How do you solve $3 (\sqrt{12}) + 4 (\sqrt{18})$ $3(sqrt12)+4(sqrt18)$ ?

Question

1469 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-13T14:24:38

See a solution process below:

First, use this rule of radicals to rewrite the expression:

$\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$ $sqrt(a * b) = sqrt(a) * sqrt(b)$

$3 (\sqrt{12}) + 4 (\sqrt{18}) \Rightarrow$ $3(sqrt(12)) + 4(sqrt(18)) =>$

$3 (\sqrt{4 \cdot 3}) + 4 (\sqrt{9 \cdot 3}) \Rightarrow$ $3(sqrt(4 * 3)) + 4(sqrt(9 * 3)) =>$

$3 (\sqrt{4} \cdot \sqrt{3}) + 4 (\sqrt{9} \cdot \sqrt{3}) \Rightarrow$ $3(sqrt(4) * sqrt(3)) + 4(sqrt(9) * sqrt(3)) =>$

$3 (2 \cdot \sqrt{3}) + 4 (3 \cdot \sqrt{3}) \Rightarrow$ $3(2 * sqrt(3)) + 4(3 * sqrt(3)) =>$

$6 \sqrt{3} + 12 \sqrt{3}$ $6sqrt(3) + 12sqrt(3)$

We can now factor a $\sqrt{3}$ $sqrt(3)$ out of each term and calculate the result:

$6 \sqrt{3} + 12 \sqrt{3} \Rightarrow$ $6sqrt(3) + 12sqrt(3) =>$

$(6 + 12) \sqrt{3} \Rightarrow$ $(6 + 12)sqrt(3) =>$

$18 \sqrt{3}$ $18sqrt(3)$

Or

$31.177$ $31.177$ rounded to the nearest thousandth.

How do you solve 3(√12)+4(√18)3(sqrt12)+4(sqrt18)?