Question

x ao quadrado -10x +25=0

Answer 1

Resposta:

x=5, raiz dupla

Explicação passo-a-passo:

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}-10x+25=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=-10~e~c=25\\\\\Delta=(b)^{2}-4(a)(c)=(-10)^{2}-4(1)(25)=100-(100)=0\\\\x^{'}=x^{''}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-10)-\sqrt{0}}{2(1)}=\frac{10-0}{2}=\frac{10}{2}=5$

Tem raiz dupla

Answer 2

x²—10x+25=0

Onde'

a = 1 ; b = —10 e c = 25

∆ = b²—4*a*c

∆ = (—10)²—4*1*25

∆ = 100—100

∆ = 0

x1,2 = (-b±√∆)/2*a

x1,2 = (10±√0)/2*1

x1,2 = (10±0)/2

x1,2 = 10/2 = 5

Sol: { 5 }

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