If α and β are the roots of x² + px + q = 0 and α⁴, β⁴ are the roots of x² - rx + s = 0, then the eq

Q: Wait, why do we ignore $p$ and $s$ in the final calculation?

In competitive exams, problems often include extraneous information to test your ability to filter what is necessary. The target equation only contains $q$ and $r$. Since we can express $q$ and $r$ directly in terms of the roots $\alpha$ and $\beta$, $p$ and $s$ are simply not required to determine the nature of the roots.

Q: What if $\alpha$ and $\beta$ are complex numbers? Does the discriminant still work?

Yes, absolutely! The condition for real roots, $D \ge 0$, is derived from the coefficients of the quadratic equation. As long as the coefficients of your target equation are real (which they are, given $q$ and $r$ are real), the discriminant will tell you whether the roots are real or complex, regardless of the nature of the original roots $\alpha$ and $\beta$.

Q: How did you know to factor out the 4 in the final step?

Pattern recognition is key. When you arrive at $D = 8\alpha^2\beta^2 + 4\alpha^4 + 4\beta^4$, you should notice the structure of a perfect square trinomial: $a^2 + 2ab + b^2 = (a+b)^2$. By factoring out the 4, the expression $4(\alpha^4 + 2\alpha^2\beta^2 + \beta^4)$ clearly becomes $4(\alpha^2 + \beta^2)^2$.

Question

If $\alpha$ and $\beta$ are the roots of $x^2 + px + q = 0$ and $\alpha^4, \beta^4$ are the roots of $x^2 - rx + s = 0$, then the equation $x^2 - 4qx + 2q^2 - r = 0$ has always

Options:

(A) two real roots

(B) two positive roots

(C) two negative roots

(D) one positive and one negative root

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future IITians! Today, we are embarking on a journey through the elegant world of quadratic equations. We are presented with a puzzle: two initial equations with roots $\alpha, \beta$ and $\alpha^4, \beta^4$, and a third, mysterious target equation.

Our mission is to determine the nature of the roots of this third equation. To do this, we must act like detectives, extracting only the clues that matter.

### The Power of Vieta's Relations

Let us look at our first equation, $x^2 + px + q = 0$. Its roots are $\alpha$ and $\beta$. By Vieta's relations, we know the sum of the roots is $\alpha + \beta = -p$ and the product is $\alpha\beta = q$.

Now, consider the second equation, $x^2 - rx + s = 0$, with roots $\alpha^4$ and $\beta^4$. Again, by Vieta's, the sum is $\alpha^4 + \beta^4 = r$ and the product is $\alpha^4\beta^4 = s$.

Notice something? Our target equation, $x^2 - 4qx + 2q^2 - r = 0$, contains $q$ and $r$, but not $p$ or $s$. This is our first major breakthrough: we can ignore $p$ and $s$ entirely!

### The Discriminant as the Ultimate Judge

To determine the nature of the roots of any quadratic equation $Ax^2 + Bx + C = 0$, we must consult the discriminant, $D = B^2 - 4AC$. For our target equation, the coefficients are $A = 1$, $B = -4q$, and $C = 2q^2 - r$.

Let us substitute these into our formula:
$$D = (-4q)^2 - 4(1)(2q^2 - r)$$

Expanding this, we get $D = 16q^2 - 8q^2 + 4r$, which simplifies beautifully to:
$$D = 8q^2 + 4r$$

### The Algebraic Alchemy

Now, we perform the final, magical substitution. We know $q = \alpha\beta$ and $r = \alpha^4 + \beta^4$. Plugging these into our expression for $D$, we get:
$$D = 8(\alpha\beta)^2 + 4(\alpha^4 + \beta^4)$$

Let us rearrange this:
$$D = 4\alpha^4 + 8\alpha^2\beta^2 + 4\beta^4$$

If we factor out a 4, we are left with:
$$D = 4(\alpha^4 + 2\alpha^2\beta^2 + \beta^4)$$

Look closely at the expression inside the parentheses. It is a perfect square! Specifically, it is $(\alpha^2 + \beta^2)^2$.

Thus, our discriminant is:
$$D = 4(\alpha^2 + \beta^2)^2$$

Since the square of any real number is non-negative, $D$ must be greater than or equal to zero. This guarantees that the roots are **always real**. We have successfully navigated the complexity and arrived at a definitive, elegant conclusion. Keep this analytical mindset, and you will conquer any problem the JEE throws your way!

Animated Solution for Mathematics - Quadratic Equations: If $α$ and $β$ are the roots of $x^{2} + p x + q = 0$ and $α^{4}, β^{4}$ are the roots of $x^{2} - r x + s = 0$ , then the equation $x^{2} - 4 q x + 2 q^{2} - r = 0$ has always

Select Answer:

Visualized Solution

Understanding the Goal

Extracting Information from Equation 1

Extracting Information from Equation 2

The Target Equation and its Discriminant

Substituting the Coefficients

Simplifying the Expression for $D$

Connecting Roots to the Discriminant

Factoring and Creating a Perfect Square

Analyzing the Sign of the Discriminant

Determining the Nature of Roots

The Sigma Insight: Nature of Roots

Analyzing the Setup

The Power of Vieta's Relations

The Discriminant as the Ultimate Judge

The Algebraic Alchemy

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