JIPMAT 2025 (QA) - Amit and Alok attempted to solve a quadratic equation. Amit made a mistake in writing down the constant term and ended up with roots (4, 3). Alok made a mistake in writing down coefficient of x to get roots (3, 2). The correct roots of the equation are: | PYQs + Solutions

JIPMAT 2025

Algebra

Quadratic Equations

Hard

Amit and Alok attempted to solve a quadratic equation. Amit made a mistake in writing down the constant term and ended up with roots $(4, 3)$ . Alok made a mistake in writing down coefficient of $x$ to get roots $(3, 2)$ . The correct roots of the equation are:

-4, -3

6, 1

4, 3

-6, -1

✅ Correct Option: 2

For equation:

ax^2+bx+c=0

Sum of roots

=-\frac{b}{a}

\newline

Product of roots

= \frac{c}{a}

Let's assume the equation here to be

x^2+bx+c=0

, we need not worry ourself with the value of

a

as it doesn't matter here!

Amit got the constant term wrong, but he has gotten the coefficient of

x

right.

So from his equation, we can derive

-b=4+3=7 \Rightarrow \boxed{-b=7}

Similarly, Alok has gotten the coefficient of

x

wrong, but he has gotten the constant term right.

So from his equation, we can derive

c=3\times2=6 \Rightarrow \boxed{c=6}

Now we know sum of roots is

7

and product of roots is

6

We can go through the options to see what matches these (add both the numbers for sum of roots; multiply both the numbers for product).

Roots as per Option 2:

(6,1)

\newline

Sum

= 6+1 = 7

\newline

Product

= 6 \times 1 = 6