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So I was sitting in math class and this hit me, can the square root of a quantity with a constant in it be equivalent to another square root plus an arbitrary constant?

What I did was something like $\frac{\sqrt{9x^2+5}}{2x^3}$ and turned it into $\frac{\sqrt{9x^2}+C}{2x^3}$

I was thinking this could work because each case is only in incremented by a constant, even though the first one is under a square root

I plugged in different numbers for x and then solved for $C$, assuming they would be equal at all values, and what I kept getting was close(especially when the value I chose was around 1) but not exactly the same(except at that one value of x).

Some simple algebra got me a general solution for this approximation as$$\frac{\sqrt{a(x)}+\sqrt{a(k)+C_a}-\sqrt{a(k)}}{b(x)}$$ where a(x) is the original numerator function($9x^2$), b(x) is the denominator function($2x^3$), $C_a$ is the original constant under the radical(5), and k is the chosen point to approximate off of. This approximation does change whether or not the approximation function would look like $\frac{\sqrt{a(x)}+C}{b(x)}$ or $\frac{\sqrt{a(x)}}{b(x)}+C$, this is the approximation for the former

Would it be possible to get an exact other equation, with keeping the square root term the same as before and have to modify the constant with a quantity that changes with x, or would the previous square root also have to change to accommodate the constant being removed?

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    $\begingroup$ Did you mean the first denominator to be different ($2x^3$) from the second ($2x^2$)? $\endgroup$
    – coffeemath
    Apr 26, 2017 at 16:07
  • $\begingroup$ no. proportional change under the radical sign is a different rate than outside. If $\sqrt{x + n} = \sqrt{x} + m$ it will not follow that $\sqrt{x + a*n} = \sqrt{x} + a*m$. If $\sqrt{9x^2 + 5} = \sqrt{9x^2} + C$ then either C is not a constant and is a variable function of $x$ (actually $C(x) =\sqrt{9x^2 +5} - 3|x|$; ex $C(0) = \sqrt 5;C(1)=\sqrt{14} - 3$). Of C is a constant that fixes $x$ and allows us to solve for $x$ ($x=\pm \frac{5-C^2}{6C}$). $\endgroup$
    – fleablood
    Apr 26, 2017 at 19:56

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Supposing you are working with rational numbers: If $x = \sqrt{a} = b + \sqrt{c}$ then $x$ satisfies $x^2-a=0$ and $(x-b)^2-c = x^2-2bx+b^2-c=0$. Eliminating $x^2$ leads to $2b x + c-b^2-a=0$. So either $b=0$, which is not what we want here, or $x$ is itself rational (in which case $a$ and $c$ are both squares).

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You can't do it exactly, but it can be close, particularly when the constant is small compared to the rest. In your example $\sqrt {9x^2+5}=3x\sqrt{1+\frac 5{3x^2}} \approx 3x(1+\frac 5{6x^2})=3x+\frac 5{2x}=\sqrt{9x^2}+\frac 5{2x}$ where the approximation is the first term in the Taylor series and reasonably accurate when $3x^2 \gg 5$ If the range of $x$ is not large you can take it in the middle of the range and say $C=\frac 5{2x}$

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...[C]an the square root of a quantity with a constant in it be equivalent to another square root plus a ... constant?

Absolutely yes. $\sqrt{7} = 2 + c$ for some constant $c$.

The problem here is not that we cannot perform some kind of substitution $\sqrt{a^2 + b} \rightarrow a+c$. Such a step is reasonable. The problem is that $c$ is not actually a constant independent of $a$ or $b$ which you can see directly by trying to get an expression for it. Because your constant is not independent from the variable in question, you're not actually eliminating the square root you want to eliminate but hiding it in some function $C(x)$. If there are other constraints on the problem such that the values of $a$ and $b$ are already fixed then $c$ is indeed a constant. But that's basically not saying much, since if $a$ and $b$ are fixed the entire expression is a constant. Oh well!

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