Let α, β be the roots of the equation x² - 4λx + 5 = 0 and α, γ be the roots of the equation x² - (3

Q: Wait, why can't we just use the quadratic formula to find $\alpha$ directly? It seems like a standard approach.

Ah, the temptation of the quadratic formula! While it is a powerful tool, look at the coefficients: $4\lambda$ and $3\sqrt{2} + 2\sqrt{3}$. If you plug these into the discriminant formula, you will be drowning in a sea of radicals and variables. In JEE Advanced, the 'brute force' method is often a trap. We use Vieta's formulas because they allow us to manipulate the properties of the roots without ever needing to know the specific values of the roots themselves. It is elegant, efficient, and much less prone to calculation errors.

Q: How do we know for sure that $\alpha$ is the common root? Could it be $\beta$ or $\gamma$?

Excellent observation! The problem explicitly states that $\alpha$ is a root of the first equation and $\alpha$ is also a root of the second. This is a gift from the examiner. By identifying the common root immediately, we know exactly which variable to eliminate. If the problem hadn't specified, we would have had to test cases, but here, the path is clearly laid out for us.

Q: Why did we rewrite the target expression $(\alpha + 2\beta + \gamma)^2$ as $((\alpha + \beta) + (\beta + \gamma))^2$?

This is the 'Aha!' moment of the problem. If you tried to find $\alpha, \beta,$ and $\gamma$ individually, you would have to calculate $\lambda$, then find the roots, then plug them in. That is a long, winding road. By regrouping, we utilize the values we already calculated (the sums of the roots). It is about working smarter, not harder. We transform a complex expression into a simple sum of known quantities.

Question

Let $\alpha, \beta$ be the roots of the equation $x^2 - 4\lambda x + 5 = 0$ and $\alpha, \gamma$ be the roots of the equation $x^2 - (3\sqrt{2} + 2\sqrt{3})x + 7 + 3\lambda\sqrt{3} = 0$. If $\beta + \gamma = 3\sqrt{2}$, then $(\alpha + 2\beta + \gamma)^2$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given two quadratic equations:
1. $x^2 - 4\lambda x + 5 = 0$ with roots $\alpha$ and $\beta$.
2. $x^2 - (3\sqrt{2} + 2\sqrt{3})x + 7 + 3\lambda\sqrt{3} = 0$ with roots $\alpha$ and $\gamma$.

The variable $\alpha$ acts as the common bridge between these two systems. Our objective is to determine the value of $(\alpha + 2\beta + \gamma)^2$.

Using Vieta's formulas for the first equation:
$$\alpha + \beta = 4\lambda$$
$$\alpha\beta = 5$$

For the second equation:
$$\alpha + \gamma = 3\sqrt{2} + 2\sqrt{3}$$
$$\alpha\gamma = 7 + 3\lambda\sqrt{3}$$

### The Elimination Strategy

To isolate the variables, we subtract the sum of the roots of the second equation from the first:
$$(\alpha + \beta) - (\alpha + \gamma) = 4\lambda - (3\sqrt{2} + 2\sqrt{3})$$
$$\beta - \gamma = 4\lambda - 3\sqrt{2} - 2\sqrt{3}$$

Given the problem constraints, we identify $\beta + \gamma = 3\sqrt{2}$. We now have a system of linear equations for $\beta$ and $\gamma$:
1. $\beta - \gamma = 4\lambda - 3\sqrt{2} - 2\sqrt{3}$
2. $\beta + \gamma = 3\sqrt{2}$

Adding these equations yields:
$$2\beta = 4\lambda - 2\sqrt{3} \implies \beta = 2\lambda - \sqrt{3}$$

Substituting $\beta$ back into the sum $\alpha + \beta = 4\lambda$:
$$\alpha = 4\lambda - (2\lambda - \sqrt{3}) = 2\lambda + \sqrt{3}$$

### Solving for the Parameter

We utilize the product of the roots from the first equation, $\alpha\beta = 5$:
$$(2\lambda + \sqrt{3})(2\lambda - \sqrt{3}) = 5$$

Applying the difference of squares identity:
$$4\lambda^2 - 3 = 5$$
$$4\lambda^2 = 8 \implies \lambda^2 = 2$$
$$\lambda = \sqrt{2}$$

### Final Calculation

We evaluate the target expression $(\alpha + 2\beta + \gamma)^2$ by grouping the terms strategically:
$$(\alpha + 2\beta + \gamma)^2 = ((\alpha + \beta) + (\beta + \gamma))^2$$

Substituting the known sums $\alpha + \beta = 4\lambda$ and $\beta + \gamma = 3\sqrt{2}$:
$$(4\lambda + 3\sqrt{2})^2$$

Substituting $\lambda = \sqrt{2}$:
$$(4\sqrt{2} + 3\sqrt{2})^2 = (7\sqrt{2})^2$$
$$49 	imes 2 = 98$$

The final result is **98**.

Animated Solution for Mathematics - Quadratic Equations: Let $α, β$ be the roots of the equation $x^{2} - 4 λ x + 5 = 0$ and $α, γ$ be the roots of the equation $x^{2} - (32 + 23) x + 7 + 3 λ 3 = 0$ . If $β + γ = 32$ , then $(α + 2 β + γ)^{2}$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Problem Overview

Roots of the First Equation

Roots of the Second Equation

Finding the Difference $β - γ$

Solving for $β$

Solving for $α$

Finding the Value of $λ$

Simplifying the Target Expression

Final Substitution and Calculation

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Elimination Strategy

Solving for the Parameter

Final Calculation

Similar Questions

If $α$ and $β$ are the roots of the equation $x^{2} + p x + 2 = 0$ and $\frac{1}{α}$ and $\frac{1}{β}$ are the roots of the equation $2 x^{2} + 2 q x + 1 = 0$ , then $(α - \frac{1}{α}) (β - \frac{1}{β}) (α + \frac{1}{β}) (β + \frac{1}{α})$ is equal to :

Let $λ \neq = 0$ be a real number. Let $α, β$ be the roots of the equation $14 x^{2} - 31 x + 3 λ = 0$ and $α, γ$ be the roots of the equation $35 x^{2} - 53 x + 4 λ = 0$ . Then $\frac{3 α}{β}$ and $\frac{4 α}{γ}$ are the roots of the equation :

If $α$ and $β$ be two roots of the equation $x^{2} - 64 x + 256 = 0$ . Then the value of $(\frac{α ^{3}}{β ^{5}})^{1/8} + (\frac{β ^{3}}{α ^{5}})^{1/8}$ is

If $α, β$ are the roots of the equation $x^{2} - (5 + 3^{l o g_{3} 5} - 5^{l o g_{5} 3}) x + 3 (3^{(l o g_{3} 5)^{1/3}} - 5^{(l o g_{5} 3)^{2/3}} - 1) = 0$ then the equation, whose roots are $α + \frac{1}{β}$ and $β + \frac{1}{α}$ , is

Let $α, β$ be the roots of the equation $x^{2} - a x - b = 0$ with $I m (α) < I m (β)$ . Let $P_{n} = α^{n} - β^{n}$ . If $P_{3} = - 57 i, P_{4} = - 37 i, P_{5} = 117 i$ and $P_{6} = 457 i,$ then $∣ α^{4} + β^{4} ∣$ is equal to

Let $a, b, c$ be real numbers with $a \neq = 0$ and let $α, β$ be the roots of the equation $a x^{2} + b x + c = 0$ . Express the roots of $a^{3} x^{2} + ab c x + c^{3} = 0$ in terms of $α, β$ .

If the sum of the squares of the reciprocals of the roots $α$ and $β$ of the equation $3 x^{2} + λ x - 1 = 0$ is 15, then $6 (α^{3} + β^{3})^{2}$ is equal to:

Let $a, b \in R, a \neq = 0$ be such that the equation, $a x^{2} - 2 b x + 5 = 0$ has a repeated root $α$ , which is also a root of the equation, $x^{2} - 2 b x - 10 = 0$ . If $β$ is the other root of this equation, then $α^{2} + β^{2}$ is equal to:

Let $a, b \in R, a \neq = 0$ , such that the equation, $a x^{2} - 2 b x + 5 = 0$ has a repeated root $α$ , which is also a root of the equation $x^{2} - 2 b x - 10 = 0$ . If $β$ is the root of this equation, then $α^{2} + β^{2}$ is equal to:

If $α, β$ , where $α < β$ , are the roots of the equation $λ x^{2} - (λ + 3) x + 3 = 0$ such that $\frac{1}{α} - \frac{1}{β} = \frac{1}{3}$ , then the sum of all possible values of $λ$ is