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  • Help me solve this - Complex numbers and quadratic equations - JEE Main-4

The value of 'a' for which x^{2}-ax+a= 0 has both root positive is

  • Option 1)

    a\geqslant 1

  • Option 2)

    a\geqslant 2

  • Option 3)

    a\geqslant 3

  • Option 4)

    a\geqslant 4

 

Answers (1)

best_answer

\left ( i \right )\: D\geq 0\: \Rightarrow \: a^{2}-4a\geq 0\: \Rightarrow \: a\left ( a-4 \right )\geq 0

                       \Rightarrow \: a\leq 0\, \cup \, a\geq 4\cdots \cdots \left ( 1 \right )

\left ( ii \right )\: \frac{-\left ( -a \right )}{1}> 0 \: \Rightarrow \: a> 0\cdots \cdots \left ( 2 \right )

\left ( iii \right )\: a> 0\, \cdots \cdots \left ( 3 \right )

\therefore \: a\geq 4

\therefore Option (D)

 

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)

a\geqslant 1

This is incorrect

Option 2)

a\geqslant 2

This is incorrect

Option 3)

a\geqslant 3

This is incorrect

Option 4)

a\geqslant 4

This is correct

Posted by

divya.saini

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