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garik1379 [7]
1 year ago
11

Go step by step to reduce the radical of the equation given. (Figure the step for the box)

Mathematics
1 answer:
Rama09 [41]1 year ago
7 0

\sqrt{32} ~~ \begin{cases} 32=2\cdot 2\cdot 2\cdot 2\cdot 2\\ \qquad 2^4\cdot 2\\ \qquad 2^{(2)(2)}\cdot 2\\ \qquad (2^2)^2\cdot 2 \end{cases}\implies \sqrt{(2^2)^2\cdot 2}\implies 2^2\sqrt{2}\implies 4\sqrt{2}

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Answer:

You want to solve for x, the first thing you need to do is distribute

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2(x+4)= 2x+8

2(-8-x)= -16-2x

2x+8=-16-2x-2x (combine like terms on the right side)

2x+8=-16-4x (now subtract 8 from each side)

2x=-24-4x (8-8=0, -16-8=-24) (now add 4x to each side)

6x=-24 (-4x+4x=0, 2x+4x=6x), (divide each side by 6)

x=-4 (6/6=1, -24/6=-4)

x=-4

Hope this helps ;)

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