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The value of $'a'$ for which $x^{2}-\left ( a-3 \right )x+a= 0$ has both roots negative is

Option 1)
$\left [ 0,1 \right ]$

Option 2)
$(0,1]$

Option 3)
$[0,1)$

Option 4)
$\left ( 0,1 \right )$

Answers (1)

best_answer

$\left ( i \right )\: D\geq 0\: \Rightarrow\: a^{2}-6a+9-4a\geq 0$

$\Rightarrow \: a^{2}-10a+9\geq 0$

$\Rightarrow \: a\leq 1\: \cup \: a\geq 9\cdots \cdots \left ( 1 \right )$

$\left ( ii \right )\: \frac{a-3}{1}< 0\: \Rightarrow \: a< 3\cdots \cdots \left ( 2 \right )$

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

$\left ( 1 \right )\cap \left ( 2 \right )\cap \left ( 3 \right )$ we get,

$a\, \epsilon \, (0,1]$

$\therefore$ Option (B)

Both roots of a Quadratic Equation are negative -

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

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

- wherein

$ax^{2}+bx+c=0$

is the quadratic equation

Option 1)

$\left [ 0,1 \right ]$

This is incorrect

Option 2)

$(0,1]$

This is correct

Option 3)

$[0,1)$

This is incorrect

Option 4)

$\left ( 0,1 \right )$

This is incorrect

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