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x^2+3x-2=0

I have to solve for x by completing the square

I am a bit slow and get stuck at this part:
x^2+3x-2=0
x^2+3x=2
x^2+3x+2.25=4.25
(x+ )(x+ )=4.25

I can't factor the 2.25 to get 3. Or am i doing this wrong???

nevermind i got it thanks..

I did it by the use of decimals.

x^2+3x+2.25=4.25
(x+1.5) (x+1.5)=4.25
(x+1.5)^2=4.25

sqgrt both sides

x+1.5=sgrt(4.25)
-1.5 -1.5

x= -1.5 +-sgrt(4.25)

is this correct also?

I am having some trouble with these 3 quadratic equations. I need to solve for x.

x^4+2x^2-15=0

2(x-1)^2+7(x-1)+6=0

2x-9sqrt(x)+4=0

Thanks for your help, stapel.. Here's what I got for the first one, x= i sqrt(5) and x=sqrt(3) i being the imaginary number of sqrt(-1) I'm not sure it is correct though :oops: For the second one, I can't seem to get the factors to add up to 7y (the middle term) so I'm stuck here I want to do (2z+1)...

Wait I'm confused. For 2(x-1)^2+7(x-1)+6=0 shouldn't I be looking for factors of 6 and not 12?

And for, 2x-9sqrt(x)+4=0, don't i need factors of 4 and not 8 ?

Thanks..

I am having a really hard time with this section. My teacher will be absent tomorrow and I have a test on Monday. I don't understand this. So, for practice could anyone help me out and show me step by step how to do some problems? In my book, it says: "use the Vertex Form: y - k = a (x - h)^2 W...

Thanks stapel for the simple explanation. Here's the work I did for the 1st example. y = a(x - h)^2 + k 7 = a(4 - 3)^2 + 5 7 = a(1)^2 +5 7= a + 5 2 = a And the final answer is: y = 2(x-3)^2 + 5 For the second one I got 1/2 = a and the final answer/equation is: y = 1/2(x - (-1))^2 + 1 Are these corre...

My next section is a bit different and once again I have no clue what to do. In the book it says :

"Find the coordinates of the vertex then decide if the parabola has a minimum or maximum value."

Ex.1
f(x) = x^2 - 4x + 4

Ex. 2
f(x) = -x^2 - 8x - 16

Can you tell me what to do here?

OKay Thanks You once again I got that now too.