Sure, let's fill in the blanks step-by-step so that the resulting statement is true.
1. [tex]\( a \)[/tex] is defined as [tex]\( \sqrt{20} \)[/tex].
2. [tex]\( b \)[/tex] is defined as [tex]\( \sqrt{6} \)[/tex].
The given expression is [tex]\( (\sqrt{20} + \sqrt{6})(\sqrt{20} - \sqrt{6}) \)[/tex].
We will use the difference of squares formula: [tex]\( (a + b)(a - b) = a^2 - b^2 \)[/tex].
Let's break this down:
- For [tex]\( a \)[/tex]:
[tex]\( a^2 = (\sqrt{20})^2 \)[/tex].
The square of [tex]\( \sqrt{20} \)[/tex] is [tex]\( 20 \)[/tex].
- For [tex]\( b \)[/tex]:
[tex]\( b^2 = (\sqrt{6})^2 \)[/tex].
The square of [tex]\( \sqrt{6} \)[/tex] is [tex]\( 6 \)[/tex].
Thus,
[tex]\[
a^2 = 20
\][/tex]
and
[tex]\[
b^2 = 6
\][/tex]
Now, substituting these into the difference of squares formula:
[tex]\[
(\sqrt{20}+\sqrt{6})(\sqrt{20}-\sqrt{6}) = a^2 - b^2
\][/tex]
[tex]\[
= 20 - 6
\][/tex]
[tex]\[
= 14
\][/tex]
Therefore, the filled-in statement is:
[tex]\[
(\sqrt{20}+\sqrt{6})(\sqrt{20}-\sqrt{6})=\left(20^2-(6)^2=20-6=14\right)
\][/tex]
