Question

Δ=64k^2m^2-4(3+4k^2)(4m^2-12)＞0怎样化简得到3+4K^2＞m^2

Answer 1

64k^2m^2-4(12m^2-36+16k^2m^2-48k^2)>0

64k^2m^2-48m^2+144-64k^2m^2+192k^2>0

144+192k^2>48m^2

两边除以48得,

3+3k^2>m^2