The correct statement is / are-
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If $x _ { 1 }$ $\&$ $x _ { 2 }$ are roots of the equation $2 x ^ { 2 } - 6 x - b = 0 ( b > 0 ) ,$ then
$\dfrac { x _ { 1 } } { x _ { 2 } } + \dfrac { x _ { 2 } } { x _ { 1 } } < - 2$
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Equation $a x ^ { 2 } + b x + c = 0$ has real roots if $a < 0 , c > 0$ and $b \in R$
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If $P ( x ) = a x ^ { 2 } + b x + c$ and $Q ( x ) = - a x ^ { 2 } + b x + c ,$ where $a c \neq 0$ and ${ a },{ b },{ c }\in { R },$ then ${ P }({ x })\cdot { Q }({ x })$ has at least two real roots.
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None of these