If one root is square of the other root of the equation $x^2 + px + q = 0$, then the relation between $p$ and $q$ is Options: (A) $p^3 - q(3p-1) + q^2 = 0$ (B) $p^3 - q(3p+1) + q^2 = 0$ (C) $p^3 + q(3p-1) + q^2 = 0$ (D) $p^3 + q(3p+1) + q^2 = 0$

### Analyzing the Setup We are given the quadratic equation $x^2 + px + q = 0$. We are told that one root is the square of the other. Let the roots of the equation be $\alpha$ and $\alpha^2$. By applying **Vieta's Formulas**, we establish the relationship between the roots and the coefficients: 1. Sum of roots: $\alpha + \alpha^2 = -p$ 2. Product of roots: $\alpha \cdot \alpha^2 = \alpha^3 = q$ ### The Strategy of Elimination Our goal is to eliminate $\alpha$ to find a direct relationship between $p$ and $q$. We start with the sum equation: $$\alpha + \alpha^2 = -p$$ To introduce the term $\alpha^3 = q$, we cube both sides of the equation: $$(\alpha + \alpha^2)^3 = (-p)^3$$ ### The Algebraic Dance Using the binomial expansion identity $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$, we expand the left side where $a = \alpha$ and $b = \alpha^2$: $$\alpha^3 + (\alpha^2)^3 + 3(\alpha)(\alpha^2)(\alpha + \alpha^2) = -p^3$$ Now, we perform the substitution using our known values $\alpha^3 = q$ and $\alpha + \alpha^2 = -p$: $$q + q^2 + 3(q)(-p) = -p^3$$ This simplifies to the following expression: $$q + q^2 - 3pq = -p^3$$ ### Final Calculation To reach the final form, we rearrange the terms to set the equation to zero: $$p^3 + q^2 + q - 3pq = 0$$ Grouping the terms involving $q$, we obtain: $$p^3 + q^2 - q(3p - 1) = 0$$ Thus, the final relationship is: **$p^3 - q(3p - 1) + q^2 = 0$**

JEE Advanced 2004

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Animated Solution for Mathematics - Quadratic Equations: If one root is square of the other root of the equation $x^{2} + p x + q = 0$ , then the relation between $p$ and $q$ is

Q: Why did we choose to cube the sum equation? Why not square it?

That is a fantastic question! If you square the equation $\alpha + \alpha^2 = -p$, you get $\alpha^2 + \alpha^4 + 2\alpha^3 = p^2$. While this is a valid equation, it leaves you with an $\alpha^4$ term, which is harder to simplify using only $\alpha^3 = q$. By cubing the sum, we generate an $\alpha^3$ term directly, which allows us to substitute $q$ immediately. It is all about choosing the path of least resistance!

Q: How do I know which algebraic identity to use in the expansion?

Great intuition! We used $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$ because it keeps the term $(a+b)$ intact. Since we already know that $a+b$ (which is $\alpha + \alpha^2$) equals $-p$, this form of the identity allows us to substitute $-p$ directly into the expansion, making the simplification process smooth and logical.

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Visualized Solution

The Quadratic Equation

Given equation: $x^{2} + p x + q = 0$
Condition: One root is the square of the other.

Defining the Roots

Let the first root be $α$ .
Then, the second root must be $α^{2}$ .

Vieta's Formula: Sum of Roots

Sum of roots formula: $r_{1} + r_{2} = - \frac{b}{a}$
For $x^{2} + p x + q = 0$ , $a = 1, b = p, c = q$ .
Therefore, $α + α^{2} = - p$

Vieta's Formula: Product of Roots

Product of roots formula: $r_{1} \cdot r_{2} = \frac{c}{a}$
Therefore, $α \cdot α^{2} = q$
Simplifying: $α^{3} = q$

Strategy to Eliminate $α$

We have: $α + α^{2} = - p$ and $α^{3} = q$
Goal: Find a relation between $p$ and $q$ (eliminate $α$ ).
Strategy: Cube the sum equation to generate $α^{3}$ terms.

Cubing the Sum Equation

Take the sum equation: $α + α^{2} = - p$
Cube both sides: $(α + α^{2})^{3} = (- p)^{3}$
Right side simplifies to: $- p^{3}$

Expanding the Cubic Expression

Use the algebraic identity: $(a + b)^{3} = a^{3} + b^{3} + 3 ab (a + b)$
Here, $a = α$ and $b = α^{2}$ .
Expansion: $α^{3} + (α^{2})^{3} + 3 (α) (α^{2}) (α + α^{2}) = - p^{3}$

Simplifying the Expanded Terms

Simplify $(α^{2})^{3} = α^{6} = (α^{3})^{2}$
Simplify $3 (α) (α^{2}) = 3 α^{3}$
The equation becomes: $α^{3} + (α^{3})^{2} + 3 α^{3} (α + α^{2}) = - p^{3}$

Substituting Known Values

Substitute $α^{3} = q$
Substitute $α + α^{2} = - p$
The equation becomes: $q + q^{2} + 3 q (- p) = - p^{3}$

Rearranging the Equation

Multiply the terms: $q + q^{2} - 3 pq = - p^{3}$
Bring $- p^{3}$ to the left side: $p^{3} + q^{2} + q - 3 pq = 0$

Final Factoring to Match Options

Group the terms with $q$ : $p^{3} + q^{2} - q (3 p - 1) = 0$
Rearrange to match standard form: $p^{3} - q (3 p - 1) + q^{2} = 0$
This matches Option 1.

00:00 / 00:00

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

We are given the quadratic equation

x^{2} + p x + q = 0

. We are told that one root is the square of the other.

Let the roots of the equation be

α

and

α^{2}

. By applying Vieta's Formulas, we establish the relationship between the roots and the coefficients:

1. Sum of roots:

α + α^{2} = - p

2. Product of roots:

α \cdot α^{2} = α^{3} = q

The Strategy of Elimination

Our goal is to eliminate

α

to find a direct relationship between

p

and

q

. We start with the sum equation:

α + α^{2} = - p

To introduce the term

α^{3} = q

, we cube both sides of the equation:

(α + α^{2})^{3} = (- p)^{3}

The Algebraic Dance

Using the binomial expansion identity

(a + b)^{3} = a^{3} + b^{3} + 3 ab (a + b)

, we expand the left side where

a = α

and

b = α^{2}

α^{3} + (α^{2})^{3} + 3 (α) (α^{2}) (α + α^{2}) = - p^{3}

Now, we perform the substitution using our known values

α^{3} = q

and

α + α^{2} = - p

q + q^{2} + 3 (q) (- p) = - p^{3}

This simplifies to the following expression:

q + q^{2} - 3 pq = - p^{3}

Final Calculation

To reach the final form, we rearrange the terms to set the equation to zero:

p^{3} + q^{2} + q - 3 pq = 0

Grouping the terms involving

q

, we obtain:

p^{3} + q^{2} - q (3 p - 1) = 0

Thus, the final relationship is:

$p^{3} - q (3 p - 1) + q^{2} = 0$

Common Traps & FAQs

Why did we choose to cube the sum equation? Why not square it?

That is a fantastic question! If you square the equation

α + α^{2} = - p

, you get

α^{2} + α^{4} + 2 α^{3} = p^{2}

. While this is a valid equation, it leaves you with an

α^{4}

term, which is harder to simplify using only

α^{3} = q

. By cubing the sum, we generate an

α^{3}

term directly, which allows us to substitute

q

immediately. It is all about choosing the path of least resistance!

Can we just solve for

α

using the quadratic formula and plug it in?

Technically, yes, but it would be a nightmare of radicals! You would have

α = \frac{- p \pm p ^{2} - 4 q}{2}

. Raising that to the power of three to satisfy

α^{3} = q

would lead to an incredibly messy algebraic expression. In JEE problems, we always look for the elegant, symmetric path—using Vieta's formulas and algebraic identities is much faster and less prone to calculation errors.

How do I know which algebraic identity to use in the expansion?

Great intuition! We used

(a + b)^{3} = a^{3} + b^{3} + 3 ab (a + b)

because it keeps the term

(a + b)

intact. Since we already know that

a + b

(which is

α + α^{2}

) equals

- p

, this form of the identity allows us to substitute

- p

directly into the expansion, making the simplification process smooth and logical.

Animated Solution for Mathematics - Quadratic Equations: If one root is square of the other root of the equation $x^{2} + p x + q = 0$ , then the relation between $p$ and $q$ is

Select Answer:

Visualized Solution

The Quadratic Equation

Defining the Roots

Vieta's Formula: Sum of Roots

Vieta's Formula: Product of Roots

Strategy to Eliminate $α$

Cubing the Sum Equation

Expanding the Cubic Expression

Simplifying the Expanded Terms

Substituting Known Values

Rearranging the Equation

Final Factoring to Match Options

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Strategy of Elimination

The Algebraic Dance

Final Calculation

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Animated Solution for Mathematics - Quadratic Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If one root is square of the other root of the equation x2+px+q=0, then the relation between p and q is

Select Answer:

Visualized Solution

The Quadratic Equation

Defining the Roots

Vieta's Formula: Sum of Roots

Vieta's Formula: Product of Roots

Strategy to Eliminate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Cubing the Sum Equation

Expanding the Cubic Expression

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Substituting Known Values

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Final Factoring to Match Options

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Strategy of Elimination

The Algebraic Dance

Final Calculation

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