JEE Main 2025 (Online) 23rd January Morning ShiftQuestion 1
If the equation $a(b-c)x^{2}+b(c-a)x+c(a-b)=0$ has equal roots, where $a+c=15$ and $b=\dfrac{36}{5}$, then $a^{2}+c^{2}$ is equal to $\_\_\_\_\_\_\_\_$ .
Quadratic Equation · Nature of Roots and Discriminant
Answer$117$
Solution
Concept. The coefficients add up to $0$, so $x=1$ is a root; equal roots means the other root is also $1$. Formulas used. (1) Sum of coefficients $=0\Rightarrow x=1$ is a root(2) Product of roots $=\dfrac ca$Step 1 - Use (1). $a(b-c)+b(c-a)+c(a-b)=0$, so $x=1$ is a root, and both roots are $1$. Step 2 - Product $=1$. By (2), $$\frac{c(a-b)}{a(b-c)}=1\ \Longrightarrow\ ca-bc=ab-ac\ \Longrightarrow\ b=\frac{2ac}{a+c}.$$ Step 3 - Find $ac$. $\frac{36}{5}=\frac{2ac}{15}$, so $ac=54$. Step 4 - Value. $a^2+c^2=(a+c)^2-2ac=225-108=117$. Answer: 117