Let a and b be the roots of the equation x² - 10cx - 11d = 0 and those of x² - 10ax - 11b = 0 are c,

Q: Why can't we just solve for $a, b, c,$ and $d$ using the four equations we have?

While theoretically possible, solving for four variables with a system of quadratic equations is algebraically exhausting and prone to error. The beauty of this problem lies in the symmetry. By manipulating the equations as a whole, we find the sum $a + b + c + d$ without ever needing the individual values.

Q: Why is the condition $a \neq b \neq c \neq d$ given?

This condition is crucial for the final step. When we factor the expression $(a - c)(a + c - 121) = 0$, we must divide by $(a - c)$ to isolate $(a + c)$. If $a$ were equal to $c$, we would be dividing by zero, which is undefined. The problem guarantees $a \neq c$, ensuring our division is valid.

Q: Can we use the product of roots instead of the sum?

You certainly could! Vieta's formulas tell us $ab = -11d$ and $cd = -11b$. However, this introduces products of variables, which are generally harder to work with than the linear sums. The sum of roots approach is much more direct and elegant.

Question

Let $a$ and $b$ be the roots of the equation $x^2 - 10cx - 11d = 0$ and those of $x^2 - 10ax - 11b = 0$ are $c, d$ then value of $a + b + c + d$, when $a 
eq b 
eq c 
eq d$, is .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a puzzle that seems to demand brute force. You have two quadratic equations, $x^2 - 10cx - 11d = 0$ and $x^2 - 10ax - 11b = 0$, and you are told that their roots are $(a, b)$ and $(c, d)$ respectively.

Your mission is to find the sum $a + b + c + d$. In the world of JEE Advanced, brute force is rarely the intended path; the true path is paved with symmetry.

### Phase 1: The Vieta's Insight

We begin by invoking the wisdom of Vieta. For any quadratic equation $x^2 - Bx - C = 0$, the sum of the roots is simply $B$.

Applying this to our given equations, we immediately see:
$$a + b = 10c$$
$$c + d = 10a$$

Look closely at these two equations. They are mirrors of each other. If we subtract the second from the first, we get $(a + b) - (c + d) = 10c - 10a$.

Rearranging this, we find $b - d = 11(c - a)$. This is our first major breakthrough, as we have linked the differences of the roots to the coefficients.

### Phase 2: The Algebraic Dance

Now, we must use the fact that $a$ and $c$ are roots. By definition, a root must satisfy its equation.

Therefore:
$$a^2 - 10ca - 11d = 0$$
$$c^2 - 10ac - 11b = 0$$

Subtracting these two equations is where the magic happens. The $10ac$ terms, which look so intimidating, vanish completely!

We are left with $(a^2 - c^2) - 11d + 11b = 0$. Rearranging this, we get:
$$(a^2 - c^2) + 11(b - d) = 0$$

### Phase 3: The Grand Finale

We are almost there. We know that $a^2 - c^2$ is $(a - c)(a + c)$. We also know from our earlier work that $b - d = 11(c - a)$, which is $-11(a - c)$.

Substituting these into our equation, we get:
$$(a - c)(a + c) + 11[-11(a - c)] = 0$$

This simplifies to $(a - c)(a + c - 121) = 0$. Since the problem explicitly states that $a 
eq c$, we can safely divide by $(a - c)$ to conclude that $a + c = 121$.

Finally, to find the total sum $a + b + c + d$, we simply add our two Vieta equations:
$$(a + b) + (c + d) = 10c + 10a = 10(a + c)$$

Substituting our value of $121$, we get $10 	imes 121 = 1210$. The complexity dissolves, leaving behind the final result of **1210**.

Animated Solution for Mathematics - Quadratic Equations: Let $a$ and $b$ be the roots of the equation $x^{2} - 10 c x - 11 d = 0$ and those of $x^{2} - 10 a x - 11 b = 0$ are $c, d$ then value of $a + b + c + d$ , when $a \neq = b \neq = c \neq = d$ , is .........

Enter Numerical Value:

Visualized Solution

The Given Equations

Vieta's Formulas for Sum of Roots

Relating the Variables

Roots Satisfy Their Equations

Subtracting the Root Equations

Substituting and Solving for $a + c$

Calculating the Final Sum

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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