MHB Can I Solve for c by Replacing b with a Value Greater Than 0?

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The discussion revolves around solving for variable c by substituting b with a value greater than 0. The equation (2x - b)(7x + b) is expanded and equated to 14x^2 - cx - 16, leading to two solutions for b: 4 and -4. When b equals 4, c is determined to be 20, and when b is -4, c is -20. The calculations confirm the relationships between the variables, demonstrating the validity of the solutions. The thread concludes with a lighthearted comment unrelated to the mathematical discussion.
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I replace b with any value greater than 0 and then solve for c. Right?

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What do you get when you expand the LHS and equate the resulting terms with the corresponding terms in the RHS of the given equation
 
Greg said:
What do you get when you expand the LHS and equate the resulting terms with the corresponding terms in the RHS of the given equation

Ok. I will do as you suggested and be back.
 
It's been 7 months now! Are you still working on it?

For those who were wondering, there are two solutions.
(2x- b)(7x+ b)= 14x^2+ 2bx- 7bx- b^2= 14x^3- 5bx- b^2. That is to be equal to 14x^2- cx- 16.

So we must have -5b= -c and -b^2= -16. b^2= 16 so b= 4 or b= -4.
If b= 4, -5b= -20= -c so c= 20.
If b= -4, -5b= 20= -c so c= -20.

Check:
(2x- 4)(7x+ 4)= 14x^2+ 8x- 28x- 16= 14x^2- 20x- 16.
(2x+ 4)(7x- 4)= 14x^2- 8x+ 28x- 16= 14x^2+ 20x- 16.
 
Beer soaked comment follows.
Country Boy said:
It's been 7 months now! Are you still working on it?
...
He's been banned permanently.
 
I'll bet I could drink that cask of wine in less than 20 days!
 

Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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