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QJEEM09Maf47d: JEE Main 2009 Mathematics - Class 11 JEEM09Maf47d Solution: If the roots of the equation imaginary, then for all real…

If the roots of the equation bx2+cx+a=0b{x^2} + cx + a = 0 imaginary, then for all real values of xx , the expression 3b2x2+6bcx+2c23{b^2}{x^2} + 6bcx + 2{c^2} is :

Multiple Choice Options

  • A less than 4ab4ab
  • B greater than −4ab-4ab
  • C less than −4ab-4ab
  • D greater than 4ab4ab

Step-by-Step Verified Solution

Given that roots of the equation

bx2+cx+a=0b{x^2} + cx + a = 0 are imaginary

∴\therefore {c^2} - 4ab < 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)

Let y=3b2x2+6bc x+2c2y = 3{b^2}{x^2} + 6bc\,x + 2{c^2}

⇒3b2x2+6bc x+2c2−y=0\Rightarrow 3{b^2}{x^2} + 6bc\,x + 2{c^2} - y = 0

As xx is real, D≥0D \ge 0

⇒36b2c2−12b2(2c2−y)≥0\Rightarrow 36{b^2}{c^2} - 12{b^2}\left( {2{c^2} - y} \right) \ge 0

⇒12b2(3c2−2c2+y)≥0\Rightarrow 12{b^2}\left( {3{c^2} - 2{c^2} + y} \right) \ge 0

⇒c2+y≥0\Rightarrow {c^2} + y \ge 0

⇒y≥−c2\Rightarrow y \ge - {c^2}

But from eqn. (i),(i), {c^2} < 4ab

or - {c^2} > - 4ab

∴\therefore we get y \ge - {c^2} > - 4ab

y > - 4ab

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