If a>0 and discriminant of ax^{2}+2bx+c is -ve, then

     is

  • Option 1)

    +ve

  • Option 2)

    (ac-b^{2})(ax^{2}+2bx+c)

  • Option 3)

    -ve

  • Option 4)

    0

 

Answers (1)

As we learnt in 

Value of determinants of order 3 -

-

 

\begin{vmatrix} a & b &ax+b \\ b & c &bx+c \\ ax+b& bx+c &0 \end{vmatrix}

R_{3}\rightarrow R_{3}-\left ( R_{2}+xR_{1} \right )

\begin{vmatrix} a & b &ax+b \\ b & c &bx+c \\ 0&0 &-\left ( ax^{2}+2bx+c \right ) \end{vmatrix}

\therefore -\left ( ax^{2}+2bx+c \right )\left ( ac-b^{2} \right )

Now\: 4b^{2}-4ac< 0 \:given

b^{2}-ac< 0

ac-b^{2}> 0

So that it is -ve. 


Option 1)

+ve

This option is incorrect.

Option 2)

(ac-b^{2})(ax^{2}+2bx+c)

This option is incorrect.

Option 3)

-ve

This option is correct.

Option 4)

0

This option is incorrect.

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