Question

If 2 and -3 are the zero's of the quadratic polynomial x2+ (a+1)x+ b ,then find the value of a?

Answer 1

Answer:

A = - 1, b = - 3

Step-by-step explanation:

Given,

P(x) = x^2 + ax + b

Let alfa = 2

Let beta = - 3

Now,

Sum of zeros = alfa + beta = 2 + (-3) = - 1

Product of zeros = alfa × beta = 2 × (-3) = - 6

So, the required quadratic polynomial is

X^2 - ( sum of zeros)x + (product of zeros)

X^2 - x - 6

So the values are - 1 and - 6

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Answer 2

ANSWER:

• Value of 'a' is 0 and 'b' is (-6)

GIVEN:

• First zero (α) = 2
• Second zero (β) = (-3)
• P(x) = x²+(a+1)x+b

TO FIND:

• Value of 'a'.

SOLUTION:

Finding sum of zeros (α+β)

=> (α+β) = 2+(-3)

=> (α+β) = (-1)

Finding product of zeros (αβ)

= 2(-3)

= -6

Formula:

=> Sum of zeros (α+β) = -(Coefficient of x)/Coefficient of x². .....(i)

P(x) = x²+(a+1)x+b

Here:

Coefficient of x = (a+1)

Coefficient of x² = 1

Constant term = b

Putting the values in the formula (i) we get;

=> -1 = -(a+1)/1

=> -1 = -a-1

=> -1+1 = a

=> a = 0

Formula:

=> Product of zeros (αβ) = Constant term/ Coefficient of x²

=> -6 = b/1

=> b = (-6)

Value of 'a' is 0 and 'b' is (-6)