Asked by Andrea
Solve the equation by completing the square 16x^2+1=7x!!! Please help me ASAP!!!! :(
Answers
Answered by
Bosnian
16 x² + 1 = 7 x
Subtract 7 x to both sides
16 x² + 1 - 7 x = 7 x - 7 x
16 x² - 7 x + 1 = 0
Divide both sides by 16
x² - 7 x / 16 + 1 / 16 = 0
Subtract 1 / 16 to both sides
x² - 7 x / 16 + 1 / 16 - 1 / 16 = 0 - 1 / 16
x² - 7 x / 16 = - 1 / 16
Add [ ( 7 / 16 ) / 2 ]² = ( 7 / 32 )² to both sides
x² - 7 x / 16 + [ ( 7 / 32 ) ]² = - 1 / 16 + [ ( 7 / 32 ) ]²
_______________________________________
Remark:
( x - 7 / 32 )² = x² - 2 x ∙ 7 / 32 + ( 7 / 32 )²
( x - 7 / 32 )² = x² - 2 ∙ 7 x / 2 ∙ 16 + ( 7 / 32 )²
( x - 7 / 32 )² = x² - 7 x / 16 + ( 7 / 32 )²
So:
x² - 7 x / 16 + ( 7 / 32 )² = ( x - 7 / 32 )²
______________________________________
( x - 7 / 32 )² = - 1 / 16 + 49 / 1024
( x - 7 / 32 )² = - 1 ∙ 64 / 16 ∙ 64 + 49 / 1024
( x - 7 / 32 )² = - 64 / 1024 + 49 / 1024
( x - 7 / 32 )² = - 15 / 1024
Take square root of both sides
x - 7 / 32 = ± √ ( - 15 / 1024 )
x - 7 / 32 = ± √ ( - 15 ) / √1024
x - 7 / 32 = ± √ [ ( - 1 ) ∙ √15 ] / 32
x - 7 / 32 = ± √ ( - 1 ) ∙ √15 / 32
x - 7 / 32 = ± i ∙ √15 / 32
Add 7 / 32 to both sides
x - 7 / 32 + 7 / 32 = 7 / 32 ± i ∙ √15 / 32
x = 7 / 32 ± i ∙ √15 / 32
x = ( 1 / 32 ) ∙ ( 7 ± i ∙ √15 )
x1= ( 1 / 32 ) ∙ ( 7 + i ∙ √15 )
x2 = ( 1 / 32 ) ∙ ( 7 - i ∙ √15 )
Subtract 7 x to both sides
16 x² + 1 - 7 x = 7 x - 7 x
16 x² - 7 x + 1 = 0
Divide both sides by 16
x² - 7 x / 16 + 1 / 16 = 0
Subtract 1 / 16 to both sides
x² - 7 x / 16 + 1 / 16 - 1 / 16 = 0 - 1 / 16
x² - 7 x / 16 = - 1 / 16
Add [ ( 7 / 16 ) / 2 ]² = ( 7 / 32 )² to both sides
x² - 7 x / 16 + [ ( 7 / 32 ) ]² = - 1 / 16 + [ ( 7 / 32 ) ]²
_______________________________________
Remark:
( x - 7 / 32 )² = x² - 2 x ∙ 7 / 32 + ( 7 / 32 )²
( x - 7 / 32 )² = x² - 2 ∙ 7 x / 2 ∙ 16 + ( 7 / 32 )²
( x - 7 / 32 )² = x² - 7 x / 16 + ( 7 / 32 )²
So:
x² - 7 x / 16 + ( 7 / 32 )² = ( x - 7 / 32 )²
______________________________________
( x - 7 / 32 )² = - 1 / 16 + 49 / 1024
( x - 7 / 32 )² = - 1 ∙ 64 / 16 ∙ 64 + 49 / 1024
( x - 7 / 32 )² = - 64 / 1024 + 49 / 1024
( x - 7 / 32 )² = - 15 / 1024
Take square root of both sides
x - 7 / 32 = ± √ ( - 15 / 1024 )
x - 7 / 32 = ± √ ( - 15 ) / √1024
x - 7 / 32 = ± √ [ ( - 1 ) ∙ √15 ] / 32
x - 7 / 32 = ± √ ( - 1 ) ∙ √15 / 32
x - 7 / 32 = ± i ∙ √15 / 32
Add 7 / 32 to both sides
x - 7 / 32 + 7 / 32 = 7 / 32 ± i ∙ √15 / 32
x = 7 / 32 ± i ∙ √15 / 32
x = ( 1 / 32 ) ∙ ( 7 ± i ∙ √15 )
x1= ( 1 / 32 ) ∙ ( 7 + i ∙ √15 )
x2 = ( 1 / 32 ) ∙ ( 7 - i ∙ √15 )
Answered by
Bosnian
My typo:
it is written
x - 7 / 32 = ± √ [ ( - 1 ) ∙ √15 ] / 32
should be
x - 7 / 32 = ± √ [ ( - 1 ) ∙ 15 ] / 32
it is written
x - 7 / 32 = ± √ [ ( - 1 ) ∙ √15 ] / 32
should be
x - 7 / 32 = ± √ [ ( - 1 ) ∙ 15 ] / 32
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