Question

Find the value of a and p ,
if x²-px+28 = (x-4)(x-a) identity.​

Answer 1

Answer:

it implies roots of the given quadratic polynomial are 4 and a.

$so \: 4 and \: a \: are \: the \: roots \: of \: the \: equation \: {x}^{2} - px + 28 \\ so \: product \: of \: \: the \: roots \: a \times 4 = 28 \\ a = 7 \\ now \: sum \: of \: the \: roots \: a + 4 = p \\ p = 7 + 4 = 11$

so a = 7 and p = 11

alter method

${x}^{2} - px + 28 = {x}^{2} - ax - 4x + 4a \\ {x}^{2} - px + 28 = {x}^{2} - (a + 4)x + 4a \\ on \: comparing \\ 4a \: = 28 \\ a = 7 \\ \\ also \: p = a + 4 \\ p = 7 + 4 = 11$