Question

If one root is the square of the other root of the equation \[{x^2} + px + q = 0\] then find the relation between p and q.

Answer 1

Hint: Use relation already present between the roots of a quadratic equation.
The sum of the roots of the quadratic equation is equal to the negation of the coefficient of the second term, divided by the leading coefficient (leading coefficient means coefficient of first term).
The product of the roots of the quadratic equation is equal to the constant term.

Complete step-by-step answer:
Let a and b be the roots of a given quadratic equation.
But according to given condition,
 One root is the square of the other root, \[b = {a^2}\].
Now let’s get the relation between the coefficients of the quadratic equation.
1 The sum of the roots of the quadratic equation is equal to the negation of the coefficient of the second term, divided by the leading coefficient.
\[ \Rightarrow [a(a + 1)] = - p\]
2.The product of the roots of the quadratic equation is equal to the constant term.
\[
   \Rightarrow a \times b = q \\
   \Rightarrow a \times {a^2} = q \\
\]
Now let’s find the relation between p and q.
\[
   \Rightarrow a + {a^2} = - p \\
   \Rightarrow [a(a + 1)] = - p \\
\]
Cubing both sides ,
\[
   \Rightarrow {\left[ {a\left( {a + 1} \right)} \right]^3} = {\left( { - p} \right)^3} \\
   \Rightarrow \left[ {{a^3}\left( {{a^3} + 3{a^2} + 3a + {1^3}} \right)} \right] = - {p^3} \\
   \Rightarrow \left[ {{a^3}\left( {{a^3} + 3{a^2} + 3a + 1} \right)} \right] = - {p^3} \\
   \Rightarrow \left[ {{a^3}\left( {{a^3} + 3(a + {a^2}) + 1} \right)} \right] = - {p^3} \\
\]
Replacing \[a + {a^2}\]by \[ - p\]
\[ \Rightarrow \left[ {q\left( {q - 3p + 1} \right)} \right] = - {p^3}\]
On multiplying by q,
\[ \Rightarrow {q^2} - 3pq + q = - {p^3}\]
Thus rearranging the terms,
\[ \Rightarrow {p^3} + {q^2} + q - 3pq = 0\]
This is the relation between p and q.

Note: Roots of the quadratic equation have relation between them.
Many students may have a doubt that why to take cubing of both sides not to take square. Because squaring both sides will stuck the solution at a point where terms of a will not be replaced by either p or q. So we took cubing of both sides.