Answer:
Oh! that expression is equal to 4
Step-by-step explanation:
[tex]\sqrt{2\sqrt{7} -\sqrt{12} } \cdot \sqrt{2\sqrt{7} +\sqrt{12} }[/tex]
First, rewrite [tex]\sqrt{12} =\sqrt{2 \cdot 2 \cdot 3}=2\sqrt{3}[/tex]
[tex]\sqrt{2\sqrt{7} -2\sqrt{3} } \cdot \sqrt{2\sqrt{7} +2\sqrt{3} }[/tex]
[tex]\sqrt{2\sqrt{7} -2\sqrt{3} } \cdot \sqrt{2\sqrt{7} +2\sqrt{3} }[/tex]
Now we got a difference of two squares, therefore,
[tex](2\sqrt{7} -2\sqrt{3})(2\sqrt{7} +2\sqrt{3})=(2\sqrt{7} )^2-(2\sqrt{3})^2=[/tex]
[tex]= (4 \cdot 7)-(4 \cdot3)=28-12=16[/tex]
[tex]\sqrt{2\sqrt{7} -2\sqrt{3} } \cdot \sqrt{2\sqrt{7} +2\sqrt{3} }=\sqrt{16}[/tex]
[tex]\sqrt{16} = 4[/tex]
