Another simple square root problem

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The square root of 25 is 5 because the square root function returns only the positive value. However, when solving the equation x² = 25, both x = 5 and x = -5 are valid solutions, leading to the notation x = ±5. This distinction arises because the square root function and the equation-solving process are different; the former yields only the principal root, while the latter considers both roots. The proper method to solve x² - 25 = 0 involves factoring, which clearly shows both solutions. Therefore, it's essential to recognize that equations can have multiple solutions, unlike the square root function.
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why is the square root of 25 just 5

but when the question x^2 = 25, the answer becomes +/- 5?

whats the logic here? i don't quite get it. thanks.
 
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Square root is a function that returns the positive value only.

When you have the eqn,

X² = 25

then both x=5 and x = -5 are solutions of the eqn,

Which is why you write down the answer as x = +/- 5

If you applied the square root function to x², you would get

Sqrt(x²) = |x|
 
so its basically a matter of function vs equation?
 
Solving the equation x^2=25 by taking the square root is a bit sloppy, IMO. What you are really doing is a shortcut for the following:

x^2-25=0

(x-5)(x+5)=0

x-5=0 or x+5=0

x=5[/itex] or x=-5<br /> <br /> When you solve the equation by taking the square root (I should say <b>roots</b>), you have to <i>remember</i> that there are two of them.
 
More or less, yes.
 
In general, an equation like x2= a has two solutions.
One is the square root of a: \sqrt{a}.
The other is the negative of the square root of a:-\sqrt{a}.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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