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Can someone explain - Complex numbers and quadratic equations - JEE Main-6

The number of solutions of equation \frac{\left ( x+a \right )^{2}}{\left ( a-b \right )\left ( a-c \right )}+\frac{\left ( x+b \right )^{2}}{\left ( b-c \right )\left ( b-a \right )}+\frac{\left ( x+c \right )^{2}}{\left ( c-a \right )\left ( c-b \right )}= 1  \left ( Where\; a\neq b\neq c \right )  equals

  • Option 1)

    0

  • Option 2)

    1

  • Option 3)

    infinite

  • Option 4)

    2

 
Answers (1)
121 Views
S subam

If it is equation, being a 2nd degree, maximum it can have two solutions.

on putting x = -a we get L.H.S =

\frac{\left ( b-a \right )^{2}}{\left ( b-c \right )\left ( b-a \right )}+\frac{\left ( c-a \right )^{2}}{\left ( c-a \right )\left ( c-b \right )}=\frac{b-a}{b-c}+\frac{c-a}{c-b}=\frac{b-c}{b-c}=1

= R.H.S

So x = -a is a solution.

Similarly x = -b and  x = -c are also solutions.

So, we get three different solutions of a quadratic, so it will be an identity so satisfied for large number of values.

\therefore Option (C)

 

Quadratic Equation become an Identity -

Satisfied by more than two values of x

\Leftrightarrow a=b=c=0

 

-

 

 


Option 1)

0

This is incorrect

Option 2)

1

This is incorrect

Option 3)

infinite

This is correct

Option 4)

2

This is incorrect

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