# 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.

grade 10 | Questions Set 2
grade 10 | Questions Set 1