How do you multiply $\sqrt{2} (3 \sqrt{14} - \sqrt{7})$ $sqrt2(3sqrt14 - sqrt7)$ ?

Question

1706 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-27T00:33:25

$6 \sqrt{7} - \sqrt{14}$ $6sqrt7-sqrt14$

Multiply $\sqrt{2}$ $sqrt2$ by each term in the parentheses

$\sqrt{2} (3 \sqrt{14} - \sqrt{7})$ $sqrt2(3sqrt14-sqrt7)$

$\sqrt{2} \cdot 3 \sqrt{14} - \sqrt{2} \cdot \sqrt{7}$ $sqrt2*3sqrt14-sqrt2*sqrt7$

$3 \sqrt{28} - \sqrt{14} \to 3 \sqrt{28}$ $3sqrt28-sqrt14 rarr 3sqrt28$ can be simplified

$3 \sqrt{28} = 3 \sqrt{4} \cdot \sqrt{7} \to$ $3sqrt28=3sqrt4*sqrt7 rarr$ $4$ $4$ is a perfect square

$6 \cdot \sqrt{7} = 6 \sqrt{7} = 3 \sqrt{28}$ $6*sqrt7=6sqrt7=3sqrt28$

$6 \sqrt{7} - \sqrt{14}$ $6sqrt7-sqrt14$

How do you multiply √2(3√14−√7)sqrt2(3sqrt14 - sqrt7)?