Question

If the zeroes of quadratic polynomial x^2+(a+1)x+b are 2 and -3,then what is the value of a and b? ​

Answer 1

Answer:

a=0 n b = -6

Step-by-step explanation:

x2−(α+β)x+αβ , where α and β are two zeroes of the equation.

According to the problem statement, we are given a quadratic polynomial x2+(a+1)x+b having 2 and -3 as zeroes. Therefore, α=2 and β=−3.

By using the above expression, the sum of zeroes could be expressed as

−(α+β)=a+1−(2+−3)=a+1−(−1)=a+1a=1−1a=0

The product of zeroes could be expressed

αβ=b2×−3=bb=−6

So, the respective values are a = 0 and b = -6.

Therefore, option (d) is correct.

Note: This problem could be alternatively solved by using the concept that zeros can also be expressed as factors. Now, we have -3 and 2 as zeros. Therefore, the polynomial would be represented as (x+3)⋅(x−2)=x2+x−6. Now, on comparing this equation with the given problem, we obtain a = 0 and b = -6.