3 years ago

Could I have assistance, please...? Find the domain of the square-root function.

1 answer:

5 0

Answer:

x≥-8

Step-by-step explanation:

domain of f(x) :

9(x+8)≥0

x+8≥0

x≥-8

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$x^2+10x+22=13\\x^2+10x+22-13=0\\x^2+10x+9=0\\$

Now completing the square.

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(x+5)² = 16 is the rewritten equation... For first question btw.

For the solutions to the equation -

$x+5=\sqrt{16},-\sqrt{16} \\x+5=4,-4\\x=4-5,-4-5\\x=-1,-9$

x = -1,-9 is the solution, for the second question.

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