Question

Vompley o valor de x na equação x2=7×-10

Answer 1

Resposta:

x²=7x-10

x²-7x+10=0

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}-7x+10=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=-7~e~c=10\\\\\Delta=(b)^{2}-4(a)(c)=(-7)^{2}-4(1)(10)=49-(40)=9\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-7)-\sqrt{9}}{2(1)}=\frac{7-3}{2}=\frac{4}{2}=2\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-7)+\sqrt{9}}{2(1)}=\frac{7+3}{2}=\frac{10}{2}=5\\\\S=\{2,~5\}$

Answer 2

Resposta:

S = {5; 2}

Explicação passo-a-passo:

x² = 7x - 10

x² - 7x + 10 = 0

a = 1

b = - 7

c = 10

Calculando o Δ:

Δ = b² - 4.a.c  

Δ = - 7² - 4 . 1 . 10  

Δ = 49 - 40  

Δ = 9

Há 2 raízes reais.

Aplicando Bhaskara:

x = (- b ± √Δ)/2a

x' = -(-7 ±√9)/2.1

x' = 7 + 3/2

x' = 10/2

x' = 5

x" = 7 - 3/2

x" = 4/2

x" = 2

S = {5; 2}

Bons estudos.