Question

6. If zeroes of the polynomial x^2 + ( a + 1 )x + b are 2 and -3, then find the value of ( a + b ).

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Answer 1

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Answer 2

Given Equation is x^2 + (a + 1)x + b.

On comparing with ax^2 + bx + c, we get a = 1, b = (a + 1), c = b

Let a, b be the zeroes of the Quadratic polynomial.

Given that 2 and -3 are the zeroes of the polynomial.

We know that sum of zeroes = -b/a

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

= > 2 + (-3) = -(a + 1).

= > -1 = -a - 1

= > a = 0

We know that product of zeroes = c/a

= > a * b = b/1

= > 2 * (-3) = b

= > -6  = b.


Therefore the values of a = 0, b = -6.



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