Question

Two students, while solving a quadratic equation, committed the following mistakes:
(i) One of them made a mistake in the constant term and got the roots as $5$ and $9$.
(ii) Another one committed an error in the coefficient of $x$ and got the roots as $12$ and $4$.
But, in the meantime, they realised that they are wrong and they managed to get it right jointly.

Find the correct quadratic equation.

• A. $x^2-44x+14=0$
• B. $2x^2+7x-24=0$
• C. $x^2-14x+48=0$
• D. $3x^2-17x+52=0$

Answer 1

The correct option is

C

$x^2-14x+48=0$

The roots in 1st case are $5$ and $9$. (where the person made a mistake in constant term)

So, $(x-5)=0$ and $(x-9)=0$
$\Rightarrow (x-5)(x-9)=0$
$\Rightarrow x^2-14x+45=0$

Since the first person made a mistake in constant term, the sum of roots of the correct quadratic equation $=14$

Similarily, The roots in 2nd case are $12$ and $4$. (where the person made a mistake in coefficient of x)

So,$(x-12)=0$ and $(x-4)=0$
$\Rightarrow (x-12)(x-4)=0$
$\Rightarrow x^2-16x+48=0$

Since the second person made a mistake in coefficient of $x$, the product of roots of the correct quadratic equation $=48$.

So, the required quadratic equation is $x^2-14x+48=0$