Question 3 and 4

3. If the zeroes of the quadratic polynomial x ² + (a + 1)x + b are 2 and −3, then find the value of a and b.

Dear Student,

Answer 3: 

When 2 and -3 are the zeroes of the polynomial p(x) = x² + (a + 1) x + b

Therefore,

p(2) = 2²+ (a + 1) 2 + b = 0

4 + 2a + 2 + b = 0

6 + 2a + b = 0

2a + b = -6 - - - - - - - - - - (1)

Now,

p(-3) = -3²+ (a + 1) -3 + b = 0

9 - 3a - 3 + b = 0

6 - 3a + b = 0

-3a + b = -6

3a - b = 6 - - - - - - - - -- (2)

Adding (1) and (2),

2a + b = -6

3a - b = 6

- - - - - - - - -

5a = 0
$a=\frac{0}{5}$

a = 0

Now, Substituting the value of a in 2

3 x 0 - b = 6

-b = 6

b = -6

Therefore,

a = 0 and b = -6

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