Question

" alt="\sqrt 45+\sqrt 20-\sqrt5\\" align="absmiddle" class="latex-formula">

Answer 1

Answer:

$\mathbf{4\sqrt{5}=\sqrt{80}}$

Step-by-step explanation:

one thing you should keep in mind while solving questions related to surds.

• $\mathit{\sqrt{a^{4} \cdot b}=\sqrt{a^{2}} \cdot \sqrt{a^{2}} \cdot \sqrt{b}=a \cdot a \cdot \sqrt{b}=a^{2}\sqrt{b}}$
• $\mathit{\sqrt{a^{5}}=\sqrt{a^{2}} \cdot \sqrt{a^{2}} \cdot \sqrt{a}=a \cdot a \cdot \sqrt{a}=a^{2} \cdot \sqrt{a}}$

from the above examples you can see that if you have a square of any number inside a square root then you can remove the square root and the square from that number or in other words you can take the number out of the square root.

Similarly in this question:

$\sqrt{45}=\sqrt{9 \cdot 5}=\sqrt{3^{2} \cdot 5}=\sqrt{3^{2}} \cdot \sqrt{5}=3\sqrt{5}$

$\sqrt{20}=\sqrt{4 \cdot 5}=\sqrt{2^{2} \cdot 5}=\sqrt{2^{2}} \cdot \sqrt{5}=2\sqrt{5}$

∴ $\sqrt{45}+\sqrt{20}-\sqrt{5}=3 \cdot \sqrt{5}+2 \cdot \sqrt{5}-1 \cdot \sqrt{5}$

Now take out $\sqrt{5}$ common from all the terms of the above expression.

$\sqrt{5} \cdot (3+2-1)=\sqrt{5} \cdot 4=4\sqrt{5}$

In this you can send the terms back inside the square root. For example:

• $\mathit{a=\sqrt{a^{2}}}$
• $\mathit{a^{3}\sqrt{b}=a \cdot a \cdot a \cdot \sqrt{b}=\sqrt{a^{2}} \cdot \sqrt{a^{2}} \cdot \sqrt{a^{2}} \cdot \sqrt{b}=\sqrt{a^{2} \cdot a^{2} \cdot a^{2} \cdot b}=\sqrt{a^{6}b}}$

Similarly $4\sqrt{5}=\sqrt{4^{2} \cdot 5}=\sqrt{16 \cdot 5}=\sqrt{\mathbf{80}}$