Question

Utilizando a formula de Báskhara, fatore a seguinte expressão:
X^2-(2√3+1)x+2√3=0

Answer 1

Resposta:

Explicação passo-a-passo:

X^2-(2√3+1)x+2√3=0

X²-2√3x-x+2√3=0

∆=b²-4ac

∆=(-2√3-1)²- 4(1)(2√3)

∆=13+4√3-8√3

∆=13-4√3

X=-(b)+-√∆/2a

X1=-(-2√3-1)-√13-4√3/2

X1=2√3+1-√13-4√3/2

X1=2√3+1-√(1-2√3)²/2

X1=2√3+1-1+2√3/2

X1=4√3/2

X1=2√3

X2=-(-2√3-1)+√13-4√3/2

X2=2√3+1+√(1-2√3)²/2

X2=2√3+1+1-2√3/2

X2=2/2

X2=1

Answer 2

Explicação passo-a-passo:

Equação Quadrática :

Dada a equação :

$\mathtt{ \red{ x^2 - (2\sqrt{3}+1)x + 2\sqrt{3}~=~0 } }$

$\mathtt{ Coeficientes : } \begin{cases} \mathtt{a~=~1} \\ \\ \mathtt{ b~=~ -(2\sqrt{3} + 1) } \\ \\ \mathtt{ c~=~ 2\sqrt{3} } \end{cases}$

$\mathtt{ \huge{ \red{ ~~~~~~ BHASKARA } } }$

$\mathtt{ x~=~ \dfrac{ -b \pm\sqrt{ \Delta }}{2a} }$ , Onde :

∆ = b² - 4ac

$\boxed{\boxed{\mathtt{ \blue{ x~=~ \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} } } } }$

$\mathtt{ x~=~ \dfrac{-[-(2\sqrt{3}+1)] \pm \sqrt{ [-(2\sqrt{3}+1)]^2 - 4.1.2\sqrt{3}}}{2.1} }$

$\mathtt{ x~=~ \dfrac{(2\sqrt{3}+1) \pm \sqrt{(2\sqrt{3})^2+2.1.2\sqrt{3} + 1^2 - 8\sqrt{3} }}{2} }$

$\mathtt{ x~=~\dfrac{(2\sqrt{3}+1) \pm \sqrt{13+4\sqrt{3} - 8\sqrt{3} }}{2}~=~ \dfrac{(2\sqrt{3}+1)\pm\sqrt{13-4\sqrt{3}}}{2} }$

Perceba que :

$\mathtt{ (1 - 2\sqrt{3} )^2~=~1^2 - 2.2\sqrt{3} + (2.\sqrt{3})^2 }$

$\mathtt{ (1 - 2\sqrt{3})^2~=~1 - 4\sqrt{3} + 12 }$

$\boxed{ \mathtt{ \blue{ (1 - 2\sqrt{3})^2~=~13 - 4\sqrt{3} } } }$ ✅✅

Logo :

$\mathtt{ x~=~ \dfrac{(2\sqrt{3} +1) \pm \sqrt{ (1 - 2\sqrt{3})^2 }}{2}~=~\dfrac{(2\sqrt{3}+1) \pm (1 - 2\sqrt{3}) }{2} }$

$\mathtt{x :} \begin{cases} \mathtt{ x_{1}~=~\dfrac{2\sqrt{3} + 1 + 1 - 2\sqrt{3} }{2}~=~\dfrac{2}{2} } \\ \\ \mathtt{ x_{2}~=~\dfrac{2\sqrt{3} + 1 - 1 + 2\sqrt{3} }{2}~=~\dfrac{4\sqrt{3}}{2} } \end{cases}$

$\mathtt{x :} \begin{cases} \boxed{\mathtt{\green{ x_{1}~=~1}}} \\ \\ \boxed{\mathtt{\green{x_{2}~=~2\sqrt{3} } } } \end{cases}$

Espero ter ajudado bastante!)