Question

(2x-3)(2x+3)>4x-9 in multimea nr reale

Answer 1

(2x-3)(2x+3)≥4x-9⇒4x²+6x-6x-9≥4x-9⇒4x²-9≥4x-9⇒
4x²-4x-9+9≥0⇒4x²-4x≥0⇒
a=4
b=4
c=0
Δ=b²-4ac⇒Δ=16-0⇒√Δ=√16⇒Δ=4
x1x2=-b+-√Δ/2a⇒
x1=-4+4/8=x1=0
x2=-4-4/8⇒=-8/8⇒x2=-1
sucees

Answer 2

(2x-3)(2x+3)>4x-9
4x2+6x-6x-9-4x+9>0
4x2-4x>0
^=b2-4ac=4*4-4*4*0=16
^=16
x12=b2+-^ =  4+-4
        2*a         2*4
x1=1
x2=0   
S={1} N*-multsimea numerelor reale