completing the square?

Question

can someone help me solve these 2 questions by completing the sqaure

i) 2x
------- - 3x + 8 = 0
x - 2

ii) ( x - ¾) ² - 7/16 = 0

Answer 1

[11]
(2x/x-2)-3x+8=0
Multiplying both sides by x-2
2x-3x(x-2)+8(x-2)=0
2x-3x^2+6x+8x-16=0
-3x^2+16x-16=0
x^2-16x/3+16/3=0[dividing both sides by -3]
x^2-16x/3= -16/3
x^2-2*x*8/3+(8/3)^2=-16/3+64/9[adding
(8/3)^2 or 64/9 to both sides
(x-8/3)^2=16/9
x-8/3=+-4/3 [square-roo
ting both sides]
x=8/3 +-4/3
x=4 or 4/3
ii)(x-3/4)^2-7/16=0
(x-3/4)^2=7/16
x-3/4=+- sqrt (7/16)=+-sqrt7/4
x=3/4 +-sqrt7/4
=(3+-sqrt7)/4

Answer 2

Hi,
Let's do ii first because the square is already complete, and it'll show you how to handle the last part of the first problem.
(x-3/4)² -7/16 = 0
x-3/4 = +-√(7/16) (Take the square root of both sides.)
x = ¾+-√(7/16)
x=3/4 +-√(7)/4
x=(3+-√(7))/4

i) Now what we want to do is get the first one in the form of the beginning equation for the one we just did.
2x/(x-2) -3x +8 =0
2x -3x² +6x +8x -16 = 0 (Multiply through by LCD of (x-2).)
-3x² +16x -16 = 0 (Combine like terms.)
x² -(16/3)x +16/3 = 0
x² -(16/3)x =-16/3
Now, take ½ of the coefficient of x, (that’s 16/6), square it, and add it to both sides of the equation.
x²-(16/3)x +64/9 = -16/3 +64/9
(x -8/3)² =-16/3+64/9
(x-8/3)² =79/9
x-8/3 = +-√(16/9)
x=8/3+-4/3
x1=4/3
x2 =12/3=4
Hope this helps.
FE

Answer 3

2x -3x + 8 = 0
....
x-2

multiply with (x-2)

2x - 3x(x-2)+8(x-2) = 0

2x - 3x^2 + 6x + 8x - 16 = 0

-3x^2 + 16x - 16 = 0

divide by -3

x^2 - 16/3 x + 16/3 = 0

x^2 - 2x(8/3) + (8/3)^2 - (8/3)^2 + 16/3 = 0

(x - 8/3)^2 +16/3 - 64/9 = 0

(x-8/3)^2 - 16/9 = 0

(x-8/3)^2 = (4/3)^2

x -8/3 = +/- 4/3

x = 8/3 + 4/3 or 8/3 - 4/3

x = 4 or 4/3

ii)

(x-3/4)^2 - 7/16 = 0

(x - 3/4)^2 = [sqrt(7)/4]^2

x - 3/4 = +/- sqrt(7)/4

x = 3/4 + sqrt(7)/4 or 3/4 - sqrt(7)/4

x = [3+sqrt(7)]/4 or [3- sqrt(7)]/4