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  • Stuck here, help me understand: - Complex numbers and quadratic equations - JEE Main-5

Value of 'a' for which roots of  x^{2}+2\left ( a-1 \right )x+\left ( a+5 \right )= 0 has both roots positive is 

  • Option 1)

    \left ( -\infty ,\infty \right )

  • Option 2)

    (-5,-1]

  • Option 3)

    \left ( -5,-1 \right )

  • Option 4)

    [-5,-1]

 

Answers (1)

best_answer

For both roots positive \left ( i \right )\: D\geq 0\: \Rightarrow \: 4\left ( a-1 \right )^{2}-4\left ( a+5 \right )\geq 0

\Rightarrow \: a^{2}-3a-4\geq 0\: \Rightarrow \: \left (a- 4 \right )\left ( a+1 \right )\geq 0\: \Rightarrow \: a\leq -1\, \cup \, a\geq 4\cdots \cdots \left ( 1 \right )

\left ( ii \right )\frac{-2\left ( a-1 \right )}{1}> 0\: \Rightarrow \: a< 1\cdots \cdots \left ( 2 \right )  \therefore \: a\, \epsilon \, (-5,-1 ]

\left ( iii \right )\: a+5> 0\: \Rightarrow \: a> -5\cdots \cdots \left ( 3 \right )

\therefore Option (B)

 

Both roots of a Quadratic Equation are positive -

\frac{-b}{a}> 0,\frac{c}{a}> 0

D= b^{2}-4ac\geqslant 0

- wherein

ax^{2}+bx+c=0

is the quadratic equation

 

 


Option 1)

\left ( -\infty ,\infty \right )

This is incorrect

Option 2)

(-5,-1]

This is correct

Option 3)

\left ( -5,-1 \right )

This is incorrect

Option 4)

[-5,-1]

This is incorrect

Posted by

Himanshu

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