Question

find the value of a for which the zero of the polynomial x2 - 6x + a satisfy the relation 3 alpha + 2 beeta = 20

Answer 1

Given: alpha and beta are the zeroes of given p(x)= x²-6x+a

we know that,

◎sum of zeroes= -b/a

$\alpha + \beta = 6$
———————————————(1)
also,

◎product of zeroes= c/a

$\alpha \beta = a$
—————————————————(2)
now,

$3 \alpha + 2 \beta = 20 \: \: \: \: \: (given)$
——————————————————(3)

◎ multiply 2 to equation (1) and subtract (1) from (2)

$3 \alpha + 2 \beta = 20 \\ 2 \alpha + 2 \beta = 12 \\ ...................... \\ \: \: \alpha \: \: \: \: \: = 8 \\ ......................$
put this value in (1)

$8 + \beta = 6 \\ \\ \beta = - 2$
$substitute \: the \: value \: of \: \\ \alpha \: and \: \beta \: in \: (2)$
$a = \alpha \beta \\ \\ a = 8 \times - 2 \\ \\ a = - 16$

Answer 2

Answer:

please inbox me

Step-by-step explanation:

Sorry I didn't know the answer