Completing the Square: Tips and Tricks for Solving Quadratic Equations

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To complete the square for a quadratic expression with a leading coefficient, first divide the entire equation by that coefficient to simplify the process. For example, in the expression 4(x^2) - 4x + 3, dividing by 4 allows for standard completion of the square. When factoring by grouping, isolate the quadratic and linear terms, factor out the leading coefficient, and complete the square inside the parentheses. Adjust for any changes made by subtracting or adding the necessary values to maintain equality. This method ensures accurate transformation of the quadratic expression into vertex form.
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Hello,
I was working on a lengthy problem and got stuck on something simple. I have a quadratic expression 4(x^2) - 4x + 3 that I should have been able to complete the square on to rewrite as (2x - 1)^2 + 2
My knowledge of applying this technique when a coefficient is attached to the leading term is rusty so I got out my old algebra books. The solution they offered was to set 4(x^2) - 4x + 3 = 0
and then divide each side by four. With the coefficient removed, then the square can be completed the usual way. But this isn't giving me what I need at all.
Any advice on this technique?

Thanks mucho!
 
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When you set aside the quadratic and linear terms, you can factor by grouping as in the following example:

3x2-12x+2
(3x2-12x)+2
3(x2-4x)+2

Note that I have factored the 3 out of the first two terms only. Now, I complete the square inside the parenteses.

3(x2-4x+4[/color])+2-12[/color]

Note the parts in red[/color]. I added 4 to the expression in parentheses, but that 4 was multiplied by 3 to make 12. So, to keep the expression equal to my original expression, I subtract 12 (not 4) from it.

Finally, I factor the perfect square in parentheses and collect terms.

3(x-2)2-10

edit: typo
 
Thanks, Tom. Glad to see you are still on the homework help boards.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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