# Confusion over simple surds

### Confusion over simple surds

Why does $$\sqrt{180}$$ = 6$$\sqrt{5}$$
But $$\sqrt{48}$$ = 4$$\sqrt{3}$$

They are unrelated, but the way I remember doing these is sort of confusing me. This is how I've been asked to do it:

For example:
Why does $$\sqrt{2}$$ x $$\sqrt{2}$$ x $$\sqrt{3}$$ x $$\sqrt{3}$$ become 2 x 3 x $$\sqrt{5}$$
But then $$\sqrt{2}$$ x $$\sqrt{2}$$ x $$\sqrt{2}$$ x $$\sqrt{2}$$ x $$\sqrt{3}$$ become 4$$\sqrt{3}$$ ?

Am I making sense?
Guest

### Re: Confusion over simple surds

$$\sqrt{2} \times \sqrt{2} = (\sqrt{2})^2 = 2$$

$$\sqrt{2} \times \sqrt{2} \times \sqrt{2} \times \sqrt{2} \times \sqrt{3} = (\sqrt{2} \times \sqrt{2}) \times (\sqrt{2} \times \sqrt{2}) \times \sqrt{3} =$$

$$2 \times 2 \times \sqrt{3} = 4 \times \sqrt{3}$$ or $$4\sqrt{3}$$
Guest

### Re: Confusion over simple surds

Your math question here is absurd! Please get help at https://www.math10.com/forum/viewforum.php?f=25 or try somewhere else.
Guest

### Re: Confusion over simple surds

We are basically making use of two rules:
$$\sqrt{a}\times\sqrt{a} = (\sqrt{a})^2 = a$$ (square rooting a number then squaring it, leaves you with what you started with)
$$\sqrt{a\times b} = \sqrt{a} \times \sqrt{b}$$
To see why the second rule works consider the squares of both sides.
The square of the left hand side is $$a \times b$$ because of the first rule.
The square of the right hand side is:
$$(\sqrt{a} \times \sqrt{b})^2 = (\sqrt{a} \times \sqrt{b})\times (\sqrt{a} \times \sqrt{b})$$
$$= \sqrt{a} \times \sqrt{b}\times \sqrt{a} \times \sqrt{b}$$
$$= \sqrt{a} \times \sqrt{a}\times \sqrt{b} \times \sqrt{b}$$ (the order of the terms in a multiplication is allowed to change)
$$= (\sqrt{a})^2 \times (\sqrt{b})^2$$
$$= a \times b$$ (using the first rule to get rid of square roots)
So both sides of the second rule square to the same thing so the second rule holds.

In fact the second rule generalises a bit: Consider $$\sqrt{a\times b\times c}$$ by the second rule:
$$\sqrt{a\times b\times c} = \sqrt{a\times (b\times c)}$$
$$= \sqrt{a}\times \sqrt{b\times c}$$ (by the second rule applied to $$a$$ and $$b\times c$$)
$$= \sqrt{a}\times \sqrt{b}\times\sqrt{c}$$ (by the second rule applied to $$b$$ and $$c$$)
So $$\sqrt{a\times b\times c}= \sqrt{a}\times \sqrt{b}\times\sqrt{c}$$
and similarly we can show $$\sqrt{a\times b\times c\times d}= \sqrt{a}\times \sqrt{b}\times\sqrt{c}\times \sqrt{d}$$, etc.

Applying the rules to $$\sqrt{180}$$ we get
$$\sqrt{180} = \sqrt{2\times 2 \times 3 \times 3 \times 5}$$ (from the prime factorization of 180)
$$= \sqrt{2}\times \sqrt{2} \times \sqrt{3} \times\sqrt{3} \times \sqrt{5}$$ (by the second rule)
$$= (\sqrt{2})^2 \times (\sqrt{3})^2 \times \sqrt{5}$$
$$= 2 \times 3 \times \sqrt{5}$$ (by the first rule applied to both 2 and 3)
$$= 6 \times \sqrt{5}$$
$$= 6\sqrt{5}$$ (the multiplication symbol is usually omitted, to keep the expressions as short as possible)

Applying the rules to $$\sqrt{48}$$ we get
$$\sqrt{48} = \sqrt{2\times 2 \times 2 \times 2 \times 3}$$ (from the prime factorization of 48)
$$= \sqrt{2}\times \sqrt{2} \times \sqrt{2} \times\sqrt{2} \times \sqrt{3}$$ (by the second rule)
$$= (\sqrt{2})^2 \times (\sqrt{2})^2 \times \sqrt{3}$$
$$= 2 \times 2 \times \sqrt{3}$$ (by the first rule)
$$= 4 \times \sqrt{3}$$
$$= 4\sqrt{3}$$ (the multiplication symbol is usually omitted, to keep the expressions as short as possible)

Hope this helped,

R. Baber.
Guest

nice post
Guest

### Re: Confusion over simple surds

Guest wrote:Why does $$\sqrt{180}$$ = 6$$\sqrt{5}$$

$$180= 9(2)(10)= 9(2)(2)(2)(5)= (3^2)(2^2)(5)$$ so $$\sqrt{180}= \sqrt{3^2}\sqrt{2^2}\sqrt{5}= 3(2)\sqrt{5}= 6\sqrt{5}$$

But $$\sqrt{48}$$ = 4$$\sqrt{3}$$[/tex]I'm not sure why you say "but". The two are not really related. $$48= 16(3)$$ so $$\sqrt{48}= \sqrt{16}\sqrt{3}= 4\sqrt{3}$$.
They are unrelated, but the way I remember doing these is sort of confusing me. This is how I've been asked to do it:

For example:
Why does $$\sqrt{2}$$ x $$\sqrt{2}$$ x $$\sqrt{3}$$ x $$\sqrt{3}$$ become 2 x 3 x $$\sqrt{5}$$

??? It DOESN'T! $$\sqrt{2}$$ x $$\sqrt{2}$$ x $$\sqrt{3}$$ x $$\sqrt{3}$$= 2(3)= 6. Did you drop a "$$\sqrt{5}$$" from the left side?

But then $$\sqrt{2}$$ x $$\sqrt{2}$$ x $$\sqrt{2}$$ x $$\sqrt{2}$$ x $$\sqrt{3}$$ become 4$$\sqrt{3}$$ ?

$$\sqrt{2}\sqrt{2}= 2$$ (by the definition of "$$\sqrt{}$$)

wwAm I making sense?

Yes, but you seem to be forgetting that the square root is the "inverse" of squaring: $$\sqrt{x^2}= |x|$$ (so $$\sqrt{x^2]= x$$ as long as x is positive and $$\left(\sqrt{x}\right)^2= x$$.
HallsofIvy

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