Question

Simplify-

(a)${(10+\sqrt{2})}^2$

(b)${(1+3\sqrt{6})}^2$

Short answer

The values are-

(a)$102+20\sqrt{2}$

(b)$55+6\sqrt{6}$

Step-by-step solution

Step 1. Explanation

(a) and (b) We can identify the property of addition here, i.e., ${(a+b)}^2$.

We know that ${(a+b)}^2=a^2+2ab+b^2$.

By applying it, we will get the values.

Step 2. Solving for part (a)

We have ${(10+\sqrt{2})}^2$.

Applying the property of ${(a+b)}^2$will give us,

role="math" localid="1644765602873" $$\begin{gathered} ={(10+\sqrt{2})}^2 \\ ={\left(10\right)}^2+(2·10·\sqrt{2})+{\left(\sqrt{2}\right)}^2 \\ =100+20\sqrt{2}+2 \\ =102+20\sqrt{2} \end{gathered}$$

Thus we get the value as$102+20\sqrt{2}$.

Step 3. Solving for part (b)

We have ${(1+3\sqrt{6})}^2$.

Applying the property of ${(a+b)}^2$will give us,

role="math" localid="1644765862716" $$\begin{gathered} ={(1+3\sqrt{6})}^2 \\ ={\left(1\right)}^2+(2·1·3\sqrt{6})+{\left(3\sqrt{6}\right)}^2 \\ =1+6\sqrt{6}+54 \\ =55+6\sqrt{6} \end{gathered}$$

Thus we get the value as$55+6\sqrt{6}$.