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

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

Complex Numbers and Quadratic Equations

Solution:

For the equation to be defined, $x \geq 3$, and $x^2+7 x+10 \geq 40 \forall x \geq 3$. Therefore, the given equation has exactly one root, viz. 3 .