The number of integers n such that the quadratic equation n x2+(n+1) x+(n+2)=0 has rational roots is

Question

The number of integers n such that the quadratic equation n x2+(n+1) x+(n+2)=0 has rational roots is (A) 0 (B) 1 (C) 2 (D) 3

Tardigrade Admin · Accepted Answer

Δ=(n+1)2-4 n(n+2)=-(3 n2+6 n-1)=4-3(n+1)2 This takes positive values for n ∈ -2,-1,0 . But n=0 is rejected because above equation is given quadratic and in all these two cases, the value is a perfect square. So, the answer is 2.

Q. The number of integers $n$ such that the quadratic equation $n x^{2} + (n + 1) x + (n + 2) = 0$ has rational roots is

0

1

2

3

$Δ = (n + 1)^{2} - 4 n (n + 2) = - (3 n^{2} + 6 n - 1) = 4 - 3 (n + 1)^{2}$
This takes positive values for $n \in {- 2, - 1, 0}$ .
But $n = 0$ is rejected because above equation is given quadratic and in all these two cases, the value is a perfect square.
So, the answer is 2.

Q. The number of integers n such that the quadratic equation nx2+(n+1)x+(n+2)=0 has rational roots is

0

1

2

3

Δ=(n+1)2−4n(n+2)=−(3n2+6n−1)=4−3(n+1)2 This takes positive values for n∈{−2,−1,0}. But n=0 is rejected because above equation is given quadratic and in all these two cases, the value is a perfect square. So, the answer is 2.

Q. The number of integers $n$ such that the quadratic equation $n x^{2} + (n + 1) x + (n + 2) = 0$ has rational roots is

$Δ = (n + 1)^{2} - 4 n (n + 2) = - (3 n^{2} + 6 n - 1) = 4 - 3 (n + 1)^{2}$
This takes positive values for $n \in {- 2, - 1, 0}$ .
But $n = 0$ is rejected because above equation is given quadratic and in all these two cases, the value is a perfect square.
So, the answer is 2.