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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. $(4, 2)$

edited | 349 views

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 Veteran (118k points)
edited by
+1 vote

Ans => (C) (6, 1)

by Boss (41.9k points)