Question

If the zeros of the quadratic polynomials x2+(a+1)x+b are 2&-3 then values of a and b are

Answer 1

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Answer 2

Answer:

$a = -1$ and $b = - 6$

Step-by-step explanation:

As the quadratic polynomial is $x^2 +(a+1)x+b$.

Let its roots be α and β.

So, α = 2 and β = -3 (Given)

We know that

Sum of zeroes = α + β = 2 + (-3) = 2 - 3 = -1

Sum of zeroes = $-\cfrac{b}{a} = -\cfrac{(a+1)}{1}$

∴ $-\cfrac{a+1}{1} = -1$

⇒ $a=-1$

Now,

Product of roots = α · β = (2) (-3) = -6

Product of roots = $\cfrac{c}{a} = \cfrac{b}{1}$

∴ $\cfrac{b}{1} = -6$

⇒ $b = -6$