# Sum And Product Of Roots Of Quadratics

Go back to  'Quadratic Equations'

In the last section we saw the quadratic formula that gives the roots of a quadratic expression. From now on we will be denoting the two roots by $$\alpha$$  and $$\beta$$.

$\alpha ,\beta = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Equation (3) tells us that a quadratic expression can be written in terms of its roots as follows:

\begin{align}&f\left( x \right) = a{x^2} + bx + c = \,\,a\left( {x - \alpha } \right)\left( {x - \beta } \right)\\\\ \,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\quad\quad\;\;\;= \,\,a\left( {{x^2} - \left( {\alpha + \beta } \right)x + \alpha \beta } \right)\\\\ \,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\quad\quad\;\;\;= \,\,a{x^2} - a\left( {\alpha + \beta } \right)x + a\alpha \beta \\\\ \,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\quad\quad\;\;\;= \,\,a{x^2} - aSx + aP = a\left( {{x^2} - Sx + P} \right)\end{align}

where $$S$$ denotes the sum and  $$P$$  the product of the roots. Comparing the coefficients of  $$‘x’$$ on both sides gives:

\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - a\left( {\alpha + \beta } \right)\;=\,\, b\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&\qquad\qquad\qquad\;\alpha \beta \,\,\,\,\,\,\,\, = c\end{align}

\begin{align}&\Rightarrow\qquad S = {\rm{Sum}}\,{\rm{of}}\,{\rm{roots}} = \alpha + \beta = \frac{{ - b}}{a}\\&\Rightarrow\qquad P = {\rm{Product}}\,{\rm{of}}\,{\rm{roots}} = \alpha \beta = \frac{c}{a}\end{align}

These are important results. They can of course be derived directly from the expression for the roots given by the quadratic formula. For example

$\alpha + \beta = \frac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} + \frac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}} = \frac{{ - b}}{a}$

Now lets see some examples that use these results:

Example- 13

If $$\alpha ,\;\beta$$ are the roots of $$a{x^2} + bx + c = 0,$$  find the quadratic equation whose roots are $${\left( {a\alpha + b} \right)^{ - 1}}{\rm{and}}\;\; {\left( {a\beta + b} \right)^{ - 1}}$$

Solution: We have

\begin{align}&\alpha + \beta = \frac{{ - b}}{a}\\&\quad\alpha \beta = \frac{c}{a}\end{align}

for the required quadratic equation,

\begin{align}&\quad S = \frac{1}{{a\alpha + b}} + \frac{1}{{a\beta + b}} = \frac{{a\left( {\alpha + \beta } \right) + 2b}}{{{a^2}\alpha \beta + ab\left( {\alpha + \beta } \right) + {b^2}}}\\\\\,\,\,\, &\quad= \frac{{a \times \frac{{ - b}}{a} + 2b}}{{{a^2} \times \frac{c}{a} + ab \times \frac{{ - b}}{a} + {b^2}}} = \frac{b}{{ac}}\\\\&P = \frac{1}{{a\alpha + b}}\, \cdot \frac{1}{{a\beta + b}} = \frac{1}{{{a^2}\alpha \beta + ab\left( {\alpha + \beta } \right) + {b^2}}} = \frac{1}{{ac}}\end{align}

The required equation is

\begin{align}{x^2} - Sx + P = 0 \,\,\,\,\,\,\, &\Rightarrow \,\,{x^2} - \frac{b}{{ac}}x + \frac{1}{{ac}} = 0 \\\\&\Rightarrow \;\;ac{x^2} - bx + 1= 0\end{align}

Example- 14

If $$\alpha ,\beta$$ are the roots of $$a{x^2} + bx + c = 0,$$ find the values of:

(a) $${\alpha ^2} + {\beta ^2}$$         (b) $${\alpha ^3} + {\beta ^3}$$         (c) $${\alpha ^4} + {\beta ^4}$$         (d) \begin{align}\frac{\alpha }{\beta } + \frac{\beta }{\alpha }\end{align}

Solution: This question is very simple since all you have to do is rearrange the given expression in terms of what you already know, i.e $$\left( {\alpha + \beta } \right)$$  and  $$\left( {\alpha \beta } \right)$$.

(a) \begin{align}{\alpha ^2} + {\beta ^2} = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta = \frac{{{b^2}}}{{{a^2}}} - \frac{{2c}}{a} = \frac{{{b^2} - 2ac}}{{{a^2}}}\end{align}

(b)\begin{align}&{\alpha ^3} + {\beta ^3} = {\left( {\alpha + \beta } \right)^3} - 3\alpha \beta \left( {\alpha + \beta } \right) = \frac{{ - {b^3}}}{{{a^3}}} + \frac{{3bc}}{{{a^2}}} = \frac{{3abc - {b^3}}}{{{a^3}}}\end{align}

(c) \begin{align}&{\alpha ^4} + {\beta ^4} = {\left( {{\alpha ^2} + {\beta ^2}} \right)^2} - 2{\left( {\alpha \beta } \right)^2} = {\left( {\frac{{{b^2} - 2ac}}{{{a^2}}}} \right)^2}\;-\;\frac{{2{c^2}}}{{{a^2}}}\left({{\rm{From}}\,{\rm{part}}\left( a \right)} \right)\end{align}

\begin{align}&= \frac{{{b^4} - 4a{b^2}c + 2{a^2}{c^2}}}{{{a^4}}}\end{align}

(d) \begin{align}\frac{\alpha }{\beta } + \frac{\beta }{\alpha } = \frac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }} = \frac{{{b^2} - 2ac}}{{{a^2}}} \cdot \frac{a}{c} = \frac{{{b^2} - 2ac}}{{ac}}\end{align}

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