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solve 2x^2 + 3 = x by completing the square

2007-05-31 14:45:40 · 1 answers · asked by Anonymous in Science & Mathematics Mathematics

1 answers

OK... first let's move everything to the left side of the equation:
2x^2 - x +3 = 0

Then divide everything by 2, to give the first term a coefficient of 1:
x^2 - (1/2)x + (3/2) = 0

Now look at the -(1/2) coefficient in front of the middle term.

When you foil out (x + a)^2, you get x^2 + 2ax + a^2.
You need to figure out what value of "a" gives you a middle term coefficient of -(1/2).
2ax = -(1/2)x
2a = -(1/2)
a = -(1/4)

Hence, the square part will be (x - (1/4))^2
When you FOIL that, you get x^2 - (1/2)x + (1/16).

Now remember that, and go back to the beginning:
x^2 - (1/2)x + (3/2) = 0

On the left side, add +(1/16) - (1/16).
That gives you:
x^2 - (1/2)x + (1/16) - (1/16).+ (3/2) = 0

Now substitute in the (x-(1/4))^2 to get:
(x-(1/4))^2 - (1/16) + (3/2) = 0
(x - (1/4))^2 - (1/16) + (24/16) = 0
(x - (1/4))^2 + (23/16) = 0
(x - (1/4))^2 = -(23/16)

From this, you see that the equation has no solution.
The left hand side is something squared, so it can't be negative. The right hand side is negative. There's no value of x that'll make that work.

Hopefully this at least helps you see the method, though!

2007-05-31 15:01:29 · answer #1 · answered by Bramblyspam 7 · 0 0

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