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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 \(b\).

 Oct 23, 2019
edited by Guest  Oct 23, 2019
edited by Guest  Oct 23, 2019
 #1
avatar+8652 
+3

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 \(b \).

 

Hello Guest!

 

 

\(-\frac{b}{2}+\sqrt{(\frac{b}{2})^2-c}-(-\frac{b}{2}-\sqrt{(\frac{b}{2})^2-c})\\ \color{blue}=2\cdot \sqrt{(\frac{b}{2})^2-c}\)

 

The difference of the roots is

\(2\cdot \sqrt{(\frac{b}{2})^2-c}\)

 

\(c\geq (\frac{b}{2})^2\ |b\in \mathbb{R}\)

laugh  !

.
 Oct 23, 2019
edited by asinus  Oct 23, 2019
 #2
avatar+46 
+2

From #1, the difference between the roots 

\(\displaystyle = 2\sqrt{(b/2)^{2}-c}=\sqrt{b^{2}-4c}= |b-2c|.\)

Squaring,

\(\displaystyle b^{2}-4c=b^{2}-4bc+4c^{2},\)

from which

\(\displaystyle c(c-b+1)=0,\\ \text{so that} \\c=0, \text{ or }c=b-1.\)

 

\(\displaystyle c=0 \text{ is excluded, so }c=b-1.\)

 

The equation can be written as 

\(\displaystyle x^{2}+bx+b-1=(x+1)(x+(b-1))=0.\)

The roots will be

\(\displaystyle -1 \text{ and }1-b,\\\text{ so the modulus of the difference will be }\\|2-b|=|b-2(b-1)|=|b-2c|.\)

.
 Oct 23, 2019
 #3
avatar+23358 
+2

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 \(b\).

 

\(\begin{array}{|lrcll|} \hline & x^2 + \underbrace{ b }_{-(x_1+x_2)}x + \underbrace{ c }_{x_1x_2} &=& 0 \\ \hline & b &=& -(x_1+x_2) \\ & -b &=& x_1+x_2 \\ (1) & \mathbf{x_2} &=& \mathbf{-x_1-b} \\ \hline (2) & \mathbf{c} &=& \mathbf{x_1x_2} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline | x_1-x_2| &=& |b - 2c| \quad &| \quad \mathbf{x_2=-x_1-b} \\ | x_1-(-x_1-b)| &=& |b - 2c| \\ | 2x_1+b)| &=& |b - 2c| \quad &| \quad \text{compare}\\\\ 2x_1 &=& -2c \quad &| \quad : 2 \\ \mathbf{x_1} &=& \mathbf{-c} \\\\ \mathbf{x_2} &=& \mathbf{-x_1-b} \quad (1) \quad & | \quad \mathbf{x_1=-c} \\ x_2 &=& \mathbf{-(-c)-b} \\ \mathbf{x_2} &=& \mathbf{c-b} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{c} &=& \mathbf{x_1x_2} \quad (2) \quad & | \quad \mathbf{x_1=-c},\quad \mathbf{x_2} = \mathbf{c-b} \\ c &=& (-c)(c-b)\quad &| \quad : c \qquad c \neq 0\ ! \\ 1 &=& -(c-b) \\ 1 &=& -c+b \\ \mathbf{c} &=& \mathbf{b-1} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{x_1} &=& \mathbf{-c} \quad & | \quad \mathbf{c=b-1} \\ x_1 &=& -(b-1) \\ \mathbf{x_1} &=& \mathbf{1-b} \\ \\ \mathbf{x_2} &=& \mathbf{c-b} \quad & | \quad \mathbf{c=b-1} \\ x_2 &=& b-1-b \\ \mathbf{x_2} &=& \mathbf{-1} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline | x_1-x_2| &=& |1-b-(-1)| \\ | x_1-x_2| &=& |1-b+1| \\ | x_1-x_2| &=& |2-b| \\ \hline |b-2c| &=& |b-2(b-1)| \\ |b-2c| &=& |b-2b+2)| \\ |b-2c| &=& |2-b|\ \checkmark \\ \hline \end{array}\)

 

laugh

 Oct 24, 2019

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