# Examples On Sum And Product Of Roots Of Quadratics

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Example- 15

If $$\alpha ,\beta$$ be the roots of $${m^2}\left( {{x^2} - x} \right) + 2mx + 3 = 0$$and $${m_1},{m_2}$$ be the two values of m for which $$\alpha ,\beta$$ are connected by the relation \begin{align}&\frac{\alpha }{\beta } + \frac{\beta }{\alpha } = \frac{4}{3}\end{align} find the value of \begin{align}&\frac{{m_1^2}}{{{m_2}}} + \frac{{m_2^2}}{{{m_1}}}\end{align}

Solution: Since  $$\alpha ,\beta$$ are the roots of

\begin{align}{}f\left( x \right) &= {m^2}{x^2} + \left( {2m - {m^2}} \right)x + 3 = 0\\\alpha + \beta &= \frac{{{m^2} - 2m}}{{{m^2}}} = 1 - \frac{2}{m}\\\alpha \beta &= \frac{3}{{{m^2}}}\end{align}

\text{Now},\qquad\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{\alpha }{\beta } + \frac{\beta }{\alpha } = \frac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }} = \frac{{{{\left( {\alpha + \beta } \right)}^2}}}{{\alpha \beta }} - 2 = \frac{4}{3}\end{align}

\begin{align}&\\ &\Rightarrow\quad {\left( {\alpha + \beta } \right)^2} = \frac{{10}}{3}\alpha \beta \\\\ &\Rightarrow\quad {\left( {1 - \frac{2}{m}} \right)^2} = \frac{{10}}{{{m^2}}}\\\\ &\Rightarrow\quad {m^2} - 4m - 6 = 0\\\\ &\Rightarrow\quad {m_1} + {m_2} = 4 {m_1}{m_2} = - 6\\\\ &\Rightarrow\quad \frac{{m_1^2}}{{{m_2}}} + \frac{{m_2^2}}{{{m_1}}} = \frac{{m_1^3 + m_2^3}}{{{m_1}{m_2}}} = \frac{{{{\left( {{m_1} + {m_2}} \right)}^3} - 3{m_1}{m_2}\left( {{m_1} + {m_2}} \right)}}{{{m_1}{m_2}}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\qquad\;\;\;\quad\quad= \frac{{64 + 3 \times 4 \times 6}}{{ - 6}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\qquad\;\;\;\quad\quad= \frac{{ - 68}}{3}\end{align}

Example- 16

If $$\alpha ,\;\beta$$ are the roots of $${x^2} - px + q = 0$$find the quadratic equation whose roots are $$\left( {{\alpha ^2} - {\beta ^2}} \right)\left( {{\alpha ^3} - {\beta ^3}} \right)$$and $$\left( {{\alpha ^2}{\beta ^2} + {\alpha ^3}{\beta ^3}} \right)$$.

Solution: Knowing that  $$\alpha + \beta = p,\;\;\alpha \beta = - q$$ , we need to find the sum $$S$$ and product  $$P$$ of the roots of the quadratic we want to form.

$\begin{array}{l}S = \left( {{\alpha ^2} - {\beta ^2}} \right)\left( {{\alpha ^3} - {\beta ^3}} \right) + \left( {{\alpha ^2}{\beta ^3} + {\alpha ^3}{\beta ^2}} \right)\\\,\,\,\, = \left( {\alpha + \beta } \right)\left\{ {{\alpha ^4} + {\beta ^4} - \alpha \,\beta \left( {{\alpha ^2} + {\beta ^2} - \alpha \,\beta } \right)} \right\}\end{array}$

Now,

$\begin{array}{l}{\alpha ^4} + {\beta ^4} = {\left( {{\alpha ^2} + {\beta ^2}} \right)^2} - 2{\left( {\alpha \beta } \right)^2}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta } \right)^2} - 2{\left( {\alpha \beta } \right)^2}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {{p^2} - 2q} \right)^2} - 2{q^2}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {p^2} + 2{q^2} - 4{p^2}q\\\end{array}$

and

$\begin{array}{l}{\alpha ^2} + {\beta ^2} - \alpha \,\beta = {\left( {\alpha + \beta } \right)^2} - 3\,\alpha \,\beta \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {p^2} - 3q\end{array}$

Thus,

$\begin{array}{l}S = p\left( {{p^2} + 2{q^2} - 4{p^2}q - q\left( {{p^2} - 3q} \right)} \right)\\\,\,\,\,\, = p\left( {{p^2} + 5{q^2} - 5{p^2}q} \right)\end{array}$

Also,

$\begin{array}{l}P = \left( {{\alpha ^2} - {\beta ^2}} \right)\left( {{\alpha ^3} - {\beta ^3}} \right)\left( {{\alpha ^2}{\beta ^3} + {\alpha ^3}{\beta ^2}} \right)\\\,\,\,\,\, = {\left( {\alpha - \beta } \right)^2}{\left( {\alpha + \beta } \right)^2}{\left( {\alpha \,\beta } \right)^2}\left( {{\alpha ^2} + {\beta ^2} + \alpha \,\beta } \right)\\\,\,\,\,\, = {p^2}{q^2}\left( {{p^2} - q} \right)\left( {{p^2} - 4q} \right)\end{array}$

The required quadratic is $${x^2} - Sx + P = 0$$

Example- 17

Find all values of the parameter $$a$$ such that the roots $$\alpha ,\;\beta$$ of the equation $$2{x^2} + 6x + a = 0$$ satisfy $$\frac{\alpha }{\beta } + \frac{\beta }{\alpha } < 2$$.

Solution: We have $$\alpha + \beta = - 3,\;\;\;\alpha \,\beta = \frac{a}{2}$$. Thus,

\begin{align}&\frac{\alpha }{\beta } + \frac{\beta }{\alpha } < 2 \Rightarrow \frac{{{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta }}{{\alpha \beta }} < 2\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\quad\Rightarrow \frac{{9 - a}}{{\left( {\frac{a}{2}} \right)}} < 2\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\quad\Rightarrow \frac{{9 - a}}{a} < 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...{\rm{ }}\left( 1 \right)\end{align}

It would be a mistake to cross-multiply $$a$$ at this stage and obtain $$9 - a < a$$. The correct method is outlined below,and will be made more clear subsequently.

\begin{align}{}\frac{{9 - a}}{a} - 1 < 0 &\Rightarrow \frac{{9 - 2a}}{a} < 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 &\Rightarrow a\left( {2a - 9} \right) > 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\Rightarrow a < 0\;{\rm{or}}\;\;a > \frac{9}{2}\end{align}

The reason we cannot cross multiply $$a$$ in (1) is that we are not sure about the sign of $$a$$ . We cannot multiply a negative quantity across an inequality without reversing the sign of the inequality. These points will become clearer as we encounter more questions of this sort.

Example- 18

If $${x^2} + px + q = 0$$ and $${x^2} + p'x + q' = 0$$ have one root in common, find its value.

Solution: Let the common root be $$\alpha$$.Thus,

\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\alpha ^2} + p\,\alpha + q = 0\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&\qquad\;\;{\alpha ^2} + p'\,\alpha + q' = 0\\\\ &\Rightarrow \frac{{{\alpha ^2}}}{{pq' - p'q}} = \frac{\alpha }{{q - q'}} = \frac{1}{{p' - p}}\\\\ &\Rightarrow \alpha = \frac{{q - q'}}{{p' - p}} = \frac{{pq' - p'q}}{{q - q'}}\end{align}

Example- 19

If the distinct equations $${x^2} + bx + ca = 0$$ and $${x^2} + cx + ab = 0$$ have a non-zero common root, prove that $$a + b + c = 0$$ and their other roots satisfy $${x^2} + ax + bc = 0$$.

Solution: Let the roots of the two equations be $$\alpha ,\;{\beta _1}$$ and $$\alpha ,\;{\beta _2}$$ respectively. Thus,

$$\begin{array}{*{20}{c}}{{\alpha ^2} + b\alpha + ca = 0}\\{{\alpha ^2} + c\alpha + ab = 0}\end{array} \quad\Rightarrow \quad \left( {b - c} \right)\alpha = a\left( {b - c} \right) \Rightarrow \alpha = a$$

The common root is therefore $$\alpha = a$$. Substituting this into the first quadratic, we obtain $$a + b + c = 0$$. Now,

\begin{align}\alpha \,{\beta _1} = ca\;{\rm{and}}\;\alpha \,{\beta _2} = ab \quad \Rightarrow \quad {\beta _1} = c\;\,{\rm{and}}\;\,{\beta _2} = b\end{align}

Thus $${\beta _1} + {\beta _2} = b + c = - a\;{\rm{and}}\;{\beta _1}\,{\beta _2} = bc,$$ which means that the other two roots satisfy $${x^2} + ax + bc = 0$$.

Example- 20

If $$a{x^2} + bx + c = 0$$ has one root equal to the nth power of the other root, prove that

\begin{align}&{\left( {{a^n}c} \right)^{\frac{1}{{n + 1}}}} + {\left( {a{c^n}} \right)^{\frac{1}{{n + 1}}}} + b = 0\end{align}

Solution: Let the roots be $$\alpha ,\;{\alpha ^n},$$so that

\begin{align}&\alpha + {\alpha ^n} = - \frac{b}{a},\;\;{\alpha ^{n + 1}} = \frac{c}{a} \Rightarrow \alpha = {\left( {\frac{c}{a}} \right)^{\frac{1}{{n + 1}}}}\end{align}

Substituting this value of $$\alpha \,$$back into the first relation, we have

\begin{align}&{\left( {\frac{c}{a}} \right)^{\frac{1}{{n + 1}}}} + {\left( {\frac{c}{a}} \right)^{\frac{n}{{n + 1}}}} = - \frac{b}{a}\end{align}

From this, the required relation follows immediately.

## TRY YOURSELF - II

1. If a, b are roots of $${x^2} + px + 1 = 0$$ and c, d are the roots of $${x^2} + qx + 1 = 0$$, show that

${q^2} - {p^2} = \left( {a - c} \right)\left( {b - c} \right)\left( {a + d} \right)\left( {b + d} \right)$

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

3. In the equation $${x^2} + px + q = 0$$, the coefficient of x was incorrectly written as 17 instead of 13 and the roots were found to be –2 and –15. Find the correct equation.

4. If $$\alpha ,\;\beta$$ are the roots of $${x^2} + px + q = 0$$, express $${\alpha ^5} + {\beta ^5}$$ in terms of $$p$$  and $$q$$ .

5. Find m so that $$3{x^2} - 2mx - 4 = 0$$and $${x^2} - 4mx + 2 = 0$$ may have a common root.

6. Find a: b: c if $$a{x^2} + bx + c = 0$$and $${x^2} + 2x + 3 = 0$$ have a common root.

7. Prove that the equations

$\begin{array}{l}\left( {q - r} \right){x^2} + \left( {r - p} \right)x + p - q = 0\\\left( {r - p} \right){x^2} + \left( {p - q} \right)x + q - r = 0\end{array}$

have a common root

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• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
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