Question

if -2 and 3 are the zeroes of the quadratic polynomial x²+(a+1)x+b, ​

Answer 1

Step-by-step explanation:

p(-2) = x²+ax + x + b

0 = (-2)²+a(-2)+(-2)+b

= 4-2a-2+b

= 2-2a+b

2a-b = 2 ---->(1)

p(3) = (3)²+a(3)+(3)+b

= 9 +3a+3+b

-12= 3a+b ----> (2)

comparing (1)&(2)

2a-b = 2

3a+b = -12

----------------

5a = 10

a = 2....

from (1) => 2a-b = 2

2(2) - b = 2

4-2 = b

b = 2....

Answer 2

Step-by-step explanation:

Let -3 be alphgussa and -7 be beta

Alpha + beta = -3 + (-7)

= -3-7

= -10

Alpha*beta = -3*-7

= -21

x² - ( alpha + beta )x + alpha* beta = 0

x² + 10x - 21 = 0.

Therefore, The required quadratic equation is x² + 10x - 21 = 0

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