if the zeros of quadratic polynomial x^2+(a+1)x+b are 2 and -3 then fi...
X^2 + ( a +1) x + b
alpha = 2
beeta = -3
Sum of zeroes = - b / a
alpha + beeta = - ( a + 1) / 1
2 + ( -3) = - a - 1
-1 = - a - 1
-1 +1 = - a
-a = 0
a = - 0 i.e , 0
Product of zeroes = c / a
2 × ( -3 ) = c /1
-6 = c / 1
c = - 6
if the zeros of quadratic polynomial x^2+(a+1)x+b are 2 and -3 then fi...
Given:
Zeros of quadratic polynomial x^2 (a 1)x b are 2 and -3.
To find:
The value of a and b.
Solution:
Given that the zeros of the quadratic polynomial are 2 and -3. We know that the sum of the roots of a quadratic polynomial is given by -b/a and the product of the roots is given by c/a. Therefore,
Sum of roots = 2 + (-3) = -1 = -b/a
Product of roots = 2 x (-3) = -6 = c/a
Multiplying both sides of the first equation with 'a', we get:
-b = a
Substituting this value of 'b' in the second equation, we get:
-6 = c/(-b)
-6 = c/(-a) [Substituting the value of b]
6a = c
Therefore, the value of a is -b and the value of c is 6a.
Substituting the value of a and c in the given quadratic polynomial, we get:
x^2 (-b 1)x 6a
= x^2 -bx 6a
= x^2 -(-a) x 6a
= x^2 + ax - 6a
Therefore, the required quadratic polynomial is x^2 + ax - 6a, where a = -b and c = 6a.
Hence, the value of a and b are unknown unless given the value of c.
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