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 the variable c in the equation derived from the expression (2x - b)(7x + b) = 14x^2 - cx - 16. The analysis reveals two solutions for b: 4 and -4, leading to corresponding values for c of 20 and -20, respectively. The calculations confirm that substituting these values into the expanded equation yields consistent results, validating the solutions. The conversation also highlights the importance of correctly equating terms from both sides of the equation.

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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!
 

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