If the zeroes of quadratic polynomial x2+(a+1)x+b are 2 and -3 then
Answers
Step-by-step explanation:
zero = 2
2^2 + (a+1)2 + b = 0
4+2a + 2 + b = 0
2a + b = -6....…......(1)
zero = -3
-3^2 + (a+1)-3 + b = 0
9 -3a -3 + b = 0
b - 3a = -6..............(2)
by subtracting (2) from (1) we get...
5a = 0
a = 0
b = -6
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Answer: a = 0, b = -6
Step-by-step explanation:
p(x) = x² + (a + 1)x + b
Zeroes of this polynomial are 2 and (-3). This means if we put 2 and (-3) as x in the above equation, we should get 0.
p(2) = 2² + (a + 1)2 + b = 0
4 + 2a + 2 + b = 0
2a + b + 6 = 0 (First Equation)
p(-3) = (-3)² + (a + 1)(-3) + b = 0
9 - 3a - 3 + b = 0
-3a + b + 6 = 0 (Second Equation)
Now, we have two equations and two variables. So we can find the values of a and b.
First Equation: 2a + b + 6 = 0
b = -2a - 6
Put the value of b obtained here in the second equation.
Second Equation: -3a + b + 6 = 0
-3a + (-2a - 6) + 6 = 0
-3a - 2a - 6 + 6 = 0
-5a = 0
a = 0
Now, find the value of b.
b = -2a - 6
b = -2(0) - 6
b = 0 - 6
b = -6