Question

$hello \: mates \: solve \: this$
[tex]If 2 \: and -3 are \: the \: zeroes \: of \: the \: quadratic \: polynomial \: x2
\ \textless \ br /\ \textgreater \ +(a+ 1) x+
\ \textless \ br /\ \textgreater \ b; \: \: then find the values of a.[/tex]
If 2 and -3 are the zeroes of the quadratic polynomial x2

+(a+ 1) x+

b; then find the values of a. ​

Answer 1

Step-by-step explanation:

If 2 and -3 are the zeros of

$f(x) = {x}^{2} + (a + 1)x + b$

then

$f(2) = 0 \\ \\f(2) = {(2)}^{2} + (a + 1)2 + b = 0 \\ \\4 + 2a +2 + b = 0 \\ \\ 2a + b + 6 = 0 \\ \\ 2a + b = - 6...eq1 \\ \\ and \: f( - 3) = 0 \\ \\ f( - 3) = ( { - 3)}^{2} + (a + 1)( - 3) + b = 0 \\ \\ 9 - 3a - 3 + b = 0 \\ \\ - 3a + b = - 6 \\ \\ 3a - b = 6...eq2$

Since eq1 and eq2 are pair of Linear equations in two variables,then

add eq1 and eq2

$2a + b = - 6 \\ 3a - b = 6 \\ - - - - - - \\ 5a = 0 \\ \\ a = 0 \\ \\ put \: the \: value \: of \: a \: in \: eq1 \\ \\ b = - 6$

Thus the value of a = 0 and b = -6.

Hope it helps you.

Answer 2

Answer:

value of a=0 and b=-6

.......