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A student attempted to solve a quadratic equation in $x$ twice. However, in the first attempt, he incorrectly wrote the constant term and ended up with the roots as $(4, 3)$. In the second attempt, he incorrectly wrote down the coefficient of $x$ and got the roots as $(3, 2)$. Based on the above information, the roots of the correct quadratic equation are

1. $(-3, 4)$
2. $(3, -4)$
3. $(6, 1)$
4. $(4, 2)$

If equation is $ax^2 + bx+c=0$ and roots of the equation are $x_1$ and $x_2$

then  $x_1 + x_2 = -b/a$
and  $x_1 . x_2 =c/a$

in 1st attempt $c$ is incorrect, So, we can say  $x_1 + x_2 = 4+3=7\quad \to(i)$ [as here $-b/a$ is correct]

in 2nd attempt $b$ is incorrect. So, we can say  $x_1 . x_2 =3.2=6\quad \to (ii)$ [as here $c/a$ is correct]

Now solving $(ii)$ we get,

$x_2 =6 / x_1\quad \to (iii)$

Putting it in eqn. $(i)$

$x_1 + \frac{6}{ x_1} = 7$

$\implies x_1^2 -7x_1 +6 =0$

$\implies x_1 =1$ or $x_1 = 6$

by

Correct $qe\Rightarrow ax^2+bx+c=0$

However, in the first attempt, he incorrectly wrote the constant term and ended up with the roots as (4,3).

$qe_1\Rightarrow(x-4)(x-3)=x^2-7x+12$

$\therefore a=1,b=-7$

In the second attempt, he incorrectly wrote down the coefficient of x and got the roots as (3,2).

$qe_2\Rightarrow(x-3)(x-2)=x^2-5x+6$

$\therefore c=6$

$qe\Rightarrow x^2-7x+6=0$

$\Rightarrow x^2-6x-x+6=0$

$\Rightarrow x(x-6)-1(x-6)=0$

$\Rightarrow (x-1)(x-6)=0$

$\Rightarrow x=1,x=6$

Correct answer$:C$

Ans => (C) (6, 1)