The Gateway to Computer Science Excellence
+9 votes

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. $(4, 2)$
in Numerical Ability by Boss (41.9k points)
edited by | 349 views

2 Answers

+10 votes
Best answer
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 $x_1 = 6$

Answer is (C)
by Veteran (118k points)
edited by
+1 vote

Ans => (C) (6, 1)

by Boss (41.9k points)

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
50,737 questions
57,272 answers
104,792 users