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Last question thanks guys!! (only 1!)

Last question thanks guys!! (only 1!)-example-1

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x=\pm4 √(2) \\ x^2=(4 √(2) )^2 \\ x^2=4^2\cdot(√(2) )^2 \\ x^2=16\cdot (2) \\ x^2=32
User Ieggel
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The correct answer is: [C]: " x² = √32 " .
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Step-by-step explanation:
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Note: Working backward, "(± 4√2)" , squared, equals what value(s)?
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Note: We are actually given 2 (TWO) solutions:

" +4√2" and "–4√2" ;
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√32 = √8 √4 = √4 √2 √4 = 4√2 ;

- √32 = -1 * √32 = - 4√2
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or:

√32 = √16 √2 = 4√2 ;

-1 * √32 = - 4√2
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However, BOTH these values, when "squared" (i.e. raised to the exponential power of "two"; will result in the same value— which is: "32" ;
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is equal to: "(4√2)² = (4)² *(√2)² = (4*4) (√2*√2) = 16*2 = " 32 " .

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√32 = √8 √4 = √4 √2 √4 = 4√2 ;

- √32 = -1 * √32 = - 4√2
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or:

√32 = √16 √2 = 4√2 ;

-1 * √32 = - 4√2
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" (4√2)² = (4)² * (√2)² = (4 * 4) (√2*√2) = 16 * 2 = " 32 " .

" (-4√2)² = (-4)² * (√2)² = (-4 *-4) (2) = 16 * 2 = " 32 " .
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The correct answer is: " 32 " ; which is: Answer choice: [D]: " x² = 32 " .

We know that the answer is: "
" {" ± 8 "}; since we are dealing with equations that contain "x SQUARED"; that is, " x²"; and when we solve for the 'square root of all values of a [variable raised to an even positive integer] ;

we take the "plus or minus" square root values ; since the value, "x" could be plus or minus; since a "negative value" multiplied by a "negative value" is a "positive value" ;
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So, the correct answer is: Answer choice: [B]: " x² = ± 8 " .
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Let us check our answer:

√32 = √16*√2 = 4√2 ;

– √32 = – 4√2 ;
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Also, note:
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x² = 32 ; Solve for "x" ;

→ Take the square root of "each side" of the equation ;
to isolate "x" on one side of the equation; & to solve for "x" ;

→ √(x²) = √32 ;

→ | x | = √32 ;

→ x = ± √32 . Yes!
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User Kiryl Tkach
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