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lyudmila [28]
3 years ago
14

What is the solution for the square root of x<9

Mathematics
1 answer:
nydimaria [60]3 years ago
3 0

The square root only accepts non-negative inputs, so the solution will surely be a non-negative value: x \geq 0

Also, the square root of 81 is exactly 9: \sqrt{81} = 9.

Finally, we have to observe that the square root grows with its inputs:

x > y > 0 \implies \sqrt{x} > \sqrt{y}

So, you want the square root of a number to be less than 9, and you know that the square root of 81 is exactly 9, and that the square root gets smaller as the inputs get smaller, and larger as the inputs get larger.

So, the answer is 0 \leq x < 81

If you want to be more rigorous, you can observe that since both sides of the inequality are positive you can square them:

\sqrt{x}

But you can't accept this solution, because you have to remember that x has to be positive, so you have to correct it into

0 \leq x < 81

Just as before

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If -1 &lt; x &lt; 0 which of the following has the smallest value?
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