# How do you simplify sqrt(75/4 )?

Jun 11, 2018

$\setminus \frac{5 \sqrt{3}}{2}$

#### Explanation:

When you need to simplify numerical root, you should look for these two key facts:

• $\sqrt{{a}^{2}} = a$
• $\sqrt{a b} = \sqrt{a} \sqrt{b}$

So, anytime you have a number inside a root, you should try to write it as a product of other numbers, of which at least one is a perfect square. Let's analyze your case.

First of all, using the second property, we can write

$\sqrt{\frac{75}{4}} = \frac{\sqrt{75}}{\sqrt{4}}$

In fact, every fraction can be read as multiplication using

$\setminus \frac{a}{b} = a \cdot \setminus \frac{1}{b}$

Now let's deal with the two roots separately: I'd start with the denominator, since we already have a perfect square under a square root, so they simplify (see first rule above): we have

$\frac{\sqrt{75}}{\sqrt{4}} = \frac{\sqrt{75}}{2}$

As for $\sqrt{75}$, we can see that $75 = 25 \cdot 3$, and $25 = {5}^{2}$ is a perfect square. So, by the second rule above, we have

$\sqrt{75} = \sqrt{25 \cdot 3} = \sqrt{25} \cdot \sqrt{3} = 5 \sqrt{3}$

$\sqrt{\frac{75}{4}} = \setminus \frac{5 \sqrt{3}}{2}$

But how do we find the most appropriate way to rewrite our number, in this case $75 = 25 \cdot 3$? You can use the prime factorization:

$75 = 3 \cdot {5}^{2}$

and select only the primes with even exponent. In this case, ${5}^{2}$.

Jun 11, 2018

$\pm \frac{5 \sqrt{3}}{2}$

#### Explanation:

If you are stuck it is always worth have in a 'play' with numbers and see what comes up.

Consider the 75. This is exactly divisible by 5 so $75 \div 5 = 15$

Thus we have: $\sqrt{\frac{5 \times 15}{4}}$

But 15 is the same as $3 \times 5$ giving:

$\sqrt{\frac{5 \times 5 \times 3}{4}}$

$\frac{\sqrt{{5}^{2} \times 3}}{\sqrt{4}} = \pm \frac{5 \sqrt{3}}{2}$