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I have one equation I can't seem to solve. Maybe I'm just tired and making a simple error but I keep getting a negative and you can't square root those.
The equation is 2x^2 - x - 2 = 0

2006-11-15 14:38:24 · 6 answers · asked by Teh Link 1 in Science & Mathematics Mathematics

6 answers

Completing the square involves adding constants to both sides of your equation so you have a form:

(x - a)^2 = b

Once you get this form, you can take the square root of both sides:

x - a = sqrt(b)

x = a + sqrt(b)

Since a square root can be either positive or negative, it is just as correct to write:

x = a - sqrt(b)

Now, back to the original "perfect square"

(x - a)^2 = b

Expanding this out:

x^2 - 2*a + a^2 = b

Moving b to the other side:

x^2 - 2*a + a^2 - b = 0

Your equation is 2*x^2 - x - 2 = 0

Forst thing is to get plain x^2, so divide by 2:

x^2 - (1/2) * x - 1 = 0

Now from the form above, a = 1/.4 because 2*a = (1/2)*x
Next, look at the constant term:

-1 = a^2 - b
-1 = (1/4)^2 - b

Solving for b: b = 17/16

So your perfect square equation is:

(x - 1/4)^2 = 17/16

Solving for x:

x - 1/4 = sqrt(17/16)

x = 1/4 + sqrt(17/16)
x = 1/4 - sqrt(17/16)

Simplifying slightly:

x = (1 + sqrt(17))/4
x = (1 - sqrt(17))/4

2006-11-15 17:17:45 · answer #1 · answered by Pretzels 5 · 0 0

2x² - x - 2 = 0
Divide by 2:
x² - x/2 - 1 = 0
Add 17/16:
x² - x/2 + 1/16 = 17/16
Factor:
(x-1/4)² = 17/16
Take the square root:
x-1/4 = ±√17/4
Add 1/4:
x = (1±√17)/4

Edit: for some reason I got lazy and stopped writing the explanations of each step halfway through. Those have been added. Also, while we did not need it for this equation, there is a number system, called the complex numbers, in which negative numbers do have square roots. You should learn about it soon.

Edit 2: And I royally screwed up the factoring. What's with me this week? I seem to be making the most basic errors. Oh well, it's fixed now, FWIW.

2006-11-15 22:48:54 · answer #2 · answered by Pascal 7 · 0 0

2x^2 - x - 2 = 0
x^2 - (x/2) - 1 = 0
(x-1/4)^2 - 1 - (1/16) = 0
(x-1/4)^2 = 17/16
(x-1/4) = sqrt(17/16)
x = 1/4 +- sqrt(17/16)

2006-11-15 22:50:54 · answer #3 · answered by original_giraffe 1 · 0 0

2x^2 - x - 2 = 0

2x^2-x=2

2x^2-x+__ =2+__
b/2 = -1/2^2 = 1/4

Urggg.... Can't remember next steps...

I hate algebra two with a passion

EDITED: Oh, have you not learned the imaginary numbers? If haven't then I don't know what to tell you...

2006-11-15 22:47:05 · answer #4 · answered by ANIME FREAK >_< 1 · 0 1

all i can tell u is, x=2 is not right, make sure whenever u think u have an answer u plug it back into the equation wherever there was an x

2006-11-15 22:51:23 · answer #5 · answered by Chikk230 2 · 0 0

4-2=X
X=2

2006-11-15 22:47:02 · answer #6 · answered by Diana T 2 · 0 2

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