Square Root of a Quantity of X

  • #1
How does one calculate the square root of a quantity of x e.g. [itex]\sqrt{25x}[/itex]?

Also could you equate 5x[itex]^{2}[/itex] to 25x?

Just a thew random queries going around my head.

NOTE: I'm 14.
 

Answers and Replies

  • #2
mathman
Science Advisor
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The square root of an unknown (x) is simply another unknown (√x).

You could write 5x2 = 25x. This will hold for x = 0 or x = 5.
 
  • #3
35,062
6,796
How does one calculate the square root of a quantity of x e.g. [itex]\sqrt{25x}[/itex]?
There are properties of square roots that can be used here; namely that for nonnegative real numbers a and b,
[itex]\sqrt{ab} = \sqrt{a}\sqrt{b}[/itex]
So, [itex]\sqrt{25x} = \sqrt{25}\sqrt{x} = 5\sqrt{x}[/itex]
Also could you equate 5x[itex]^{2}[/itex] to 25x?
mathman already gave an answer to this, but it's worthwhile to find out exactly what you're asking.

When you ask about equating the two quantities, yes, there are a couple of values that make the equation true.

If you're asking whether you can simplify 5x2 to 25x, then no, these two quantities are not identically equal. I'm not sure that this is what you're asking, so I thought I would check.
Just a thew random queries going around my head.

NOTE: I'm 14.
 
  • #5
Another question.

How would I work out [itex]\frac{p}{4}[/itex]=[itex]\frac{y}{2}[/itex] in terms of y?
 
  • #6
FeDeX_LaTeX
Gold Member
437
13
Another question.

How would I work out [itex]\frac{p}{4}[/itex]=[itex]\frac{y}{2}[/itex] in terms of y?

What can you do to both sides to get y on its own?

Multiply both sides by 2.
 
  • #7
What can you do to both sides to get y on its own?

Multiply both sides by 2.

Times by 2?
 

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