Hello!
To solve this exercise, we must simplify these square roots until we have the same square root in both numbers (by the factorization process):
[tex]3\sqrt{98}-\sqrt{128}[/tex]
First, let's factorize the square root of 98:
So, we know that:
[tex]\begin{gathered} 3\sqrt{98}=3\sqrt{7^2\times2}=3\sqrt[\cancel{2}]{7\cancel{^2}\times2}=3\times7\sqrt{2}=21\sqrt{2} \\ \\ 3\sqrt{98}=21\sqrt{2} \end{gathered}[/tex]
Now, let's do the same with the square root of 128:
So:
[tex]\sqrt{128}=\sqrt{2^2\times2^2\times2^2\times2}^1[/tex]
Notice that it also could be written as:
[tex]\begin{gathered} \sqrt{128}=\sqrt{2\times2\times2\times2\times2\times2\times2} \\ \text{ or also} \\ \sqrt{128}=\sqrt{2^7} \end{gathered}[/tex]
As we are talking about square roots, it will be easier if we group them in pairs of powers of 2, as I did:
[tex]\sqrt[2]{128}=\sqrt[2]{2^2\times2^2\times2^2\times2^1}[/tex]
Now, let's analyze it:
If the number inside the root has exponent 2, we can cancel this exponent and remove the number inside the root. Then, we can write it outside of the root, look:
[tex]\begin{gathered} \sqrt[2]{128}=\sqrt[2]{2^{\cancel{2}}\times2^{\cancel{2}}\times2^{\cancel{2}}\times2^1} \\ \sqrt[2]{128}=2\times2\times2\sqrt[2]{2^1} \\ \sqrt[2]{128}=8\sqrt[2]{2} \end{gathered}[/tex]
Now, let's go back to the exercise:
[tex]\begin{gathered} 3\sqrt{98}-\sqrt{128}\text{ is the same as } \\ 21\sqrt{2}-8\sqrt{2} \end{gathered}[/tex]
So, we just have to solve it now:
[tex]21\sqrt{2}-8\sqrt{2}=\boxed{13\sqrt{2}}[/tex]
