Resposta:
[tex]\textsf{Leia abaixo}[/tex]
Explicação passo a passo:
[tex]\textsf{Pelo teorema da bissetriz interna, temos que:}[/tex]
[tex]\sf{\dfrac{4}{2} = \dfrac{\overline{\rm AC}}{x}}[/tex]
[tex]\sf{\overline{\rm AC} = 2x}[/tex]
[tex]\sf{(\overline{\rm AC})^2 = (\overline{\rm AB})^2 + (\overline{\rm BC})^2 }[/tex]
[tex]\sf{(2x)^2 = 4^2 + (2 + x)^2}[/tex]
[tex]\sf{4x^2 = 16 + (4 + 4x + x^2)}[/tex]
[tex]\sf{3x^2 - 4x - 20 = 0}[/tex]
[tex]\mathsf{\Delta = b^2 - 4.a.c}[/tex]
[tex]\mathsf{\Delta = (-4)^2 - 4.3.(-20)}[/tex]
[tex]\mathsf{\Delta = 16 + 240}[/tex]
[tex]\mathsf{\Delta = 256}[/tex]
[tex]\sf{x = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{4 \pm \sqrt{256}}{6} \rightarrow \begin{cases}\sf{x' = \dfrac{4 + 16}{6} = \dfrac{20}{6} = \dfrac{10}{3}}\\\\\sf{x'' = \dfrac{4 - 16}{6} = -\dfrac{12}{6} = -2}\end{cases}}[/tex]
[tex]\boxed{\boxed{\sf{x = \dfrac{10}{3}}}}[/tex]
