How do you find the domain of $\sqrt{x^{2} + 2 x - 24}$ $sqrt(x^2+2x-24)$ ?

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Answer 1 · 2017-05-24T14:29:44

the domain is: $] - \infty, - 6] \cup [4, + \infty [$ $]-oo,-6]uu[4,+oo[$

You cannot have negative values under a square root, then it must be:

$x^{2} + 2 x - 24 \geq 0$ $x^2+2x-24>=0$

The zeros of the polynomial $x^{2} + 2 x - 24$ $x^2+2x-24$ are

$x = - 1 \pm \sqrt{1 + 24} = - 1 \pm 5$ $x=-1+-sqrt(1+24)=-1+-5$

that are $x = - 6$ $x=-6$ and $x = 4$ $x=4$

then the domain is:

$] - \infty, - 6] \cup [4, + \infty [$ $]-oo,-6]uu[4,+oo[$

How do you find the domain of sqrt(x^2+2x-24)√x2+2x−24 sqrt(x^2+2x-24)?