Question

Can someone please show me their working on these on how to work out?​

Answer 1

To rationalize essentially means to get rid of the radical.

The trick to all of these it to multiply the top and bottom by a radical that will turn the radical you have into the square root of a perfect square.

Part a: $\dfrac{1}{2\sqrt{5}}$

One approach that will always work, but not be the smallest radical, is to just multiply by the radical that is already there.

In the case of $\dfrac{1}{2\sqrt{5}}$, you can multiply the top and bottom by $\sqrt{5}$, so you'll really do this:

     $\dfrac{1}{2\sqrt{5}}\cdot\dfrac{\sqrt{5}}{\sqrt{5}} = \dfrac{1\cdot\sqrt{5}}{2\sqrt{5}\sqrt{5}}$

But in the bottom, $\sqrt{5}\cdot\sqrt{5} = \sqrt{25}=5$, so your work finishes out like this:

    $\dfrac{1\cdot\sqrt{5}}{2\sqrt{5}\sqrt{5}} = \dfrac{\sqrt{5}}{2\sqrt{25}} = \dfrac{\sqrt{5}}{2\cdot5} = \dfrac{\sqrt{5}}{10}$

Part b: $\dfrac{4}{5\sqrt{2}}$

In the case of $\dfrac{4}{5\sqrt{2}}$, you can multiply the top and bottom by $\sqrt{2}$, so you'll really do this:

     $\dfrac{4}{5\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}$

But in the bottom, $\sqrt{2}\cdot\sqrt{2} = \sqrt{4}=2$, so your work finishes out like this:

    $\dfrac{4}{5\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{4\sqrt{2}}{5\sqrt{4}}=\dfrac{4\sqrt{2}}{5\cdot{2}}=\dfrac{4\sqrt{2}}{10}$

Sometimes the fraction can be reduced, so this should be reduced to $\dfrac{2\sqrt{2}}{5}$.

Looking at all the others, they'll follow this same approach and they will all reduce at the end.

Here's part c, for one more example.

Part c: $\dfrac{3}{3\sqrt{2}}$

In the case of $\dfrac{3}{3\sqrt{2}}$, you can multiply the top and bottom by $\sqrt{2}$, but you should also reduce the initial fraction first.

     $\dfrac{3}{3\sqrt{2}} = \dfrac{1}{1\cdot\sqrt{2}} = \dfrac{1}{\sqrt{2}}$

And now proceed with rationalizing:

    $\dfrac{1}{\sqrt{2}} \cdot\dfrac{\sqrt{2}}{\sqrt{2}}$

But in the bottom, $\sqrt{2}\cdot\sqrt{2} = \sqrt{4}=2$, so your work finishes out like this:

    $\dfrac{1}{\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{1\cdot\sqrt{2}}{\sqrt{4}}=\dfrac{\sqrt{2}}{{2}}$

And you're done.

Hopefully that helps you finish the rest of those.

To check your answers, here is the answer key for the rest:

part d:$\dfrac{2\sqrt{6}}{5}$

part e:$\dfrac{4\sqrt{2}}{3}$

part f:$\dfrac{4\sqrt{5}}{7}$

part g:$\dfrac{5\sqrt{10}}{3}$

part d:$\dfrac{5\sqrt{2}}{3}$

Answer 2

Answer:

Step-by-step explanation:

(a) 1/(2 sqrt5) x (sqrt 5)/(sqrt5) = (sqrt 5)/10

(b) 4/(5bsqrt2) x (sqrt2)/(sqrt2) = (4sqrt2)/10 = (2sqrt2)/5

(c) 3/(3 sqrt2) x (sqrt2)/(sqrt2) =(sqrt2)/2

(d) 12 /(5 sqrt6) x (sqrt6)/(sqrt6) = (12sqrt6)/30 = (2sqrt6)5

(e) 8/(3 sqrt2) x (sqrt2)/(sqrt2) = (8 sqrt2)/6 = (4 sqrt2)/3

(f) 20/(7 sqrt5) x (sqrt5)/(sqrt5) = (20 sqrt5)/35 = (4 sqrt5)/7

(g) 50/(3 sqrt10) x (sqrt10)(sqrt10) = (50 sqrt10)/30 = (5sqrt10)/3

(h) 10/(3 sqrt2) x (sqrt2)/(sqrt2) = (10 sqrt2)6 = (5 sqrt2)/3