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  • Solve it, - Complex numbers and quadratic equations - JEE Main-5

f\left ( x \right )= -x^{2}+ax-12 and g\left ( x \right )= -x^{2}+bx-15 if both f\left ( x \right )= 0 and g\left ( x \right )= 0 have no imaginary root then minimum value of a^{2}+b^{2} equals \left ( a,b\, \epsilon \, R \right )

  • Option 1)

    90

  • Option 2)

    108

  • Option 3)

    120

  • Option 4)

    144

 

Answers (1)

Both don't have imaginary root so D\geq 0  for both

\\*\Rightarrow a^{2}-48\geq 0\; \; \; and\; \; \; b^{2}-60\geq 0\\*\Rightarrow a^{2}\geq 48\; \; \; and\; \; \; b^{2}\geq 60\\*\therefore minimum\; \; (a^{2}+b^{2})=48+60=108

 

Quadratic Expression Graph when a< 0 & D >0 -

Real and distinct roots of

f\left ( x \right )= ax^{2}+bx+c

& D= b^{2}-4ac

- wherein

 

 

 


Option 1)

90

This is incorrect

Option 2)

108

This is correct

Option 3)

120

This is incorrect

Option 4)

144

This is incorrect

Posted by

Vakul

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