Quadratic: Solve x(sqrd) = 4x - 2

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nickilin

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Oct 28, 2007
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I have this eqn.
x(sqrd)=4x-2

I set up as 4x-2-x(sqd)=0
-(4)+-(sq root) -4(sqd)-4(-1)(-2)

My final answers are .6 and 3.4
When I put those back in they don't work. My study partner got the same thing as well. Can you please tell me what we are doing wrong?
Thanks
 
nickilin said:
I have this eqn.
x(sqrd)=4x-2

I set up as 4x-2-x(sqd)=0
-(4)+-(sq root) -4(sqd)-4(-1)(-2)
Click to expand...
I think you mean the following:

. . . . .I need to solve this equation: x^2 = 4x - 2

. . . . .I set it up like this: x^2 - 4x + 2 = 0

. . . . .Then I applied the Quadratic Formula, with a = 1,
. . . . .b = -4, and c = 2:

. . . . .x = (-(-4) +/- sqrt[(-4)^2 - 4(1)(-2)]) / 2(1)

. . . . .x = (4 +/- sqrt[16 + 8]) / 2

. . . . .x = (4 +/- sqrt[24]) / 2

...and so forth. But I'm not seeing how you get your final values...?

Please reply clearly showing all of your steps. Thank you! :D

Eliz.
 
im pretty sure standard form is ax^2 +bx+c

im not sure if that will help but u can try it...

4x^2-4x-2=0
 
See where you are going with everything, and I see where my mistake was made. However, how can the 2 be a negative when it goes under the sq root when before it was positive?

x=-(-4)+/-(sqrt)-4(sqd)-4(1)(-2)/2(1)

when did the 2 go from - to +?
 
There's a "subtract two" on the right-hand side. How would one move that over to the left-hand side, if not by adding...?

Eliz.
 
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