MATH SOLVE

2 months ago

Q:
# Let a = x^2 + 4. Rewrite the following equation in terms of a and set it equal to zero. ( x^2 + 4 )^2 +32 = 12x^2 + 48In resulting equation what is the coefficient of the a term ?In the resulting equation what is the constant?

Accepted Solution

A:

Answer: In the resulting equation: " a² - 12a + 32 = 0 " ;

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The "coefficient" of the "a" term is: " - 12" .

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The "constant" is: " 32 " .

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Explanation:

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Let: "a = x² + 4 " .

Given: (x² + 4)² + 32 = 12x² + 48 ;

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Factor: "12x² + 48" into " (x² + 4) " ;

"12x² + 48" = 12 (x² + 4) " ;

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Given: (x² + 4)² + 32 = 12x² + 48 ;

rewrite as; "a² + 32 = 12a " ;

Subtract "12a" from each side of the equation;

"a² + 32 - 12a = 12a - 12a ;

to get:

" a² - 12a + 32 = 0 " .

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The coefficient of the "a" term; that is:

The "coefficient" of " -12a" ; is: "- 12" .

The constant is: "32" .

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________________________________________________________

The "coefficient" of the "a" term is: " - 12" .

________________________________________________________

The "constant" is: " 32 " .

________________________________________________________

Explanation:

________________________________________________________

Let: "a = x² + 4 " .

Given: (x² + 4)² + 32 = 12x² + 48 ;

______________________________________________________

Factor: "12x² + 48" into " (x² + 4) " ;

"12x² + 48" = 12 (x² + 4) " ;

______________________________________________________

Given: (x² + 4)² + 32 = 12x² + 48 ;

rewrite as; "a² + 32 = 12a " ;

Subtract "12a" from each side of the equation;

"a² + 32 - 12a = 12a - 12a ;

to get:

" a² - 12a + 32 = 0 " .

___________________________________________________

The coefficient of the "a" term; that is:

The "coefficient" of " -12a" ; is: "- 12" .

The constant is: "32" .

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