Animated Solution for Mathematics - Quadratic Equations: If the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ has equal roots, where $a+c=15$ and $b=\frac{36}{5}$ then $a^2+c^2$ is equal to

Welcome, future engineer. When you first glance at the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$, it is perfectly natural to feel a surge of intimidation. It looks like a tangled mess of variables, a quadratic equation where the coefficients themselves are algebraic expressions.

But here is the secret of JEE Advanced: complexity is often a mask for elegance. Whenever you see a structure like $a(b-c)$, $b(c-a)$, and $c(a-b)$, you are looking at cyclic symmetry. It is a pattern that repeats, a circle of variables. In the world of competitive mathematics, this is a massive, flashing neon sign telling you to pause and look for a simplification.

Before you reach for the quadratic formula—that heavy hammer that crushes everything but often leaves you with a mess of algebra—let us try something lighter. Let us sum the coefficients. Let $A=a(b-c)$, $B=b(c-a)$, and $C=c(a-b)$.

When we add them:

Look closely. The $ab$ cancels with $-ba$, the $-ac$ cancels with $ca$, and the $bc$ cancels with $-cb$. The sum is exactly zero!

Why does this matter? Because there is a golden rule in algebra: for any quadratic equation $A{x}^2+Bx+C=0$, if $A+B+C=0$, then $x=1$ is guaranteed to be a root. Think about it—if you substitute $x=1$ into the equation, you are literally just calculating the sum of the coefficients. Since that sum is zero, the equation is satisfied. You have just unlocked the first root without doing any heavy lifting!

The problem explicitly states that the equation has equal roots. This is the second piece of the puzzle. We know one root is $\alpha =1$. Because the roots are equal, the second root $\beta$ must also be $1$.

Now, we have the roots, and we have the coefficients. We can use the relationship between the roots and the coefficients. The product of the roots is given by $\alpha \beta =\frac{C}{A}$, where $C$ is the constant term and $A$ is the coefficient of $x^2$.

Substituting our values, we get:

This simplifies to $c(a-b)=a(b-c)$. Let us expand this: $ac-bc=ab-ac$. Rearranging the terms gives us $2ac=ab+bc$. If we factor out the $b$ on the right side, we arrive at the beautiful relation:

We are almost there. We have the relation $2ac=b(a+c)$, and we are given $a+c=15$ and $b=\frac{36}{5}$. Let us substitute these values into our relation:

The $15$ and $5$ cancel perfectly to give $3$, so $2ac=36\times 3=108$.

Now, we need to find $a^2+c^2$. We know the fundamental algebraic identity:

We have everything we need. We know $a+c=15$, so $(a+c{)}^2=15^2=225$. We know $2ac=108$. Substituting these into the identity, we get $225=a^2+c^2+108$. Subtracting $108$ from $225$ gives us $117$.

Look at what we have achieved. We did not get lost in the forest of variables. We identified the cyclic symmetry, used the property of coefficients to find the root, and applied the product of roots to bridge the gap to our final answer.

The answer, $117$, is not just a number; it is the result of recognizing the hidden order in the chaos. Keep this mindset—look for the pattern, trust the properties, and the math will always reveal its secrets.