# Find the range of $y=\sqrt {x^2+2x+3}$

I want to find the range of: $$y=\sqrt {x^2+2x+3}$$

I would like to know if we can solve this by writing $$x$$ in terms of $$y$$ and then finding the domain of that? If so how?

• Range in real numbers for square root requires the expression to be greater or equal to zero. Oct 20, 2019 at 14:36
• Where is the square root negative? When $x^2+2x+3<0$....there is the restriction for the inputs. Which then helps with the restrictions for the output... Oct 20, 2019 at 14:36

Note

$$y = \sqrt{ (x+1)^2 +2}\ge \sqrt2$$

for all real $$x$$.

Since $$B^2 < 4AC$$ and $$A>0$$ so the quadratic is positive definite for all real values of $$x$$ so the domain of the funtion is $$(-\infty, \infty)$$ and the $$f(x)=\sqrt{(x+1)^2+2}$$ so the range is $$[\sqrt{2}, \infty)$$.

It’s clear that as $$x$$ approaches infinity, the function approaches infinity. So we have to find the minimum of the quadratic.

The axis of symmetry is $$\frac{-b}{2a}$$ which is equal to $$-1$$. Then plug in $$-$$ into your quadratic and get $$2$$. So the minimum of your function is $$2^{0.5}$$

Assuming $$x \in \mathbb{R}$$, we have $$x^2 + 2x + 3 \geq 2$$ for all $$x \in \mathbb{R}$$.

$$\implies$$ $$y \geq \sqrt{2}$$.

The range is $$[\sqrt{2}, +\infty)$$.

Or you can start by writing $$x^2 + 2x + 3 = y^2$$

$$\implies$$ $$(x+ 1)^2 + 2 - y^2 = 0$$ $$\implies$$ $$x = -1 \pm \sqrt{y^2 -2}$$.

Let $$f(y) = -1 \pm \sqrt{y^2 - 2}$$.

The set of values of $$y$$ for which $$f(y) \in \mathbb{R}$$ is $$y^2 \geq 2$$ $$\implies$$ $$y \geq \sqrt{2}$$.

• why >= 2? Why not >=0? Oct 20, 2019 at 14:37
• But I want to know if we can write x in terms of y and then finding the domain of that? Oct 20, 2019 at 14:43
• But did you write y≥ √2? Oct 20, 2019 at 14:56

$$y = \sqrt {x^2+2x+3} \text{ iff } y^2 = x^2+2x+3 \land y \ge 0 \text{ iff }$$

$$\quad x^2+2x+3 - y^2 = 0 \land y \ge 0 \text{ iff } (x+1)^2 + 2-y^2 = 0 \land y \ge 0 \text{ iff }$$

$$\quad (x+1)^2 = y^2 - 2 \land y \ge \sqrt 2 \text{ iff } x + 1 = \pm(\sqrt{y^2-2}) \land y \ge \sqrt 2 \text{ iff }$$

# $$\quad y \ge \sqrt 2 \land \big[ x = \sqrt{y^2-2} - 1 \text{ or } x = -\sqrt{y^2-2} - 1 \big]$$

• What's the vertical symbol you used "^".what does that mean? Oct 20, 2019 at 16:13
• Logical symbol for conjunction $\quad \land = \text{and}$ Oct 20, 2019 at 16:18