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The difference of the roots of the quadratic equation $$x^2 + bx + c = 0$$ is$$|b - 2c|$$ . If $$c \neq 0$$, then find $$c$$ in terms of $$b$$.

Sep 6, 2020

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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

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

So either  c = 0  or  c = b - 1

We are given that  c ≠ 0,

So it must be that  c = b - 1

Sep 6, 2020