Question

Quadratic inequality X²-9+18>0

Answer 1

Answer:

x sq2 -9x + 18 = (x-3) (x-6) so

x sq2 - 9x + 18=0 when x= 3 and x = 6. since the coefficientif the x sq2 term is positive, x sq2 - 9x + 18 >0 when x E ( - 3) U (6)

Answer 2

$\implies \sf \: {x}^{2} -9x+18 > 0$

Now , -6x-3x = - 9x and (-6x)×(-3x)= 18x²

$\implies \sf \: {x}^{2} -6x - 3x+18 > 0$

$\implies \sf \: x({x} -6)- 3(x - 6) > 0$

$\implies \sf \: ({x} -3)(x - 6) > 0$

$\implies \sf \: ({x} -3)> 0 \: \: or \: (x - 6) > 0$

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Therefore,

$\sf \: x \in ( - \infty ,6) \cup(3 , \infty)$

Answer :

$\bf \: x \in ( - \infty ,6) \cup(3 , \infty)$