Question

$x {}^{2} + 9x + 9 = 0 \\ \\ x { }^{2} + 4x + 4 = 0$
me ajudem por favor!​

Answer 1

Resposta:

Explicação passo-a-passo:

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}+9x+9=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=9~e~c=9\\\\\Delta=(b)^{2}-4(a)(c)=(9)^{2}-4(1)(9)=81-(36)=45\\\\\ \sqrt{45} =\sqrt{5.9} =\sqrt{5}.\sqrt{9} =3\sqrt{5}\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(9)-\sqrt{45}}{2(1)}=\frac{-9-3\sqrt{5}}{2}\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(9)+\sqrt{45}}{2(1)}=\frac{-9+3\sqrt{5}}{2}\\\\\\S=\{\frac{-9-3\sqrt{5}}{2}, \frac{-9+3\sqrt{5}}{2}\}$

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}+4x+4=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=4~e~c=4\\\\\Delta=(b)^{2}-4(a)(c)=(4)^{2}-4(1)(4)=16-(16)=0\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(4)-\sqrt{0}}{2(1)}=\frac{-4-0}{2}=\frac{-4}{2}=-2\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(4)+\sqrt{0}}{2(1)}=\frac{-4+0}{2}=\frac{-4}{2}=-2\\\\S=\{-2,~-2\}$