Suppose $p=\sqrt{3}+\sqrt{7}$, $q=\sqrt{8}+\sqrt{2}$ and $r=2\sqrt{3}+2$. Which of the following is true?

The correct option is q < p < r.

To compare the given values $p$, $q$, and $r$, let us first write down their expressions clearly:

$p=\sqrt{3}+\sqrt{7}$

$q=\sqrt{8}+\sqrt{2}$

$r=2\sqrt{3}+2=\sqrt{12}+\sqrt{4}$

Since all three values are positive, comparing $p$, $q$, and $r$ is equivalent to comparing their squares $p^2$, $q^2$, and $r^2$.

Step 1: Calculate the square of each expression

Using the algebraic identity ${(a+b)}^2=a^2+b^2+2ab$:

For $p$:
$p^2={(\sqrt{3}+\sqrt{7})}^2=3+7+2\sqrt{3\times 7}=10+2\sqrt{21}$

For $q$:
$q^2={(\sqrt{8}+\sqrt{2})}^2=8+2+2\sqrt{8\times 2}=10+2\sqrt{16}$

For $r$:
$r^2={(\sqrt{12}+\sqrt{4})}^2=12+4+2\sqrt{12\times 4}=16+2\sqrt{48}$

Alternatively, computing $r^2$ directly from $2\sqrt{3}+2$:
$r^2={(2\sqrt{3}+2)}^2=12+4+8\sqrt{3}=16+8\sqrt{3}$

Step 2: Compare $q^2$ and $p^2$

We have:
$q^2=10+2\sqrt{16}$
$p^2=10+2\sqrt{21}$

Since $16<21$, it follows that $\sqrt{16}<\sqrt{21}$, which means:
$q^2<p^2\Rightarrow q<p$

Step 3: Compare $p^2$ and $r^2$

We have:
$p^2=10+2\sqrt{21}$
Since $21<\sqrt{25}=5$, we know that $p^2<10+2\left(5\right)=20$.

For $r$:
$r=2\sqrt{3}+2\approx 2\left(1.732\right)+2=3.464+2=5.464$
$r^2=16+8\sqrt{3}\approx 16+8\left(1.732\right)=16+13.856=29.856$

Clearly, $p^2<20<r^2$, so $p<r$.

Conclusion:
Combining the inequalities gives:
$q<p<r$