+0  
 
0
121
6
avatar

\({-0.1x^{2} - 0.6x + 0.4} = 0\)

 

I grouped the variables together and factored out a common term: \({-0.1(x^2 + 6x) + 0.4} = 0\)

I complete the square within the brackets: \({-0.1(x^2 + 6x + ({6\over2})^2 + 0.4} = 0\)

\({-0.1(x^2 + 6x + 9) + 0.4 - 0.9} = 0\) -> i balanced the equation

\({-0.1(x + 3)^2 - 0.5} = 0\)

and now i cant isolate the squared term:

\({-0.1(x + 3)^2} = 0.5\)

\({(x + 3)^2} = -5\)

which isn't possible.

 

however, the answer in the textbook is \(x = {-3 \pm \sqrt{13}}\). what did i do wrong?

Guest Jul 13, 2018
 #1
avatar+64 
0

which book? AoPS or McDOUGAl or stuff?

Dr.Amione  Jul 13, 2018
 #2
avatar
0

mcgraw-hill ryerson pre-calculus 11...

Guest Jul 13, 2018
 #3
avatar+64 
0

mcgraw?

Dr.Amione  Jul 13, 2018
 #4
avatar
0

i'm not mistaken.

Guest Jul 13, 2018
 #5
avatar
+1

Look at the 0.9's on the third line down.

The first one, coming from when the bracket is multiplied out, will be negative, so the one at the end has to be positive.

Guest Jul 13, 2018
 #6
avatar+90023 
+1

-0.1x^2  - 0.6x  + 0.4    =  0      

 

First off....the easiest thing to  do  is to multiply  everything through  by  -10....this gives us

 

x^2  + 6x - 4  = 0        take  1/2 of 6  = 3....square it  =  9....add and subtract it

 

x^2   + 6x  + 9  - 4  - 9  = 0        factor the first three terms and simplify  the rest

 

(x + 3)^2  - 13  = 0       add 13 to both sides

 

(x + 3)^2  = 13           take both roots

 

x + 3  = ±√13       subtract 3  from  both sides

 

x  = ±√13  - 3    or    -3 ±√13   if you prefer!!!!

 

 

cool cool cool

CPhill  Jul 13, 2018

12 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.