Question

Quadratic inequality x²+6×+5>0

Answer 1

$\implies \sf {x}^{2} + 6x + 5 > 0$

Here, 5x + x = 6x and 5x × x = 5x² . Therefore ,

$\implies \sf {x}^{2} + 5x + x + 5 > 0$

$\implies \sf {x}(x + 5) +1( x + 5) > 0$

Taking out (x+5) common :-

$\implies \sf( {x + 1})(x + 5) > 0$

Therefore now :-

$\sf \: x > - 1$

or

$\sf \: x > -5$

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$\sf \: x \in ( - \infty , - 5) \cup ( - 1 ,\infty )$

Answer :

$\bf \: x \in ( - \infty , - 5) \cup ( - 1 ,\infty )$