Question

B. x2 - 16x + 64 = 20

Answer 1

`x_1 = 8+2√(5)=2(4+√(5) )`

`x_2 = 8-2√(5)=2(4-√(5) )`

Explanation:

`x^2 - 16x + 64 = 20`

I will solve it completing the square

`x^2 - 16x + 64 = 20`

`(x-8)^2 = 20`

`x-8= \pm√(20)`

`x=8 \pm√(20)`

`x=8 \pm2√(5)`

`x_1 = 8+2√(5)=2(4+√(5) )`

`x_2 = 8-2√(5)=2(4-√(5) )`

Answer 2

Answer: The answer is: " x = 8 ± 2√5 " .

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Explanation:

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Given:

x² - 16x + 64 = 20 ; Solve for "x" ;

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Subtract "20" from each side of the equation:

x² - 16x + 64 - 20 = 20 -20 ;

to get:

x² - 16x + 44 = 0 ;

(i.e. to get an equation in "quadratic equation format" ;

that is; in the format of: ax² + bx + c = 0 ; (a ≠ 0) ;

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We cannot factor the "left-hand side" of the equation;

so, we can solve for "x" by using the quadratic equation formula;

Note that our equation: " x² - 16x + 44 = 0 " ;

is in "quadratic equation format" ;

that is: " ax² + bx + c = 0 " ; (a ≠ 0) ;

in which:

a = 1 ;

(Note: The implied coefficient of: " ax² " is: "1" ;

→ since "1" ; multiplied by any value, results in that exact value.

This is known as the "identity property" of multiplication.}.

b = -16 ;

c = 44 .

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To solve for "x" ; we use the quadratic equation formula:

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→ x = [ - b ± √(b² - 4ac) ] / [2a] ;

We solve for "x" by plugging in our values for "a" ; "b" ; and "c" ;

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→ x = { - [-16) ± {√[(-16)² - 4(1)(44) ] } / [2 * 1] ;

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→ x = [ 16 ± √ (16² - 4*44) ] / [2] ;

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→ x = [ 16 ± √ (16² - 4*44) ] / [2] ;

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→ x = [ 16 ± √ (256 - 176 )] / [2] ;

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→ x = [ 16 ± √ (80 )] / [2] ;

Now, let us rewrite: " √80 " ;

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" √80 " = √16 * √5 = 4*√5 ;

→ write as: " 4√5 " .

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Now, take:

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→ x = [ 16 ± √ (80 )] / [2] ;

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And replace: " √80 " ; with: " 4√5 " ; and rewrite:

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→ x = [ 16 ± 4√5 ) / 2 ;

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Now, divide the numerator by "2" to simplify, and rewrite:

→ Note: 16 ÷ 2 = 8 ; and: 4 ÷ 2 = 2 ;

So, we have:

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→ x = 8 ± 2√5

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