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  • Try this! - Complex numbers and quadratic equations - JEE Main-14

If the roots of the equation bx^{2}+cx+a=0 be imaginary, then for all real values of x . The expression 3b^{2}x^{2}+6bcx+2c^{2} is

  • Option 1)

    less than 4 ab

  • Option 2)

    greater than  -4 ab

  • Option 3)

    less than -4 ab

  • Option 4)

    greater than 4 ab

 

Answers (2)

best_answer

As we leant

Complex Roots with non - zero Imaginary part -

D= b^{2}-4ac< 0

- wherein

ax^{2}+bx+c= 0

is the quadratic equation

 

 

 

bx2 + cx + a = 0 

Since roots are imaginary

then c2 - 4ab < 0 

\therefore\ \; c^{2}<4ab                    (i)

So that 3b^{2}x^{2}+6bcx+2c^{2}=3\left[b^{2}x^{2}+2bxc+\frac{2c^{2}}{3} \right ]

    =3\left[(bx)^{2}+2bxc+c^{2}+\frac{2c^{2}}{3}-c^{2} \right ]

    =3\left[(bx+c)^{2}-\frac{c^{2}}{3} \right ]

    =3(bx+c)^{2}-c^{2}=3(bx+c)^{2}+(-c^{2})

But c2 < 4 ab

\therefore\ \;3(bx+c)^{2}+(-4ab)

Hence, It will be greatest than - 4ab

Correct option is 2.


Option 1)

less than 4 ab

This is an incorrect option.

Option 2)

greater than  -4 ab

This is the correct option.

Option 3)

less than -4 ab

This is an incorrect option.

Option 4)

greater than 4 ab

This is an incorrect option.

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