Quadratic Equation Roots Condition Analysis

The problem asks for the condition on the coefficients \(p\) and \(q\) of the quadratic equation \(x^2 + px + q = 0\) such that one root is the square of the other root.

Understanding the Roots of a Quadratic Equation

A quadratic equation of the form \(ax^2 + bx + c = 0\) has roots, say \(\alpha\) and \(\beta\). According to Vieta's formulas, the sum and product of the roots are related to the coefficients:

• Sum of roots: \(\alpha + \beta = -\frac{b}{a}\)
• Product of roots: \(\alpha \beta = \frac{c}{a}\)

Applying Vieta's Formulas to the Given Equation

For the equation \(x^2 + px + q = 0\), the coefficients are \(a=1\), \(b=p\), and \(c=q\). Let the roots be \(\alpha\) and \(\beta\). Vieta's formulas give us:

• Sum of roots: \(\alpha + \beta = -\frac{p}{1} = -p\)
• Product of roots: \(\alpha \beta = \frac{q}{1} = q\)

Setting Up the Root Relationship

The problem states that one root is the square of the other. Let's assume \(\beta = \alpha^2\). We can substitute this relationship into Vieta's formulas:

1. Sum of roots: \(\alpha + \alpha^2 = -p\)
2. Product of roots: \(\alpha \cdot \alpha^2 = q \implies \alpha^3 = q\)

Deriving the Condition on p and q

We have the equations:

\(\alpha + \alpha^2 = -p\)    (Equation 1)

\(\alpha^3 = q\)    (Equation 2)

We need to eliminate \(\alpha\) to find a relationship between \(p\) and \(q\). From Equation 1, we can cube both sides to introduce the term \(\alpha^3\):

\((\alpha + \alpha^2)^3 = (-p)^3\)

Using the algebraic identity \((a+b)^3 = a^3 + b^3 + 3ab(a+b)\):

\((\alpha)^3 + (\alpha^2)^3 + 3(\alpha)(\alpha^2)(\alpha + \alpha^2) = -p^3\)

This simplifies to:

\(\alpha^3 + \alpha^6 + 3\alpha^3(\alpha + \alpha^2) = -p^3\)

Now, substitute \(\alpha^3 = q\) (from Equation 2) and \(\alpha + \alpha^2 = -p\) (from Equation 1) into this equation:

\(q + (q)^2 + 3(q)(-p) = -p^3\)

This simplifies to:

\(q + q^2 - 3pq = -p^3\)

To get the condition in a standard form, move all terms to one side:

\(p^3 + q + q^2 - 3pq = 0\)

Rearranging the terms:

\(p^3 + q + q^2 = 3pq\)

This is the required condition on \(p\) and \(q\) for one root of the equation \(x^2 + px + q = 0\) to be the square of the other root.

Comparing with Options

Let's check the derived condition \(p^3 + q + q^2 = 3pq\) against the given options:

• Option 1: \(1 + q + q^2 = 3pq\) - Does not match.
• Option 2: \(1 + p + p^2 = 3pq\) - Does not match.
• Option 3: \(p^3 + q + q^2 = 3pq\) - Matches.
• Option 4: \(q^3 + p + p^2 = 3pq\) - Does not match.

Therefore, the correct condition is \(p^3 + q + q^2 = 3pq\).

Revision Table: Quadratic Equation Roots

Additional Information: Types of Roots Conditions

Understanding conditions on roots is important in quadratic equations. Here are a few other common conditions and how they relate to the coefficients \(p\) and \(q\) for \(x^2 + px + q = 0\):

• Roots are equal: The discriminant is zero. \(D = b^2 - 4ac = 0\). For \(x^2 + px + q = 0\), this is \(p^2 - 4q = 0\), or \(p^2 = 4q\).
• Roots are opposite in sign and unequal magnitude: Product of roots is negative. \(\alpha \beta < 0\). For \(x^2 + px + q = 0\), this is \(q < 0\).
• Roots are opposite in sign and equal magnitude (additive inverses): Sum of roots is zero. \(\alpha + \beta = 0\). For \(x^2 + px + q = 0\), this is \(-p = 0\), or \(p = 0\). The equation becomes \(x^2 + q = 0\), roots are \(x = \pm \sqrt{-q}\). This requires \(q < 0\) for real roots, or \(q > 0\) for purely imaginary roots.
• Roots are reciprocal of each other: Product of roots is 1. \(\alpha \beta = 1\). For \(x^2 + px + q = 0\), this is \(q = 1\).

These conditions are derived by setting up relationships between the roots based on the given condition and then using Vieta's formulas to express that relationship in terms of the coefficients \(p\) and \(q\).