Answer:
See below.
Explanation:
So we want to prove that:
![\sqrt8+\sqrt2=3\cdot 2^{(1)/(2)}](https://img.qammunity.org/2021/formulas/mathematics/college/w8cdlqrupmcglf3fmnalzm95nm149or2l4.png)
First, simplify √8. This is the same as:
![\sqrt8=√(4\cdot 2)=\sqrt4\cdot\sqrt2=2\sqrt2](https://img.qammunity.org/2021/formulas/mathematics/college/t01dzcj6ppmy1qjlfqone8xefp6tqlmxff.png)
Therefore, our equation is now:
![2\sqrt2+\sqrt2=3\cdot2^{(1)/(2)}](https://img.qammunity.org/2021/formulas/mathematics/college/qh1gwl3vfsord44lkjqvhcmg1shd5pycrm.png)
Combine like terms on the left:
![3\sqrt2=3\cdot 2^(1)/(2)}](https://img.qammunity.org/2021/formulas/mathematics/college/47k5cxwl5hwgrkr1hzhpx4g9uz1kwwo35q.png)
The square root of something is the same as taking that number to the one-half power. Thus:
![3(2)^(1)/(2)}=3\cdot 2^(1)/(2)}](https://img.qammunity.org/2021/formulas/mathematics/college/olpwxfgmr5gmvlgnggtf84kt4qhlp79nmi.png)
Rewrite:
![3\cdot2^(1)/(2)}\stackrel{\checkmark}{=}3\cdot2^(1)/(2)}](https://img.qammunity.org/2021/formulas/mathematics/college/gsjppqdp395ssa85y84rh2ynnej36p615o.png)
And we're done!