Examples On Sum And Product Of Roots Of Quadratics

Go back to  'Quadratic Equations'

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

Learn from the best math teachers and top your exams

  • Live one on one classroom and doubt clearing
  • Practice worksheets in and after class for conceptual clarity
  • Personalized curriculum to keep up with school