How do you add $3 \sqrt{50} + \sqrt{242}$ $3\sqrt { 50} + \sqrt { 242}$ ?

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Answer 1 · 2017-12-19T17:34:45

$26 \sqrt{2}$ $26sqrt(2)$

$3 \sqrt{50} + \sqrt{242}$ $3sqrt50+sqrt(242)$

The best first step is to decompose the radicand into prime factors:
$3 \sqrt{5^{2} \cdot 2} + \sqrt{2 \cdot 11^{2}}$ $3sqrt(5^2*2)+sqrt(2*11^2)$

THen you pass all the squared prime numbers outside

$3 \cdot 5 \sqrt{2} + 11 \sqrt{2}$ $3*5sqrt(2)+11sqrt(2)$

Use $\sqrt{2}$ $sqrt(2)$ as an incognite:

$15 \sqrt{2} + 11 \sqrt{2}$ $15sqrt(2)+11sqrt(2)$

$26 \sqrt{2}$ $26sqrt(2)$

How do you add 3\sqrt { 50} + \sqrt { 242}3√50+√2423\sqrt { 50} + \sqrt { 242}?