Question

If alpha and beta are the zeroes of the quadratic polynomial p(x)=ax2+bx+c then evaluate alpha minus beta.

Answer 1

From quadratic formula we have
$\alpha=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\beta=\dfrac{-b\mp\sqrt{b^2-4ac}}{2a}$

subtracting you get
$\alpha-\beta=\dfrac{\pm2\sqrt{b^2-4ac}}{2a} = \dfrac{\pm\sqrt{b^2-4ac}}{a}$

Answer 2

Answer:

The value of $\alpha - \beta = \frac { \sqrt { b ^ { 2 } - 4 a c } } {a}$

To find:

$\alpha-\beta$

Solution:

Given that

α and  β are zeros

$p(x)=ax^2+bx+c$

Here a, b and c are the known values but x cannot be zero

x is the variable.

For every quadratic equation, there may be one or more than one solution. Those are called as roots of the quadratic equation.

We know that from standard quadratic equation

The sum of roots is

$\alpha + \beta = \frac { - \mathrm { b } } { \mathrm { a } }$

The product of roots is

$\alpha \beta = \frac { c } { a }$

Square the $\alpha-\beta$

Will get,

$( \alpha - \beta ) ^ { 2 } = ( \alpha + \beta ) ^ { 2 } - 4 \alpha \beta$

Substitute the α+β and αβ values in the above equation

$\begin{aligned} & = \left( \frac { - b } { a } \right) ^ { 2 } - 4 \left( \frac { c } { a } \right) \\\\ & = \frac { b ^ { 2 } } { a ^ { 2 } } - \frac { 4 c } { a } \\\\ & = \frac { \mathrm { b } ^ { 2 } - 4 \mathrm { ac } } { \mathrm { a } ^ { 2 } } \end{aligned}$

$\begin{array} { l } { ( \alpha - \beta ) ^ { 2 } = \frac { \mathrm { b } ^ { 2 } - 4 \mathrm { ac } } { \mathrm { a } ^ { 2 } } } \\\\ { \alpha - \beta = \sqrt { \frac { \mathrm { b } ^ { 2 } - 4 \mathrm { ac } } { \mathrm { a } ^ { 2 } } } } \end{array}$

$\begin{array} { l } { = \frac { \sqrt { b ^ { 2 } - 4 a c } } { a } } \\\\ { \alpha - \beta = \frac { \sqrt { b ^ { 2 } - 4 a c } } { a } } \end{array}$