The answer is: " √3 " .
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→ The simplified expression is: " √3 " .
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Step-by-step explanation:
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We are given:
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→ 
;
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Note the "imaginary number" — " √(-1) " —
which is represented by the symbol: " *i* " .
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So, let us factor out Both the numerator And the denominator:
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Start with the "numerator" ; as follows:
The numerator is: √(-30) ; which factors into:
→ √-1 * √30 ;
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Note that: " √-1 " ; is an imaginary number; which is represented
by the symbol: " *i* " .
As such, we can rewrite the numerator as:
→ *i* √30 .
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The denominator is: √(-10) ; which factors into:
→ √-1 * √10 ;
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As noted above:
→ " √-1 " ; is an imaginary number; which is represented
by the symbol: " *i* " .
As such, we can rewrite the denominator as:
→ *i* √10 .
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Now, let us rewrite the entire expression:
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( *i* √30)
____________
( *i* √10)
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Note: The *i* symbols "cancel out" to: " 1 " ;
→ { since: any "non-zero value" ; divided by "that same value" ;
is equal to: "1 " .} ;
→ { i.e., in our case: [ *i* ÷ *i* ] ; that is; " [*i* / *i* ], equals : " 1 " .}.
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And we are left with:
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→ " 
" .
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Note that both the numerator and the denominator are "square roots" —
and furthermore — are square roots of a "positive number".
Specifically, the "numerator" is: " √30 " ;
And the "denominator is: " √10 " .
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→ Note that the numerator: " √30 " ;
→ can be factored into: "√10 * √3 " ;
→ And that the "denominator" ; which is: " √10 " ; is one of the aforementioned 'factors' of the "numerator" .
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Let us rewrite the expression — and further simplify:
→ ;
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= 
;
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Note: The "(√10)" values, "cancel out" to: "1" ;
→ {since: any "non-zero value" ; divided by "that same value" ;
is equal to: "1 " .} ;
→ { i.e., in our case: [" √10 ÷ √10 " ] ;
→ that is; [ " 
" ]" ;
→ equals : " 1 " .}.
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→ and we are left with: " (√3) / 1 " ;
→ which equals: " √3 " ;
→ which is our answer.
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Hope this is helpful to you.
Best wishes in your academic endeavors
— and within the "" community!
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