Question

determine the value of the q and p

$x^{2} -qx-28=(x+7)(x-p)$

Answer 1

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RHS;-

$(x + 7)(x + p) \\ {x}^{2} + 7x + xp + 7p \\ {x}^{2} + (7 + p)x + 7p$

LHS :-

${x}^{2} - qx - 28$

Both:-

${x}^{2} - qx - 28 = {x}^{2} - ( - 7 - p) - ( - 7p) \\ so \\ 28 = -7p \: \: \: and \: \: \: qx =( - 7 - p )x\\ p = - 4 \: \: and \: q = - 7 - p$

Put p for q'value

$q = - 7 - 4 \\ q = - 11 \: \: ans \:$

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