The values of λ such that sum of the squares of the roots of the quadratic equation x² + (3 - λ)x +

Q: Wait, why do we use Vieta's formulas instead of just solving for $x$ using the quadratic formula?

That is a great question! If you use the quadratic formula, you will end up with square roots involving $\lambda$, which makes squaring them and summing them a nightmare of algebraic complexity. Vieta's formulas allow us to bypass the roots themselves and work directly with the coefficients, which is the elegant, professional way to handle symmetric functions of roots.

Q: Do I need to check if the roots are real by ensuring the discriminant $D \ge 0$?

In many JEE problems, you should check the discriminant. However, the expression $S = \alpha^2 + \beta^2$ is a symmetric polynomial of the roots. Even if the roots are complex, the identity $S = (\alpha + \beta)^2 - 2\alpha\beta$ holds true. Since the question asks for the value of $\lambda$ that minimizes this sum, we can proceed with the algebraic minimization. In this specific case, the vertex of our parabola gives us the global minimum for the expression.

Q: Is there a faster way to find the minimum of $\lambda^2 - 4\lambda + 5$ without completing the square?

Absolutely! You can use calculus. If you have a function $f(\lambda) = \lambda^2 - 4\lambda + 5$, its minimum occurs where the derivative $f'(\lambda) = 0$. Here, $2\lambda - 4 = 0$, which immediately gives $\lambda = 2$. Both methods are perfectly valid, but completing the square is often faster for simple quadratics.

Question

The values of $\lambda$ such that sum of the squares of the roots of the quadratic equation $x^2 + (3 - \lambda)x + 2 = \lambda$ has the least value is :

Options:

(A) 2

(B) $\frac{4}{9}$

(C) $\frac{15}{8}$

(D) 1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a quadratic equation; we are uncovering the hidden geometry of roots.

When you look at the equation $x^2 + (3 - \lambda)x + 2 = \lambda$, I want you to see a system that breathes and changes as we tune the parameter $\lambda$. Our mission is to find the specific value of $\lambda$ that brings the sum of the squares of the roots to its absolute lowest point.

### The Standard Form

Before we can dance with the roots, we must set the stage. The equation is currently scattered: $x^2 + (3 - \lambda)x + 2 = \lambda$.

To see its true nature, we must bring it into the standard form $ax^2 + bx + c = 0$. By subtracting $\lambda$ from both sides, we reveal the structure:

$$x^2 + (3 - \lambda)x + (2 - \lambda) = 0$$

Now, the coefficients are clear: $a = 1$, $b = (3 - \lambda)$, and $c = (2 - \lambda)$. This is the foundation upon which we will build our solution.

### Vieta's Magic

Here is where the beauty of algebra shines. We do not need to know the individual values of the roots $\alpha$ and $\beta$ to understand their behavior.

Vieta's formulas are our bridge. We know that the sum of the roots is:

$$\alpha + \beta = -\frac{b}{a} = -(3 - \lambda) = \lambda - 3$$

Similarly, the product of the roots is:

$$\alpha\beta = \frac{c}{a} = 2 - \lambda$$

These two expressions are the DNA of our quadratic equation.

### The Algebraic Transformation

We are tasked with minimizing $S = \alpha^2 + \beta^2$. We have a powerful identity in our toolkit:

$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$$

By substituting our Vieta expressions into this identity, we transform a problem about roots into a problem about $\lambda$:

$$S = (\lambda - 3)^2 - 2(2 - \lambda)$$

### The Parabolic Insight

Now, let us expand this carefully. $( \lambda - 3 )^2$ becomes $\lambda^2 - 6\lambda + 9$. Distributing the $-2$ into $(2 - \lambda)$ gives us $-4 + 2\lambda$.

Combining these, we get:

$$S = \lambda^2 - 6\lambda + 9 - 4 + 2\lambda$$

This simplifies beautifully to:

$$S = \lambda^2 - 4\lambda + 5$$

Look at this! We have a parabola opening upwards. The minimum of this parabola is its vertex. By completing the square, we write:

$$S = (\lambda^2 - 4\lambda + 4) + 1 = (\lambda - 2)^2 + 1$$

The square of any real number is at least zero, so the smallest $S$ can ever be is when $(\lambda - 2)^2 = 0$. This happens precisely when **$\lambda = 2$**. We have arrived at our destination, and the elegance of the result is truly satisfying.

Animated Solution for Mathematics - Quadratic Equations: The values of $λ$ such that sum of the squares of the roots of the quadratic equation $x^{2} + (3 - λ) x + 2 = λ$ has the least value is :

Select Answer:

Visualized Solution

Analyze the Quadratic Equation

Standard Form Conversion

Vieta's Formulas: Sum and Product

The Sum of Squares Identity

Substitution of $λ$ Expressions

Expansion of the Square

Simplifying the Expression

Strategy for Minimization

Completing the Square

Finding the Minimum Value

The Sigma Insight: Maximum and Minimum Values of Quadratic Expressions

Analyzing the Setup

The Standard Form

Vieta's Magic

The Algebraic Transformation

The Parabolic Insight

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