Answer :

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]

Thanks for taking the time to read Add and subtract square roots that need simplification Number 186. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!

Rewritten by : Barada