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 \(b\).

CapriSun Sep 6, 2020

#1**0 **

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 b^{2} 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

hectictar Sep 6, 2020