Solve by completing the square. 2x^2-8x-9=0?

Answer 1

2x^2 - 8x - 9 = 0

add 9 for both sides
2x^2 - 8x = 9

divide everything by 2
x^2 - 4x = 4.5

prepare for completing square
x^2 - 4x + ? = 4.5 + ?

use (b/2)^2
x^2 - 4x + (4/2)^2 = 4.5 + (4/2)^2

x^2 - 4x + 4 = 4.5 + 4

factor and combine like terms
(x-2)^2 = 8.5

take a square root
x-2 = +/- sqr(8.5)

add 2 for both sides
x = 2 +/- sqr(8.5)

hope this helps

Answer 2

2x^2 – 8x – 9 = 0
Multiply by 2
4x^2 – 16x – 18 = 0
threfore 4x^2 – 16x = 18
To make square no. on LHS
4x^2 – 16x + 16 = 18 +16 = 34
(2x – 4)^2 = 34
2x – 4 = sqrt.34 therefore x = (sqrt.34 +4)/2 = (5.8 +4)/2 = 4.9
or 2x – 4 = – sqrt.34 therefore x = – (sqrt.34 +4)/2 = (– 5.8 +4)/2 = – 0.9
Therefore x = 4.9 or – 0.9

Answer 3

2(x² - 4x - 9/2) = 0
[ (x² - 4x + 4) - 4 - 9/2 ] = 0
[ (x - 2)² - 17/2 ] = 0
(x - 2)² = 17/2
x - 2 = ±√(17/2)
x = 2 ± √(17/2)

Answer 4

2x^2-8x-9=0
2x^2-8x=9
2(x^2-4x[-4])=9-8
2(x-2)^2=1
2(x-2)^2-1=0

I completed the square....

Answer 5

2(x^2 - 4x + 4 ) - 17= 0

2(x-2)^2 = 17

(x-2)^2 = 17/2

x-2 = +-sqrt(17/2)

x = 2 +- sqrt(17/2)

Answer 6

2x^2-8x=9
2x^2-8x+(-4)^2=9+(-4)^2
(2x-4)^2=25

2x-4=5
2x=9
x=9/2

or

2x-4=-5
2x=-1
x=-1/2

so, x=9/2 or -1/2

Answer 7

Well, let's see... divide by 2x, carry the 1, square root of 50, and add 9 and you get... hmm, yep. The answer is NO. Do your OWN homework!

Answer 8

2x^2-8x-9=2(x^2-4x+4)-8-9=2(x-2)^2-17= (sqrt(2)*(x-2)^2-sqrt(17)^2)=[(sqrt(2)*(x-2)+sqrt(17)]*[(sqrt(2)*(x-2)-sqrt(17)]=0
x1-2=-sqrt(17/2) ;x1=-sqrt(17/2)+2
x2-2=+sqrt(17/2); x2=sqrt(17/2)+2