How do you simplify $7 \sqrt{28}$ $7sqrt(28)$ ?

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Answer 1 · 2018-03-05T19:19:05

You find perfect squares that are factors in the radical.

$\sqrt 28$ $√28$

$\sqrt 4 = 2$ $√4=2$

$7 \cdot 2 \sqrt 7$ $7*2√7$

$14 \sqrt 7$ $14√7$

Answer 2 · 2018-03-05T19:53:24

$14 \sqrt{7}$ $14sqrt(7)$

$28$ $28$ is the same as $2 \times 14$ $2xx14$ witch in turn is the same as $2^{2} \times 7$ $2^2xx7$

Write as $7 \sqrt{2^{2} \times 7}$ $7sqrt(2^2xx7)$

Taking the $2^{2}$ $2^2$ 'outside' the square root changes it from $2^{2} to 2$ $2^2" to "2$

$7 \sqrt{2^{2} \times 7} = 7 \times 2 \times \sqrt{7}$ $7sqrt(2^2xx7) = 7xx2xxsqrt(7)$

$14 \sqrt{7}$ $14sqrt7$

How do you simplify 7sqrt(28)7√287sqrt(28)?