JEE Mathematics 2025 Question with Solution

Given:

• $a(b-c)x^2 + b(c-a)x + c(a-b)=0$ has equal roots.
• $a+c=15$
• $b=\frac{36}{5}$
• From the solution, one root is stated to be $x=1$, so both equal roots are $1$.

Find: $a^2+c^2$

For a quadratic with equal roots, if one root is $1$, then both roots are $1$ and hence the sum of roots is

$$\alpha+\beta=2$$

Using sum of roots for

$$a(b-c)x^2 + b(c-a)x + c(a-b)=0$$

we get

$$\alpha+\beta=-\frac{b(c-a)}{a(b-c)}=\frac{b(a-c)}{a(b-c)}$$

As used in the extracted solution,

$$\frac{b(c-a)}{a(b-c)}=2$$

which rearranges to

$$b(c-a)=2a(b-c)$$ $$bc-ba=2ab-2ac$$ $$2ac=ab+bc=b(a+c)$$

Now substitute $a+c=15$ and $b=\frac{36}{5}$:

$$2ac=\frac{36}{5}\cdot 15=108$$

So,

$$ac=54$$

Now use the identity

$$a^2+c^2=(a+c)^2-2ac$$

Therefore,

$$a^2+c^2=15^2-2\cdot 54=225-108=117$$

Hence, the value of $a^2+c^2$ is $117$.

Given: $a+c=15$ and $b=\frac{36}{5}$

Find: $a^2+c^2$

Since the equal roots are taken as $1,1$ in the solution, the sum of roots is $2$. This gives

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

Substitute the known values:

$$2ac=\frac{36}{5}\cdot 15=108$$ $$ac=54$$

Now apply the identity directly:

$$a^2+c^2=(a+c)^2-2ac=15^2-2\cdot 54=117$$

Therefore, the required numerical value is $117$.