Question

If 2 and -3 are the roots of the equation x2 + (a+1)x+b=0,
C
then find a and b.​

Answer 1

Answer:

a = 0 , b = -6

Step-by-step explanation:

Firstly let's find the real equation.

Let α and β be the two roots of this equation.

for any two roots α and β, the equation is given by :

$x^{2}$ - (α + β)$x$ + αβ

Since here α = 2 and β = -3 is given, we can simply insert these value equation.

⇒ $x^{2}$ - (2 - 3)$x$ + 2(-3) = $x^{2}$ + 1$x$ - 6  [This is the original Equation]

now,  $x^{2}$ + (a+1)x + b will be equal to $x^{2}$ + 1$x$ - 6 .

now equate the second term of both of these equations :

⇒ (a+1) = 1    (we only take the coefficient of x)

⇒ a = 0

Similarly we equate the third terms of both the equations:

⇒ b = -6 [Here third term is constant so the entire value will be taken]

Hence,

a = 0 and b = -6

Hope it helps