If the Equation a(b−c)x² + b(c−a)x + c(a−b) = 0 Has Equal Roots, and a + c = 5 and b = 16/5, Then Find the Value of a² + c²

JEE Main 2025. Previous Year Question.

Problem

If the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ has equal roots, and if $a+c=5$ and $b=\dfrac{16}{5}$, then find the value of $a^2+c^2$.

Key insight. The three coefficients $a(b-c)$, $b(c-a)$, $c(a-b)$ always add up to exactly zero — a pure algebraic identity, true for any $a,b,c$. That means $x=1$ is always a root of this equation, no matter what. Combined with “equal roots,” both roots must then be $1$, which turns the discriminant condition into a much simpler relationship between $a$, $b$, and $c$.

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Approach

Let $A=a(b-c)$, $B=b(c-a)$, $C=c(a-b)$ be the equation’s coefficients, and verify $A+B+C=0$ as a pure identity — this shows $x=1$ is always a root. Since the equation has equal roots, both roots must equal $1$, so the product of roots gives $A=C$. Translate $A=C$ into a relationship between $a,b,c$, substitute the given values of $b$ and $a+c$, and use $(a+c)^2=a^2+2ac+c^2$ to finish.

Solution

Step 1 — Verify A + B + C = 0 as an identity

$$A+B+C = a(b-c)+b(c-a)+c(a-b) = ab-ac+bc-ab+ac-bc = 0$$

This holds for any $a,b,c$ — confirming $x=1$ is always a root of the equation.

Step 2 — Use the equal-roots condition

Since the equation has equal roots and one root is always $1$, both roots must be $1$. Using $A=a(b-c)$ as the coefficient of $x^2$ and $C=c(a-b)$ as the constant term, the product of roots is $\dfrac{C}{A}$:

$$\frac{C}{A} = 1\times1 = 1 \implies A=C$$

Step 3 — Translate A = C into a relation between a, b, c

$$a(b-c) = c(a-b)$$

$$ab-ac = ac-bc$$

$$ab+bc = 2ac$$

$$b(a+c) = 2ac$$

Step 4 — Substitute the given values

$$\frac{16}{5}\times5 = 2ac \implies 16 = 2ac \implies ac=8$$

Step 5 — Use (a + c)² = a² + 2ac + c² to find a² + c²

$$a^2+c^2 = (a+c)^2-2ac = 5^2-2(8) = 25-16 = 9$$

Answer

Common mistakes

• Trying to compute the discriminant $B^2-4AC=0$ directly in terms of $a,b,c$ without first spotting $A+B+C=0$. This is technically possible but leads to a much messier expansion — recognising the coefficient-sum identity first, and using it to conclude both roots are $1$, sidesteps almost all of that algebra.
• Using $A=-C$ instead of $A=C$. Since both roots equal $1$ (not one root $1$ and the other $-1$), the product of roots is $+1$, giving $C/A=1$, i.e. $A=C$ — not $A=-C$, which would correspond to a product of $-1$.

Practise next

• If $a(b-c)x^2+b(c-a)x+c(a-b)=0$ has equal roots, with $a+c=15$ and $b=\dfrac{36}{5}$, find $a^2+c^2$, using the same identity-and-equal-roots approach.