What is the right method for solving a problem like this:
”$4\sqrt{5}$ is the same as which square root?"
Possible answers are:
- $\sqrt{20}$
- $\sqrt{10}$
- $\sqrt{40}$
- $\sqrt{80}$
I have been informed that $\sqrt{80}$ is the right answer, but I do not understand why.
$\endgroup$ 56 Answers
$\begingroup$Because $4=\sqrt{16}$. Then $\sqrt{16}\cdot \sqrt{5} = \cdots$
$\endgroup$ 3 $\begingroup$I prefer this formulation, particularly for beginners:
$x=4 \times \sqrt{5} = \sqrt{16} \times \sqrt{5} = \sqrt{16\times5} = \sqrt{80}$
This is how I did it in my head when reading your question.
It's also good practice for solving proofs when you have to appear to go in the wrong direction for a moment (making $4$ lexically "larger", as $\sqrt{16}$) in order to bring terms together.
$\endgroup$ 2 $\begingroup$$4\sqrt{5} = \sqrt{x} \Leftrightarrow \sqrt{4^25} = \sqrt{x}$
since $\sqrt{x}$ is injective $x = 4^25 = 80$
$\endgroup$ 3 $\begingroup$$x=4\sqrt 5\implies x^2=16\cdot 5=80\implies x=\sqrt{80}$
$\endgroup$ 2 $\begingroup$Hint:
use: $a\sqrt{b}=\dfrac{a}{|a|} \sqrt {a^2b}= \mbox{sign}(a)\sqrt {a^2b}$
Here $|a|$ is the absolute value of $a$, so that $\dfrac{a}{|a|}= 1$ if $a>0$ and $\dfrac{a}{|a|}= -1$ if $a<0$. This same result can be represented wit the function "sign (a)" that is simply the sign of $a$.
The use of this function is very important when we work with radicals because the square of a number is always positive, so, squaring $a$ we can forget that the basis of the square was negative.
In you case $a=4$, so sign$(4)=+1$ and you have no problem writing: $$ 4 \sqrt{5}=\sqrt{16 \times 5} $$
but if you have , e.g. : $a=-4$ the correct result is:
$$ -4 \sqrt{5}=-\sqrt{16 \times 5} $$
$\endgroup$ 4 $\begingroup$Notice, in general, for any two positive real numbers $a$ & $b$ $$a\sqrt{b}=\sqrt{a^2b}$$ Hence, we have $$4\sqrt{5}=\sqrt{(4)^2(5)}$$ $$=\sqrt{16\times 5}=\sqrt{80}$$ Yes $\sqrt{80}$ is right answer.
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