × 6 = \displaystyle\frac{-7p-3}{1}$

$\green{\implies}-24 = -7p -3$

$\green{\implies}7p = 24-3$

$\green{\implies}7p = 21$

$\green{\boxed{\implies{\boxed{p =3}}}}$

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