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Galina-37 [17]
2 years ago
14

Can some one solve this (squar roots) ​

Mathematics
2 answers:
Inessa05 [86]2 years ago
6 0

Answer:

2\sqrt{6}

Step-by-step explanation:

Using the rule of radicals

\sqrt{a} × \sqrt{b} ⇔ \sqrt{ab}

Simplifying the radicals

\sqrt{96}

= \sqrt{16(6)}

= \sqrt{16} × \sqrt{6} = 4\sqrt{6}

\sqrt{24}

= \sqrt{4(6)}

= \sqrt{4} × \sqrt{6} = 2\sqrt{6}

Thus

\sqrt{96} + 2\sqrt{6} - 2\sqrt{24}

= 4\sqrt{6} + 2\sqrt{6} - 4\sqrt{6} ← collect like terms

= 2\sqrt{6}

12345 [234]2 years ago
3 0

Answer:

<h2>2 \sqrt{6}</h2>

Step-by-step explanation:

<h3>to understand the solving steps</h3><h3>you need to know about</h3>
  • simplifying redical
  • PEMDAS

first thing

in your question about 70% is done so need to simplify it a little

\sqrt{96}  + 2 \sqrt{6}  -  2\sqrt{24}

\sqrt{ {4}^{2} \times 6 }  +  2\sqrt{6}  - 2 \sqrt{2 ^{2} \times 6 }

4 \sqrt{6}  + 2 \sqrt{6}  - 4 \sqrt{6}

<h2>2 \sqrt{6}</h2>

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