Question

let
A ={m:m an integer and the roots of $x^{2} +mx+2020=0$ are positive integers }
and
B ={n:n an integer and the roots of $x^{2} +2020x+n=0$ are positive integers }

Suppose a is the largest element of A and b is the smallest element of B. Find the sum of digits a+b.

Answer 1

Answer:

a+b = -121

Step-by-step explanation:

Given: A = {m:m an integer and the roots of x²+mx+2020=0 are positive integers}  and  B = {n:n an integer and the roots of x²+2020x+n=0 are positive integers}

Consider, x²+mx+2020 = 0

⇒ α+β = -b/a = -m and αβ = c/a = 2020

Also, roots of the equation x²+mx+2020 = 0 are positive integers i.e., α and β are positive integers.

So, m wil be definitely negative.

So, possible pairs of (α,β) can be (20,101) , (4,505) , (2,1010) and (1,2020).

⇒ Possible values of m (=α+β) can be -121 , -509 , -1012 and -2011.

⇒ A = {-2011 , -1012 , -509 , -121}

⇒ Largest element of set A = -121 = a

Now consider, x²+2020x+n = 0

⇒ α+β = -b/a = -2020 and αβ = c/a = n

Also, roots of the equation x²+2020x+n = 0 are positive integers i.e., α and β are positive integers.

But, sum of two positive integers can never be a negative number.

⇒ There is no possible values of n.

⇒ B = {Ф}

⇒ Smallest element of set B = Ф = b

⇒ a+b = -121+Ф

⇒ a+b = -121