A good place to start is to set $\sqrt{x}$ to y. That would mean we are looking for $\sqrt{120-y}$ to be an integer. Clearly, $y\leq 120$ , because if y were greater the part under the radical would be a negative, making the radical an imaginary number, not an integer. Also note that since $\sqrt{x}$ is a radical, it only outputs values from $[0,\infty]$ , which means y is on the closed interval: $[0,120]$ .

With that, we don't really have to consider y anymore, since we know the interval that $\sqrt{x}$ is on.

Now, we don't even have to find the x values. Note that only 11 perfect squares lie on the interval $[0,120]$ , which means there are at most 11 numbers that x can be which make the radical an integer. All of the perfect squares are easily constructed. We can say that if k is an arbitrary integer between 0 and 11 then:

$\sqrt{120-\sqrt{x}}=k \implies \\ \sqrt{x}=k^2-120 \implies\\ x=(k^2-120)^2$

Which is strictly positive so we know for sure that all 11 numbers on the closed interval will yield a valid x that makes the radical an integer.

5 0

3 years ago

What's the answer to this absolute value problem? |w-3|=4

ikadub [295]

Find all solutions for w by breaking the absolute value into the positive and negative components.
w = 7, -1
Hope this helps! :)

6 0

3 years ago

Are the ratios 3:6 and 6:3 equivalent? Why or why not?

Leno4ka [110]

Answer:
Yes!
Step-by-step explanation:
It is because your still getting the same number 3 because even though your multiplying and dividing it still works

3 0

3 years ago

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