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Let a_{1}< a_{2}< a_{3}< a_{4}  then the number of real roots of equation  \left ( x-a_{1} \right )\left ( x-a_{3} \right )+\left ( x-a_{2} \right )\left ( x-a_{4} \right )= 0  equals  

  • Option 1)

    0

  • Option 2)

    1

  • Option 3)

    2

  • Option 4)

    infinite

 

Answers (1)

best_answer

Let f\left ( x \right )=\left ( x-a_{1} \right )\left ( x-a_{3} \right )+\left ( x-a_{2} \right )\left (x-a_{4} \right )

f\left ( -\infty \right )=\infty

f\left ( a_{1} \right )=+ve

f\left ( a_{2} \right )=-ve

f\left ( a_{3} \right )=-ve

f\left ( a_{4} \right )=+ve

f\left ( \infty \right )=\infty

f\left ( a_{1} \right ) & f\left ( a_{2} \right ) are of opposite sign, So atleast one root between them.

& f\left ( a_{3} \right ), f\left ( a_{4} \right ) are of opposite sign so atleast one root between them.

But being a quadratic can't have more than two roots , So exactly one in 

\left ( a_{1},a_{2} \right ) and exactly one in \left ( a_{3},a_{4} \right ) so both roots are real.

\therefore Option (C)

 

Number of roots of polynomial equation -

For a polynomial equation  P\left ( x \right )= 0 if  P\left ( a \right )  and  P\left ( b \right )  are of opposite sign then odd number of roots lie between a and b, if they are of same sign then either no root or even number of roots lie between them.

-

 

 


Option 1)

0

This is Incorrect

Option 2)

1

This is Incorrect

Option 3)

2

This is correct

Option 4)

infinite

This is Incorrect

Posted by

Himanshu

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