Are These Quadratic Solution Steps Correct?

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The discussion revolves around verifying the correctness of quadratic solution steps presented by multiple users. The first user provides a quadratic equation derived from manipulating the equation 1-x=0.5x^2 but encounters algebraic errors in subsequent steps. Another user points out mistakes in the third equation, suggesting that the original equation was misinterpreted, leading to incorrect coefficients. The corrections indicate that the proper solutions should yield x = ±2 instead of the previously calculated values. Overall, the thread highlights the importance of careful algebraic manipulation in solving quadratic equations.
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Just check these over and seen if there correct, thanks

#1
1-x=0.5x^2
Times everything by 2
2-2x=1x^2
Move it over
-1x^2 - 2x + 2=0
a=-1
b=-2
c=2

x=2 +/- root -2 - 4(-1)(2) all over -2

x=2 t/- root 6 all over -2

x=-1 +/- root -3

#2
2-x=x^2 + x
-x^2 -x -x + 2=0
a=-1
b=-2
c=2
same thing as #1 so if that correct then I guess this will be :rolleyes:

#3
6(2-x)=3x^2 + 6x
I warn you, this could be wrong
12-6x=yada yada
-3x^2 -6x -6x + 12=0
a=-3
b=-6
c=12

x=6 +/- root -6 - 4(-3)(12) all over -6

x=6 +/- root 138 all over -6

x= -1 +/- root 23

Thanks for any help
 
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FAQ said:
#3
6(2-x)=3x^2 + 6x

I think you've got some algebra errors in there:
6(2-x)=3x^2-6x
Divide both sides by 3
2(2-x)=x^2-2x
multiply out
4-2x=x^2-2x
add 2x-4 to both sides
0=x^2-4
x = \pm 2
 
FAQ said:
#3
6(2-x)=3x^2 + 6x
I warn you, this could be wrong
12-6x=yada yada
-3x^2 -6x -6x + 12=0
a=-3
b=-6
c=12
Since you have -6x-6x, b=-12

NateTG said:
I think you've got some algebra errors in there:
6(2-x)=3x^2-6x
Divide both sides by 3
2(2-x)=x^2-2x
multiply out
4-2x=x^2-2x
add 2x-4 to both sides
0=x^2-4
x = \pm 2
I think the original equation was 6(2-x)=3x^2+6x, not 6(2-x)=3x^2-6x
 
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