不等式的問題

since the curve y = 2x^2 + x + k doesn't cut the x-axis as 2x^2 + x + k > 0 for all real value of x. the equation 2x^2 + x + k = 0 doesn't have a real solution. therefore, Δ<0

對於所有實數 x means 2x^2 + x + k > 0 is true for whatever real value we put in to x.

since Δ<0,

1 - 4(2)(k) <0
k > 1/8

當 y =2x^2 + x + k
因為對於任何1個實數 x , y 既數值永遠大過0
即係the graph of curve y 永遠都唔會同x-axis 有touching point/cutting points
即係no real roots.
而Δ就係代表有無實數根 , 如果Δ>0 , 即係有兩個唔同數值既實數根 (即係curve y cuts x-axis at two distinct points)
如果Δ=0 , 即係curve y 岩岩掂住x-axis , 即係有兩個相同數值既實數根
如果Δ<0 , 就即係curve y 永遠唔會掂到x-axis , 即係無實數根
所以無論條題目係「對於所有實數 x ,2x^2 + x + k > 0」定係「對於所有實數 x ,2x^2 + x + k < 0」 , 都係用Δ<0去計

如果佢唔指明係對於所有實數x 的話 , 咁就即係唔係全部y 既數值都係大過0 , 咁就唔可以計到k值既範圍in terms of 數目字