If $α$ and $β$ are the zeroes of the quadratic polynomial $f(x)\ =\ x^2\ +\ x\ –\ 2$, find the value of $\frac{1}{α}\ –\ \frac{1}{β}$.
 Given:
$α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ x^2\ +\ x\ -\ 2$.
To do:
Here, we have to find the value of $\frac{1}{α} - \frac{1}{β}$.
Solution:  
We know that,
The standard form of a quadratic equation is $ax^2+bx+c=0$, where a, b and c are constants and $a≠0$.
Comparing the given equation with the standard form of a quadratic equation,
$a=1$, $b=1$ and $c=-2$
Sum of the roots $= α+β = \frac{-b}{a} = \frac{– 1}{1} = -1$.
Product of the roots $= αβ = \frac{c}{a} = \frac{-2}{1} = -2$.
Also,
$(a-b) ^2=(a+b) ^2-4ab$
$(a-b) =\sqrt{(a+b) ^2-4ab}$
Therefore,
$\frac{1}{α} -\frac{1}{β}=\frac{(β-α)}{αβ}$
$=-(\frac{α-β}{αβ})$
$=-\frac{\sqrt{(α+β) ^2-4αβ}}{αβ}$
$=-(\frac{\sqrt{(-1) ^2-4(-2)}}{(-2) })$
$=-(\frac{\sqrt{1+8}}{-2})$
$=-(\frac{\sqrt{9}}{-2})$
$=\frac{3}{2}$
The value of $\frac{1}{α}-\frac{1}{β}$ is $\frac{3}{2}$.
Related Articles
- If $α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ 6x^2\ +\ x\ –\ 2$, find the value of $\frac{α}{β}\ +\ \frac{β}{α}$.
- If $α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ x^2\ –\ x\ –\ 4$, find the value of $\frac{1}{α}\ +\ \frac{1}{β}\ –\ αβ$.
- If $α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ x^2\ -\ 2x\ +\ 3$, find a polynomial whose roots are $\frac{α\ -\ 1}{α\ +\ 1},\ \frac{β\ -\ 1}{β\ +\ 1}$.
- If $α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ x^2\ –\ 5x\ +\ 4$, find the value of $\frac{1}{α}\ +\ \frac{1}{β}\ –\ 2αβ$.
- If $α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ x^2\ -\ 3x\ -\ 2$, find a quadratic polynomial whose zeros are $\frac{1}{2α\ +\ β}$ and $\frac{1}{2β\ +\ α}$.
- If $α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ x^2\ -\ 1$, find a quadratic polynomial whose zeros are $\frac{2α}{β}$ and $\frac{2β}{α}$.
- If $α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ ax^2\ +\ bx\ +\ c$, then evaluate: $\frac{1}{α}\ -\ \frac{1}{β}$.
- If $α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ ax^2\ +\ bx\ +\ c$, then evaluate: $a\left(\frac{α^2}{β}\ +\ \frac{β^2}{α}\right)\ +\ b\left(\frac{α}{β}\ +\ \frac{β}{α}\right)$
- If $α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ ax^2\ +\ bx\ +\ c$, then evaluate: $\frac{1}{α}\ +\ \frac{1}{β}\ -\ 2αβ$
- If $α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ x^2\ -\ p(x\ +\ 1)\ –\ c$, show that $(α\ +\ 1)(β\ +\ 1)\ =\ 1\ -\ c$.
- If $α$ and $β$ are the zeros of the quadratic polynomial $p(s)\ =\ 3s^2\ -\ 6s\ +\ 4$, find the value of $\frac{α}{β}\ +\ \frac{β}{α}\ +\ 2\left(\frac{1}{α}\ +\ \frac{1}{β}\right)\ +\ 3αβ$.
- If $α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ x^2\ -\ px\ +\ q$, prove that $\frac{α^2}{β^2}\ +\ \frac{β^2}{α^2}\ =\ \frac{p^4}{q^2}\ -\ \frac{4p^2}{q}\ +\ 2$.
- If $α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ x^2\ -\ 2x\ +\ 3$, find a polynomial whose roots are $α\ +\ 2,\ β\ +\ 2$.
- If $α$ and $β$ are the zeros of the polynomial $f(x)\ =\ x^2\ +\ px\ +\ q$, form a polynomial whose zeros are $(α\ +\ β)^2$ and $(α\ -\ β)^2$.
- If $α$ and $β$ are the zeros of the quadratic polynomial $p(y)\ =\ 5y^2\ –\ 7y\ +\ 1$, find the value of $\frac{1}{α}\ +\ \frac{1}{β}$.
Kickstart Your Career
Get certified by completing the course
Get Started
Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies