Question

x² + 5x + 6=
x² + 9x + 20=
x² + 8x +15 =
x² + 9x + 14 =
x² + 15x + 56=

Answer 1

X² + 5x + 6=(x+2)(x+3)
6=2·3
5=2+3
x² + 9x + 20=(x+4)(x+5)
20=4·5
9=4+5
x² + 8x +15 =(x+3)(x+5)
15=3·5
8=3+5
x² + 9x + 14 =(x+2)(x+7)
14=2·7
9=2+7

x² + 15x + 56=(x+7)(x+8)
56=7·8
15=7+8

Answer 2

Afla coeficientii a.b.c,pe urma Δ=b²-4ac,daca delta Δ>0 ⇒are doua solutii x1 si x2,care se afla tot dupa formulele opusul lui b + radical din delta/2a si opusul lui b -radical din delta/pe 2a.In cazul cind Δ=0 ecuatia are o singura solutie x
=-b/2a,cind Δ<0 ecuatia nu are solutii.