Would you help me solve 2x^2 - 3x + 1 = 0 by completing the square?

Answer 1

2 (x² - (3/2)x + 1/2) = 0
x² - (3/2)x + 1/2 = 0
x² - (3/2)x + 9 /16 = -1/2 + 9 /16
(x - 3/4)² = 1/16
x - 3/4 = ± (1/4)
x = 3/4 ± 1/4
x = 1 , x = 1/2
Check
(2x - 1)(x - 1) = 0
x = 1/2 , x = 1

Answer 2

2x^2 - 3x + 1 = 0
x^2 - 3/2x = -1/2
x^2 - 3/2x + (3/4)^2 = -1/2 + (3/4)^2
(x - 3/4)^2 = -1/2 + 9/16
(x - 3/4)^2 = -8/16 + 9/16
(x - 3/4)^2 = 1/16
sqrt [(x - 3/4)^2] = sqrt (1/16)
x - 3/4 = +- 1/4
x = +- 1/4 + 3/4
x = 1
or
x = 1/2

Answer 3

2x^2 - 3x + 1 = 0

subtrac 1 for both sides
2x^2 - 3x = -1

before complete the square, you need to make sure the leading coefficient is 1. Divide 2 for both sides
x^2 - 1.5x = -0.5

add (b/2)^2 for both sides
x^2 - 1.5x + .5625 = -0.5 + .5625

factor
(x - .75)^2 = .0625

take a square root
x - .75 = -0.25 or 0.25

add .75 for both sides
x = 0.5 or 1

Answer 4

2x^2-3x=-1
or 2(x^2-3/2)=-1
or x^2-3/2=-1/2
or {x^2-2.x.3/4+(3/4)^2} =-1/2+(3/4)^2
or(x-3/4)^2=-1/2+9/16=(9-8)/16=1/16=(1/4)^2
so we have x-3/4=+-1/4
or x=1/4+3/4 or x=-1/4+3/4
so x=1 or 1/2 ans

Answer 5

for these kinds of equation you can always use delta
delta=b^2-4ac (formula)
delta=9-8=1
because
1>0
so...
x1=-b +sqr root delta divide by 2a
x2=-b -sqr root delta divide by 2a

so x1=1 x2=0.5