The condition for (a - 2) $x^{2}$ + 2ax + (a + 3) = 0 to have both roots to be real is

a [2, )

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a (- , -3]

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a (-, 6]

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a (-2, 3]

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Solution

The correct option is C
a (-, 6]

For roots to be real

Discriminant $\geq$ 0

$b^{2}$ - 4ac $\geq$ 0

${(2 a)}^{2}$ - 4 (a - 2)(a + 3) $\geq$ 0

4 $a^{2}$ - 4( $a^{2}$ + a - 6) $\geq$ 0

24 - 4a $\geq$ 0

a $\leq$ 6

So, a $\in$ (- $\infty$ , 6]

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The condition for (a - 2) x2 + 2ax + (a + 3) = 0 to have both roots to be real is

The correct option is C a (-, 6] For roots to be real Discriminant ≥ 0 b2 - 4ac ≥ 0 (2a)2 - 4 (a - 2)(a + 3) ≥ 0 4a2 - 4(a2 + a - 6) ≥ 0 24 - 4a ≥ 0 a ≤ 6 So, a ∈ (-∞ , 6]

The condition for (a - 2) $x^{2}$ + 2ax + (a + 3) = 0 to have both roots to be real is

The correct option is C
a (-, 6]

For roots to be real

Discriminant $\geq$ 0

$b^{2}$ - 4ac $\geq$ 0

${(2 a)}^{2}$ - 4 (a - 2)(a + 3) $\geq$ 0

4 $a^{2}$ - 4( $a^{2}$ + a - 6) $\geq$ 0

24 - 4a $\geq$ 0

a $\leq$ 6

So, a $\in$ (- $\infty$ , 6]