Sum and Product of Roots - Examples

Solved Example 1: Write the sum and the product of the roots of \(5{x^2} + 6x - 1 = 0\).

Solution: We have:

\[S =  - \frac{b}{a} =  - \frac{6}{5},\,\,\,P = \frac{c}{a} =  - \frac{1}{5}\]

Solved Example 2: If the sum and product of the roots of \(p{x^2} + 2x + 3p = 0\) are equal, then the value of \( - 3p\) is _____.

Solution: We have:

\[\begin{align}&S = P\\& \Rightarrow \,\,\, - \frac{2}{p} = \frac{{3p}}{p}\\&\Rightarrow \,\,\,p =  - \frac{2}{3}\,\,\, \Rightarrow \,\,\, - 3p = 2\end{align}\]

Solved Example 3: If \(\alpha ,\,\,\beta \) are the roots of \({x^2} + 4x + 6 = 0\), find the equation whose roots are \(\frac{1}{\alpha },\,\,\frac{1}{\beta }\).

Solution: We note that:

\[\alpha  + \beta  =  - 4,\,\,\,\alpha \beta  = 6\]

Now, let us evaluate the sum and product of roots of the equation we are looking for.

\[\begin{align}&S = \frac{1}{\alpha } + \frac{1}{\beta } = \frac{{\alpha  + \beta }}{{\alpha \beta }} = \frac{{ - 4}}{6} =  - \frac{2}{3}\\&P = \left( {\frac{1}{\alpha }} \right)\left( {\frac{1}{\beta }} \right) = \frac{1}{{\alpha \beta }} = \frac{1}{6}\end{align}\]

The required equation is:

\[\begin{align}&{x^2} - Sx + P = 0\\&\Rightarrow \,\,\,{x^2} - \left( { - \frac{2}{3}} \right)x + \frac{1}{6} = 0\\&\Rightarrow \,\,\,{x^2} + \frac{2}{3}x + \frac{1}{6} = 0\\&\Rightarrow \,\,\,6{x^2} + 4x + 1 = 0\end{align}\]

Solved Example 4: Find the value of p if one root of \({x^2} + x - p = 0\) is the square of the other.

Solution: Let the two roots be \(\alpha ,\,\,{\alpha ^2}\). We have:

\[\alpha  + {\alpha ^2} =  - 1,\,\,\,\alpha  \times {\alpha ^2} = {\alpha ^3} =  - p\]

From the first relation, we have:

\[{\alpha ^2} + \alpha  + 1 = 0\]

Multiplying both sides by \(\left( {\alpha  - 1} \right)\), we have:

\[\begin{align}&{\alpha ^3} - 1 = 0\,\,\, \Rightarrow \,\,\, - p - 1 = 0\\&\Rightarrow p =  - 1\end{align}\]

You may observe that for this value of p, the roots of the equation are non-real. However, one root is the square of the other (you can verify this if you know how to manipulate complex numbers).

Solved Example 5: Find the values of x for which the roots p and q of the quadratic equation

\[{t^2} - 8t + x = 0\]

satisfy the condition \({p^2} + {q^2} = 4\).

Solution: Note that the given equation is quadratic in the variable t, and not in x. Since p and q are the two roots of this equation, we have:

\[\left\{ \begin{array}{l}p + q = 8\\pq = x\end{array} \right.\]

Now,

\[\begin{align}&{p^2} + {q^2} = 4\\&\Rightarrow \,\,\,{\left( {p + q} \right)^2} - 2pq = 4\\&\Rightarrow \,\,\,64 - 2x = 4\\&\Rightarrow \,\,\,x = 30\end{align}\]

Solved Example 6: The roots of \(a{x^2} + bx + c = 0\) are in the ratio \(p:q\). Find the value of \(\frac{{{b^2}}}{{ac}}\).

Solution: Let us represent the roots by \(\alpha ,\,\,\beta \). We have:

\[\begin{align}&\frac{\alpha }{\beta } = \frac{p}{q}\,\,\, \Rightarrow \,\,\,\frac{{\alpha  + \beta }}{{\alpha  - \beta }} = \frac{{p + q}}{{p - q}}\\&\Rightarrow \,\,\,{\left( {\frac{{\alpha  + \beta }}{{\alpha  - \beta }}} \right)^2} = {\left( {\frac{{p + q}}{{p - q}}} \right)^2}\\ & \Rightarrow \,\,\,\frac{{{{\left( {\alpha  + \beta } \right)}^2}}}{{{{\left( {\alpha  + \beta } \right)}^2} - 4\alpha \beta }} = {\left( {\frac{{p + q}}{{p - q}}} \right)^2}\end{align}\]

Now, we use the relations for the sum and product of the roots:

\[\begin{align}&\frac{{{{\left( { - \frac{b}{a}} \right)}^2}}}{{{{\left( { - \frac{b}{a}} \right)}^2} - 4\frac{c}{a}}} = {\left( {\frac{{p + q}}{{p - q}}} \right)^2}\\&\Rightarrow \,\,\,\frac{{{b^2}}}{{{b^2} - 4ac}} = {\left( {\frac{{p + q}}{{p - q}}} \right)^2}\end{align}\]

Cross-multiplying, we have:

\[\begin{align}&{b^2}{\left( {p - q} \right)^2} = {b^2}{\left( {p + q} \right)^2} - 4ac{\left( {p + q} \right)^2}\\& {b^2}\left\{ {{{\left( {p + q} \right)}^2} - {{\left( {p - q} \right)}^2}} \right\} = 4ac{\left( {p + q} \right)^2}\\&\Rightarrow \,\,\,{b^2}\left( {4pq} \right) = 4ac{\left( {p + q} \right)^2}\\&\Rightarrow \,\,\,\frac{{{b^2}}}{{ac}} = \frac{{{{\left( {p + q} \right)}^2}}}{{pq}}\end{align}\]

Solved Example 7: The roots of \(6{x^2} - 5x - 3 = 0\) are \(\alpha ,\,\,\beta \). Find the quadratic equation whose roots are \(\alpha  - {\beta ^2}\) and \(\beta  - {\alpha ^2}\).

Solution: We have:

\[\alpha  + \beta  = \frac{5}{6},\,\,\,\alpha \beta  =  - \frac{1}{2}\]

Let us determine the sum and the product of the roots of the quadratic equation we are seeking.

Sum

\[\begin{align}&S = \left( {\alpha  - {\beta ^2}} \right) + \left( {\beta  - {\alpha ^2}} \right)\\& = \left( {\alpha  + \beta } \right) - \left( {{\alpha ^2} + {\beta ^2}} \right)\\&= \left( {\alpha  + \beta } \right) - \left\{ {{{\left( {\alpha  + \beta } \right)}^2} - 2\alpha \beta } \right\}\\&= \frac{5}{6} - \left\{ {{{\left( {\frac{5}{6}} \right)}^2} - 2\left( { - \frac{1}{2}} \right)} \right\}\\&= \frac{5}{6} - \frac{{25}}{{36}} - 1 =  - \frac{{31}}{{36}}\end{align}\]

Product

\[\begin{align}P &= \left( {\alpha  - {\beta ^2}} \right)\left( {\beta  - {\alpha ^2}} \right)\\&= \alpha \beta  - {\alpha ^3} - {\beta ^3} + {\alpha ^2}{\beta ^2}\\&= \alpha \beta  + {\alpha ^2}{\beta ^2} - \left( {{\alpha ^3} + {\beta ^3}} \right)\\
& = \alpha \beta\; + {\alpha ^2}{\beta ^2} - \left\{ {{{\left( {\alpha  + \beta } \right)}^3} - 3\alpha \beta \left( {\alpha  + \beta } \right)} \right\}\\&=\left( { - \frac{1}{2}} \right) + {\left( { - \frac{1}{2}} \right)^2} - \left\{ {{{\left( {\frac{5}{6}} \right)}^3} - 3\left( { - \frac{1}{2}} \right)\left( {\frac{5}{6}} \right)} \right\}\\&=  - \frac{1}{2} + \frac{1}{4} - \left\{ {\frac{{125}}{{216}} + \frac{5}{4}} \right\} =  - \frac{{449}}{{216}}\end{align}\]

Thus, the required quadratic equation is:

\[\begin{align}&{x^2} - Sx + P = 0\\&\Rightarrow \,\,\,{x^2} - \left( { - \frac{{31}}{{36}}} \right)x + \left( { - \frac{{449}}{{216}}} \right) = 0\\&\Rightarrow \,\,\,216{x^2} + 186x - 449 = 0\end{align}\]

Solved Example 8: Find the quadratic equation whose roots are double the roots of the equation \({x^2} + ax + b = 0\).

Solution: Let the roots of the given equation be \(\alpha ,\,\,\beta \). We have:

\[\alpha  + \beta  =  - a,\,\,\,\alpha \beta  = b\]

The roots of the equation we are seeking will be \(2\alpha ,\,\,2\beta \). The sum and product of the roots will be:

\[\begin{align}&S = 2\alpha  + 2\beta  = 2\left( {\alpha  + \beta } \right) =  - 2a\\&P = \left( {2\alpha } \right)\left( {2\beta } \right) = 4\alpha\,\beta  = 4b\end{align}\]

The required equation is:

\[\begin{align}&{x^2} - Sx + P = 0\\&\Rightarrow \,\,\,{x^2} + 2ax + 4b = 0\end{align}\]

Solved Example 9: Let \(a,\,\,b\) be the roots of \({x^2} + px + q = 0\) and \(c,\,\,d\) be the roots of \({x^2} + Px + Q = 0\). Find the value of

\[Z = \left( {a - c} \right)\left( {b - c} \right)\left( {a - d} \right)\left( {b - d} \right)\]

Solution: Multiply the first two brackets and the last two brackets:

\[\begin{align}&Z = \left( {ab - c\left( {a + b} \right) + {c^2}} \right)\left( {ab - d\left( {a + b} \right) + {d^2}} \right)\\&= \left( {q - c\left( { - p} \right) + {c^2}} \right)\left( {q - d\left( { - p} \right) + {d^2}} \right)\\& = \left( {q + c\left( {p + c} \right)} \right)\left( {q + d\left( {p + d} \right)} \right)\\&= \left\{ \begin{array}{l}{q^2} + qc\left( {p + c} \right) + qd\left( {p + d} \right)\\ + cd\left( {p + c} \right)\left( {p + d} \right)\end{array} \right.\end{align}\]

Now, we express this in terms of the sum and product of \(c,\,\,d\). We have:

\[\begin{align}& Z = \left\{ \begin{array}{l} {q^2} + qc\left( {p + c} \right) + qd\left( {p + d} \right)\\ + cd\left( {p + c} \right)\left( {p + d} \right) \end{array} \right.\\&\;\;\; = \left\{ \begin{array}{l} {q^2} + pq\left( {c + d} \right) + q\left( {{c^2} + {d^2}} \right)\\ + {p^2}cd + pcd\left( {c + d} \right) + {c^2}{d^2} \end{array} \right. \end{align}\]

Now we replace the values of \(c + d\) and \(cd\):

\[\begin{align}&Z\,\, = \left\{ \begin{array}{l}{q^2} + pq\left( { - P} \right) + q\left( {{P^2} - 2Q} \right)\\ + {p^2}Q + pQ\left( { - P} \right) + {Q^2}\end{array} \right.\\&\,\,\,\;\; = \left\{ \begin{array}{l}{q^2} - pqP + q{P^2} - 2qQ\\ + {p^2}Q - pPQ + {Q^2}\end{array} \right.\\&\,\,\,\,\; = \left\{ \begin{array}{l}{Q^2} + {q^2} - pP\left( {Q + q} \right)\\ + q{P^2} + Q{p^2} - 2Qq\end{array} \right.\end{align}\]

Solved Example 10: Let \(p\left( x \right)\) and \(Q\left( x \right)\) be two quadratic polynomials with integer coefficients. Suppose that they have a non-rational zero in common. Show that \(p\left( x \right) = rq\left( x \right)\), where r is some rational number.

Solution: Let the non-rational common zero be \(\alpha \). Let the other root of \(p\left( x \right)\) be \({\beta _1}\) and that of \(Q\left( x \right)\) be \({\beta _2}\). What can we say about \({\beta _1}\)? Will it be rational or irrational? If it is rational, then \(\alpha  + {\beta _1}\) and \(\alpha {\beta _1}\) will be irrational, which is not possible because \(p\left( x \right)\) has integer coefficients. Thus, \({\beta _1}\) must be irrational. Similarly, \({\beta _2}\) must be irrational.

Since \(\alpha  + {\beta _1}\) and \(\alpha  + {\beta _2}\) are both rational (why?), their difference, which is \({\beta _1} - {\beta _2}\), must be rational. Similarly, since \(\alpha {\beta _1}\) and \(\alpha {\beta _2}\) are both rational (again, why?), their ratio \(\frac{{{\beta _1}}}{{{\beta _2}}}\) must be rational.

Let\({\beta _1} - {\beta _2} = {r_1},\,\,\,\frac{{{\beta _1}}}{{{\beta _2}}} = {r_2}\), where both \({r_1},\,\,{r_2}\)are rational. Substituting \({\beta _1} = {r_2}{\beta _2}\) in the first relation, we have:

\[\begin{align}&{r_2}{\beta _2} - {\beta _2} = {r_1}\\&\Rightarrow \,\,\,{\beta _2}\left( {{r_2} - 1} \right) = {r_1}\end{align}\]

Since \({\beta _2}\) is irrational, the above equality is possible only if \({r_2} = 1\), which means that \({\beta _1} = {\beta _2}\). Thus, the other zeroes of the two quadratic polynomials are also equal. This means that their coefficients are in the same ratio. Since the coefficients are integers, this ratio must be a rational number, say r.

This means that we can write \(p\left( x \right) = rq\left( x \right)\).

Solved Example 11: If \({x^2} - 10ax - 11b = 0\) has roots \(c,\,\,d\) and \({x^2} - 10cx - 11d = 0\) has roots \(a,\,\,b\), find the value of \(a + b + c + d\).

Solution: Using the relation for the sum of roots, we have:

\[\begin{align}&c + d = 10a,\,\,\,a + b = 10c\\&\Rightarrow \,\,\,\left( {a + b} \right) - \left( {c + d} \right) = 10c - 10a\\&\Rightarrow \,\,\,\left( {a - c} \right) + \left( {b - d} \right) = 10\left( {c - a} \right)\\&\Rightarrow \,\,\,\left( {b - d} \right) = 11\left( {c - a} \right)\end{align}\]

Now, since c is a root of \({x^2} - 10ax - 11b = 0\), we have:

\[{c^2} - 10ac - 11b = 0 \quad …(1)\]      

Similarly, since a is a root of \({x^2} - 10cx - 11d = 0\), we have:

\[{a^2} - 10ac - 11d = 0 \quad …(2)\] 

By (1) – (2), we have:

\[\left( {c + a} \right)\left( {c - a} \right) = 11\left( {b - d} \right) = 121\left( {c - a} \right)\]

Thus,

\[\begin{align}&a + b + c + d = 10c + 10a\\& = 10\left( {a + c} \right) = 1210\end{align}\]

Solved Example 12: It is given that the square of the difference of the roots of the quadratic equation \({x^2} + px + 10 = 0\) is 9. Find the roots of the equation.

Solution: Let the two roots be \(\alpha ,\,\,\beta \). We have:

\[\begin{align}&{\left( {\alpha  - \beta } \right)^2} = 9\,\,\,\\&\Rightarrow \,\,\,{\left( {\alpha  + \beta } \right)^2} - 4\alpha \beta  = 9\\&\Rightarrow \,\,\,{\left( { - p} \right)^2} - 4\left( {10} \right) = 9\\&\Rightarrow \,\,\,{p^2} = 49\\&\Rightarrow \,\,\,p =  \pm 7\end{align}\]

If \(p =  + 7\), then the quadratic equation becomes

\[\begin{array}{l}{x^2} + 7x + 10 = 0\\ \Rightarrow \,\,\,\left( {x + 2} \right)\left( {x + 5} \right) = 0\end{array}\]

In this case, the roots are \( - 2,\,\, - 5\). On the other hand, if \(p =  - 7\), the quadratic equation becomes

\[\begin{array}{l}{x^2} - 7x + 10 = 0\\ \Rightarrow \,\,\,\left( {x - 2} \right)\left( {x - 5} \right) = 0\end{array}\]

In this case, the roots are \(2,\,\,5\).

Solved Example 13: Find the values of \(a\) for which one of the roots of the quadratic equation

\[{x^2} + \left( {2a + 1} \right)x + \left( {{a^2} + 2} \right) = 0\]

is twice the other root. For these values of \(a\), find the roots of the equation as well.

Solution: Since one of the roots is double the other, we can assume the two roots to be \(\alpha ,\,\,2\alpha \). Using the relations for the sum and product of roots, we have:

\[\left\{ \begin{align}S = \alpha  + 2\alpha  =  - \left( {2a + 1} \right)\\   \Rightarrow  \qquad\;\;\; \alpha  =  - \frac{{\left( {2a + 1} \right)}}{3}\\P = 2{\alpha ^2} = \left( {{a^2} + 2} \right)\end{align} \right.\]

Substituting the value of \(\alpha \) from the first relation into the second, we have:

\[\begin{align}&2{\left( { - \frac{{\left( {2a + 1} \right)}}{3}} \right)^2} = {a^2} + 2\\&\Rightarrow \,\,\,2{\left( {2a + 1} \right)^2} = 9{a^2} + 18\\&\Rightarrow \,\,\,2\left( {4{a^2} + 4a + 1} \right) = 9{a^2} + 18\\&\Rightarrow \,\,\,8{a^2} + 8a + 2 = 9{a^2} + 18\\&\Rightarrow \,\,\,{a^2} - 8a + 16 = 0\\&\Rightarrow \,\,\,{\left( {a - 4} \right)^2} = 0\\&\Rightarrow \,\,\,a = 4\end{align}\]

For this value of \(a\), the original equation becomes

\[\begin{align}&{x^2} + 9x + 18 = 0\\&\Rightarrow \,\,\,\left( {x + 3} \right)\left( {x + 6} \right) = 0\end{align}\]

Thus, the two roots are \(- 3,\,\, - 6\). Clearly, one root is twice the other.