Completing the square 3x^2 + 2x + 4=0

wannabeanerd

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Mar 11, 2010
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I can solve simple equations. However, this one is a little more complicated please help...

3x^2 + 2x + 4=0

3x^2+2x=-4
Here is where I get stuck

I think this is next

3(x+1x)^2=-4 Am I just completely lost? Please help. If you can show your work step by step that would be very useful thanks
 
wannabeanerd said:
I can solve simple equations. However, this one is a little more complicated please help...

3x^2 + 2x + 4=0

3x^2+2x=-4
Here is where I get stuck

I think this is next

3(x+1x)^2=-4 Am I just completely lost? Please help. If you can show your work step by step that would be very useful thanks

When you get to

3x[sup:bmmjggwc]2[/sup:bmmjggwc] + 2x = -4, I would divide both sides of the equation by 3 (why? So that the coefficient of x[sup:bmmjggwc]2[/sup:bmmjggwc] is 1):

x[sup:bmmjggwc]2[/sup:bmmjggwc] + (2/3)x = -4/3

Now, to complete the square, multiply the coefficient of "x" by 1/2. (1/2)*(2/3) is 1/3. SQUARE that, and add the result to both sides of the equation:

x[sup:bmmjggwc]2[/sup:bmmjggwc] + (2/3)x + (1/9) = (-4/3) + (1/9)

Write the left side as the square of a binomial, and do the arithmetic on the right side:

[x + (1/3)][sup:bmmjggwc]2[/sup:bmmjggwc] = (-12/9) + (1/9)
[x + (1/3)][sup:bmmjggwc]2[/sup:bmmjggwc] = -11/9

Then, take the square root of both sides and continue....
 
wannabeanerd said:
There is no square root of -11


x= +or- sqrt 11/3-1/3

Yes, sqrt(-11) exists...but it is not a real number.

You may want to review the sections in your textbook which cover the complex numbers...if you are expected to solve an equation like this one, you should have dealt with the complex numbers first.
 
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