How do you simplify $\sqrt{7} \cdot (2 \sqrt{3} + 3 \sqrt{7})$ $sqrt(7)*(2 sqrt(3) + 3 sqrt(7))$ ?

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Answer 1 · 2015-10-16T09:41:37

$2 \sqrt{21} + 21$ $2sqrt(21)+21$ , or if you prefer $\sqrt{21} \cdot (2 + \sqrt{21})$ $sqrt(21)*(2+sqrt(21))$ .

Expand the multiplications:

$\sqrt{7} \cdot (2 \sqrt{3} + 3 \cdot \sqrt{7}) = 2 \sqrt{3 \cdot 7} + 3 \sqrt{7 \cdot 7}$ $sqrt(7)*(2sqrt(3)+3*sqrt(7)) = 2sqrt(3*7) + 3sqrt(7*7)$

Of course, $\sqrt{7 \cdot 7} = \sqrt{49} = 7$ $sqrt(7*7)=sqrt(49)=7$ , so the expression begins

$2 \sqrt{21} + 3 \cdot 7 = 2 \sqrt{21} + 21$ $2sqrt(21)+3*7=2sqrt(21)+21$ .

Since $21 = \sqrt{21} \cdot \sqrt{21}$ $21=sqrt(21)*sqrt(21)$ , you can factor $\sqrt{21}$ $sqrt(21)$ and obtain

$\sqrt{21} \cdot (2 + \sqrt{21})$ $sqrt(21)*(2+sqrt(21))$

How do you simplify sqrt(7)*(2 sqrt(3) + 3 sqrt(7))√7⋅(2√3+3√7)sqrt(7)*(2 sqrt(3) + 3 sqrt(7))?