Question

resolva as equações a) x²+4x=5 b) 2x²-2x=14. por favor me ajudem

Answer 1

Olá.

$\boxed{\mathsf{x^{2} + 4x = 5}}$

• Passando todas as parcelas para o mesmo lado da equação:

$\mathsf{x^{2} + 4x - 5 = 0}$

• Calculando o determinante (delta):

$\mathsf{\triangle = b^{2} - 4ac \to \triangle = 4^{2} - 4 \times 1 \times (-5) \to \triangle = 16 + 20 \to \triangle = 36}$

• Calculando as raízes:

$\mathsf{x = \dfrac{-b \pm \sqrt{\triangle}}{2a}}$

$\mathsf{x_{1} = \dfrac{-4 + 6}{2} = 1} \\ \\ \mathsf{x_{2} = \dfrac{-4 - 6}{2} = -5}$

$\boxed{\boxed{\boxed{\mathsf{S = (-5, 1)}}}}}$

$\boxed{2x^{2} - 2x = 14}$

$\mathsf{2x^{2} - 2x - 14 = 0 \: \: \: \: \div 2} \\ \mathsf{x^{2} - x - 7 = 0} \\ \\ \mathsf{\triangle = 1 + 28 \to \triangle = 29}$

$\mathsf{x = \dfrac{-1 \pm \sqrt{-29}}{2}}$

$\mathsf{x_{1} = \dfrac{1 + \sqrt{29}}{2}} \\ \\ \mathsf{x_{2} = \dfrac{1 - \sqrt{29}}{2}}$

$S = \boxed{\boxed{\boxed{\mathsf{S = (\dfrac{1 + \sqrt{29}}{2} \: ; \: \dfrac{1 - \sqrt{29}}{2})}}}}}$

Bons estudos.