Question

Question #94618

Answer 1

`sqrt3900=10sqrt39`

Approximate form: `62.44997998` `larr` using calculator

Explanation:

Prime factorize `3900`.

`sqrt3900=sqrt(2xx2xx3xx5xx5xx13)`

Group the same prime factors into pairs.

`sqrt3900=sqrt((2xx2)xx(5xx5)xx3xx13)`

Rewrite each pair in exponent form.

`sqrt3900=sqrt(2^2xx5^2xx3xx13)`

Apply the rule: `sqrt(x^2)=x`

`sqrt3900=(2xx5)sqrt(3xx13)`

Simplify.

`sqrt3900=10sqrt39`

Answer 2

Simplify the square root (by "splitting" then solving), and approximate if needed:

`sqrt(3900) = 10 sqrt(39) ~~ 10 * 6.245 = 62.45`

Explanation:

A rule about exponents is that raising a product to a power, is the same as taking the product of each multiplier having been raised to the same power:

`(ab)^c = a^c b^c`

And the square root of a number can be rewritten as the same number raised to an exponent of `1/2`:

`sqrt(n) = n^(1/2)`

Why is that? Well, when we say square root, we mean a number where:

`sqrt(n) * sqrt(n) = n`

`n = n^1` and a number times itself is the number squared (`a * a = a^2`), so:

`(sqrt(n))^2 = n^1`

So if we rewrite `sqrt(n)` as `n^p` (where we don't yet know the value of `p`), we get:

`(n^p)^2 = n^1`

And `(a^b)^c = a^(bc)`:

`n^(2p) = n^1`

Shouldn't it be that `2p = 1 rarr p = 1/2`?

If `(ab)^c = a^c b^c` and `sqrt(n) = n^(1/2)`, then...

`sqrt(ab) = (ab)^(1/2) = a^(1/2) b^(1/2) = sqrt(a) * sqrt(b)`

This rule is important here! It tells us that we can somehow "split" the square root of a product!

Now back to our problem:

`sqrt(3900) = "???"`

Well, `3900 = 39 * 100`, so:

`sqrt(3900) = sqrt(39) * sqrt(100) = 10 sqrt(39)`

Unfortunately, we can't simplify this any further, but we can approximate `sqrt(39)`:

`sqrt(39) ~~ 6.245`

So:

`sqrt(3900) = 10 sqrt(39) ~~ 10 * 6.245 = 62.45`

Pay attention to the squiggly lines; it indicates an approximation! So if they ask for an exact value, just give `10 sqrt(39)`, but you may use the approximation in solving word problems.