Question

8° Encontre o valor de f(x) = x2 + 3x - 10 para que f(x) = 0​

Answer 1

Resposta:

S={-5,2}

Explicação passo-a-passo:

f(x)=x²+3x-10

para f(x)=0

x²+3x-10=0

$\displaystyle Aplicando~a~f\acute{o}rmula~de~Bhaskara~para~x^{2}+3x-10=0~~e~comparando~com~(a)x^{2}+(b)x+(c)=0,~determinamos~os~coeficientes:~\\a=1{;}~b=3~e~c=-10\\\\C\acute{a}lculo~do~discriminante~(\Delta):&\\&~\Delta=(b)^{2}-4(a)(c)=(3)^{2}-4(1)(-10)=9-(-40)=49\\\\C\acute{a}lculo~das~raizes:&\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(3)-\sqrt{49}}{2(1)}=\frac{-3-7}{2}=\frac{-10}{2}=-5\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(3)+\sqrt{49}}{2(1)}=\frac{-3+7}{2}=\frac{4}{2}=2\\\\S=\{-5,~2\}$

Answer 2

Resposta:

S={-5,2

Explicação passo-a-passo:

f(x)=x²+3x-10

para f(x)=0

x²+3x-10=0

\begin{gathered}\displaystyle Aplicando~a~f\acute{o}rmula~de~Bhaskara~para~x^{2}+3x-10=0~~e~comparando~com~(a)x^{2}+(b)x+(c)=0,~determinamos~os~coeficientes:~\\a=1{;}~b=3~e~c=-10\\\\C\acute{a}lculo~do~discriminante~(\Delta):&\\&~\Delta=(b)^{2}-4(a)(c)=(3)^{2}-4(1)(-10)=9-(-40)=49\\\\C\acute{a}lculo~das~raizes:&\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(3)-\sqrt{49}}{2(1)}=\frac{-3-7}{2}=\frac{-10}{2}=-5\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(3)+\sqrt{49}}{2(1)}=\frac{-3+7}{2}=\frac{4}{2}=2\\\\S=\{-5,~2\}\end{gathered}

Aplicando a f

o

ˊ

rmula de Bhaskara para x

2

+3x−