One more Algebra Question - Please Help!

The roots of the quadratic equation    $x^2+bx+c=0$    are:

 

$x=\frac{-b+\sqrt{b^2-4c}}{2}$          and          $x=\frac{-b-\sqrt{b^2-4c}}{2}$

 

The difference between these two roots is:

 

$\frac{-b\ +\ \sqrt{b^2-4c}}{2}-\frac{-b\ -\ \sqrt{b^2-4c}}{2}\\~\\ =\quad\frac{-b\ +\ \sqrt{b^2-4c}}{2}+\frac{b\ +\ \sqrt{b^2-4c}}{2}\\~\\ =\quad\frac{-b\ +\ \sqrt{b^2-4c}\ +\ b\ +\ \sqrt{b^2-4c}}{2}\\~\\ =\quad\frac{ \sqrt{b^2-4c}\ +\ \sqrt{b^2-4c}}{2}\\~\\ =\quad\frac{ 2\sqrt{b^2-4c}}{2}\\~\\ =\quad\sqrt{b^2-4c}$

 

which we are told must be equal to  $|b-2c|$ ...so we can make this equation:

 

$\sqrt{b^2-4c}\ =\ |b-2c|$                    Now let's solve this equation for  c

 

$(\sqrt{b^2-4c}\ )^2\ =\ (\ |b-2c|\ )^2$

 

Squaring a negative number gives the same result as squaring the positive version of that number,

so we can drop the absolute value signs

 

$b^2-4c\ =\ (b-2c)^2 \\~\\ b^2-4c\ =\ (b-2c)(b-2c) \\~\\ b^2-4c\ =\ b^2-4bc+4c^2$

                                                       Subtract  b²  from both sides of the equation

$-4c\ =\ -4bc+4c^2$

                                                       Add  4c  to both sides of the equation

$0 =\ -4bc+4c^2 + 4c$

                                                       Rearrange the terms

$0 =\ 4c^2 -4bc+ 4c$

                                             Divide through by  4

$0 =\ c^2 -bc+ c$

                                             Factor  c  out of all three terms on the right sidde

$0 =\ c(c -b+ 1)$

                                             Set each factor equal to zero and solve for  c

 

$\begin{array}{ccc} c=0&\quad\text{or}\quad&c-b+1=0\\ &&c-b=-1\\ &&c=b-1 \end{array}$

 

Check: https://www.wolframalpha.com/input/. . .

 

So either  c = 0  or  c = b - 1

 

We are given that  c ≠ 0,

 

So it must be that  c = b - 1