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if the zeros of quadratic polynomial x^2+(a+1)x+b are 2 and -3 then find the value of a and b.
Most Upvoted Answer
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
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
Community Answer
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.
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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if the zeros of quadratic polynomial x^2+(a+1)x+b are 2 and -3 then find the value of a and b.
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if the zeros of quadratic polynomial x^2+(a+1)x+b are 2 and -3 then find the value of a and b. for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about if the zeros of quadratic polynomial x^2+(a+1)x+b are 2 and -3 then find the value of a and b. covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for if the zeros of quadratic polynomial x^2+(a+1)x+b are 2 and -3 then find the value of a and b..
if the zeros of quadratic polynomial x^2+(a+1)x+b are 2 and -3 then find the value of a and b. for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about if the zeros of quadratic polynomial x^2+(a+1)x+b are 2 and -3 then find the value of a and b. covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for if the zeros of quadratic polynomial x^2+(a+1)x+b are 2 and -3 then find the value of a and b..
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