Question

Solve the equation by completing the square 16x^2+1=7x!!! Please help me ASAP!!!! :(

Answer 1

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
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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
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( 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 )

Answer 2

My typo:

it is written
x - 7 / 32 = ± √ [ ( - 1 ) ∙ √15 ] / 32

should be
x - 7 / 32 = ± √ [ ( - 1 ) ∙ 15 ] / 32