I NEED HELP RN PLEASEEEE ITS AN EMERGENCY!!

Simplify.

The square root of 24 multiplied by the square root of 12

Answer Choices:

A) 2sqroot 12
B) 6
C) 12sqroot 2
D) 288

I need the CORRECT ANSWER ASAP! (worth 10 points)

Respuesta :

Answer:

The answer is C) [tex]12\sqrt{2}[/tex].

Step-by-step explanation:

For a real number [tex]a[/tex],

[tex]\sqrt{a} \cdot \sqrt{a} = a[/tex].

In other words, multiplying a square root by itself gets rid of the square root.

How does this rule apply here?

[tex]24 = 2\times 12[/tex].

Similarly,

[tex]\sqrt{24} = \sqrt{2}\times \sqrt{12}[/tex].

That is:

[tex]\begin{aligned}\sqrt{24}\times \sqrt{12} &= (\sqrt{2}\times \sqrt{12}) \times \sqrt{12}\\&=\sqrt{2}\times (\sqrt{12}\times \sqrt{12}) && \begin{array}{l}\text{By the associative property}\\\text{of multiplication}\end{array}\\&=\sqrt{2} \times 12\\ &= 12\sqrt{2}\end{aligned}[/tex].

Answer:  The correct answer is:  [C]:

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                   →   " 12sqroot " ;   or, write as:  " 12√2 " .

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Step-by-step explanation:

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√24 * √12 = ?

Let us start by simplifying:  " √24 " ;

   24 = 4 * 6 ;

So;  √24 = √4 *√6  ;

  √4 = 2 ;

So:  " √24  = √4 *√6  =  2√6 " .

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Now, let us simplify:  " √12 " :

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   " √12  = ?  "

       12 = 4 * 3 ;

So:  "√12 = √4 *√3  " ;

    √4 * √3 = 2√3 ;

So:  "√12 = √4 *√3 = 2√3 " .

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We are asked to solve—"simplify"— "√24 * √12 " .

 √24 = 2√6 ; as we simplified above.

 √12 =  2√3  ; as we simplified above.

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So:  " √24 * √12 " ;

     = 2√6  * 2√3 ;  

     = ?  ;  Note:  "2 * 2 = 4 " ; and:  "√6 * √3 = √(6*3) = √18 ;

So;    2√6  * 2√3 ;

     =  4√18 ;

Now, we can simplify this value further:

   by simplifying:  " √18 " ;

       18 = 9 * 2 ;

So:    " √18 = √9 *√2  = 3 √2 " ;

So:  " 4√18 = 4 * (3√2) =  12√2 " .

     →  which  is our answer:  " 12√2 " .

    →  which corresponds to:  

     →  Answer choice:  [C]:  " 12 * sq root 2 " .  {or, write as:  " 12√2 ".}.

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Hope this answer and explanation is of help to you!

       Wishing you the best in your academic endeavors

                    — and within the "Brainly" community!

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