The number real of roots of √x-3(x2+7 x+10)=0, where x ∈ R is

Question

The number real of roots of √x-3(x2+7 x+10)=0, where x ∈ R is (A) 0 (B) 1 (C) 2 (D) 3

Tardigrade Admin · Accepted Answer

For the equation to be defined, x ≥ 3, and x2+7 x+10 ≥ 40 ∀ x ≥ 3. Therefore, the given equation has exactly one root, viz. 3 .

Q. The number real of roots of $x - 3 (x^{2} + 7 x + 10) = 0$ , where $x \in R$ is