Answer:
[tex]x_1 = 8+2\sqrt{5}=2(4+\sqrt{5} )[/tex]
[tex]x_2 = 8-2\sqrt{5}=2(4-\sqrt{5} )[/tex]
Step-by-step explanation:
[tex]x^2 - 16x + 64 = 20[/tex]
I will solve it completing the square
[tex]x^2 - 16x + 64 = 20[/tex]
[tex](x-8)^2 = 20[/tex]
[tex]x-8= \pm\sqrt{20}[/tex]
[tex]x=8 \pm\sqrt{20}[/tex]
[tex]x=8 \pm2\sqrt{5}[/tex]
[tex]x_1 = 8+2\sqrt{5}=2(4+\sqrt{5} )[/tex]
[tex]x_2 = 8-2\sqrt{5}=2(4-\sqrt{5} )[/tex]
Answer: The answer is: " x = 8 ± 2√5 " .
____________________________
Step-by-step explanation:
____________________________
Given:
x² - 16x + 64 = 20 ; Solve for "x" ;
____________________________
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) ;
____________________________
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 .
____________________________
To solve for "x" ; we use the quadratic equation formula:
____________________________
→ x = [ - b ± √(b² - 4ac) ] / [2a] ;
We solve for "x" by plugging in our values for "a" ; "b" ; and "c" ;
____________________________
→ x = { - [-16) ± {√[(-16)² - 4(1)(44) ] } / [2 * 1] ;
____________________________
→ x = [ 16 ± √ (16² - 4*44) ] / [2] ;
____________________________
→ x = [ 16 ± √ (16² - 4*44) ] / [2] ;
____________________________
→ x = [ 16 ± √ (256 - 176 )] / [2] ;
____________________________
→ x = [ 16 ± √ (80 )] / [2] ;
Now, let us rewrite: " √80 " ;
____________________________
" √80 " = √16 * √5 = 4*√5 ;
→ write as: " 4√5 " .
____________________________
Now, take:
____________________________
→ x = [ 16 ± √ (80 )] / [2] ;
____________________________
And replace: " √80 " ; with: " 4√5 " ; and rewrite:
____________________________
→ x = [ 16 ± 4√5 ) / 2 ;
____________________________
Now, divide the numerator by "2" to simplify, and rewrite:
→ Note: 16 ÷ 2 = 8 ; and: 4 ÷ 2 = 2 ;
So, we have:
____________________________
→ x = 8 ± 2√5
____________________________
