Animated Solution for Mathematics - Quadratic Equations: Both the roots of the equation $(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=0$ are always

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a tangled mess of variables. You see an equation like $(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=0$, and your instinct might be to panic.

You see $a,b,$ and $c$ floating around, and you wonder, 'How can I possibly find the nature of the roots when I don't even know what these constants are?'

Take a deep breath. In the world of JEE Advanced, complexity is often just a mask. Our job is to peel back that mask, simplify the expression, and reveal the elegant truth hiding underneath. This problem is not about brute force; it is about symmetry and the beauty of algebraic identities.

Let's start by expanding the terms. We have three distinct parts to this equation. The first term is $(x-b)(x-c)$, which expands to $x^2-(b+c)x+bc$.

The other two terms, $(x-a)(x-c)$ and $(x-a)(x-b)$, follow the exact same logic. They become $x^2-(a+c)x+ac$ and $x^2-(a+b)x+ab$, respectively.

When we sum these three expanded expressions, we gather our forces. We have three $x^2$ terms, which gives us $3x^2$. For the $x$ terms, we have $-(b+c)x$, $-(a+c)x$, and $-(a+b)x$, which simplifies to $-2(a+b+c)x$.

Finally, the constant terms are simply $ab+bc+ca$. Our equation now stands in its true, standard quadratic form:

This is the moment where the 'intimidating' equation becomes a standard friend. We have $A=3$, $B=-2(a+b+c)$, and $C=(ab+bc+ca)$.

The Discriminant

Now that we have our coefficients, the nature of the roots is entirely decided by the discriminant, $D=B^2-4AC$. This is the heartbeat of any quadratic equation. If $D\ge 0$, the roots are real; if $D<0$, they are imaginary.

Let's substitute our values into the formula. Be careful with the signs here—this is where most students stumble.

Squaring the negative two gives us positive four. So, we have:

We can factor out a $4$ to make our lives easier:

Now, expand the square $(a+b+c{)}^2$. It becomes $a^2+b^2+c^2+2(ab+bc+ca)$. When we subtract the $3(ab+bc+ca)$ from this, we are left with:

This is the climax of our journey. This expression, $a^2+b^2+c^2-ab-bc-ca$, is a classic in competitive mathematics. We can rewrite this by multiplying and dividing by $2$:

Which rearranges beautifully into:

The Beauty of Symmetry

Look at what we have achieved. We have expressed the discriminant as a sum of squares. Since the square of any real number is always non-negative, $(a-b{)}^2\ge 0$, $(b-c{)}^2\ge 0$, and $(c-a{)}^2\ge 0$.

Therefore, their sum must be greater than or equal to zero. This implies that $D\ge 0$ for all real values of $a,b,$ and $c$.

Because the discriminant is always non-negative, the roots of the equation are guaranteed to be real. We didn't need to know the specific values of $a,b,$ or $c$. We only needed to understand the structure of the equation.