Question

can some one solve this (squar roots) ​

Answer 1

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$\sqrt{6}$

Answer 2

Answer:

$2 \sqrt{6}$

Step-by-step explanation:

to understand the solving steps

you need to know about

• simplifying redical
• PEMDAS

first thing

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

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

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

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

$2 \sqrt{6}$