How do you simplify $\sqrt{78}-\sqrt{8}+\sqrt{128}$ ?

Hint: We know that $\sqrt{ab}$ is equal to product of $\sqrt{a}$ and $\sqrt{b}$ .By using this formula we can write square root of 78 as product of square root of 39 and 2, square root of 8 as product of square root of 4 and 2, square root of 128 as product of square root of 64 and 2. Then we can easily simplify.

Complete step by step answer:
The given equation is $\sqrt{78}-\sqrt{8}+\sqrt{128}$
We know that $\sqrt{ab}$ is equal to product of $\sqrt{a}$ and $\sqrt{b}$
So $\sqrt{78}=\sqrt{39}\times \sqrt{2}$
Similarly, we can write $\sqrt{8}=\sqrt{4}\times \sqrt{2}$ and $\sqrt{128}=\sqrt{64}\times \sqrt{2}$ , we know that square root of 4 is equal to 2 and square root of 64 is equal 8.
So $\sqrt{8}$ is equal to $2\sqrt{2}$ and $\sqrt{128}$ is equal to $8\sqrt{2}$
Replacing above value in the equation we get $\sqrt{78}-\sqrt{8}+\sqrt{128}=\sqrt{39}\sqrt{2}-2\sqrt{2}+8\sqrt{2}$
We can see that $\sqrt{2}$ is common in each term, $\sqrt{78}-\sqrt{8}+\sqrt{128}=\left( \sqrt{39}-2+8 \right)\sqrt{2}$
So $\sqrt{78}-\sqrt{8}+\sqrt{128}=\left( \sqrt{39}+6 \right)\sqrt{2}$
$\left( \sqrt{39}+6 \right)\sqrt{2}$ is the simplified form of $\sqrt{78}-\sqrt{8}+\sqrt{128}$

Note:
$\sqrt{ab}$ is not equal to the product of $\sqrt{a}$ and $\sqrt{b}$ when a and b both are negative numbers. Let’s take an example a = -1 and b = -1 . Product of -1 and -1 is equal to 1. Square root of 1 is equal to 1. So $\sqrt{ab}$ is equal to 1. The value of $\sqrt{a}$ is complex number i and $\sqrt{b}$ is equal to i . product of $\sqrt{a}$ and $\sqrt{b}$ is equal to square of i which is equal to -1. So $\sqrt{ab}$ is not equal to $\sqrt{a}$ $\sqrt{b}$ when a and b both are negative numbers.