Question

Verifique se os numero abaixo sao raiz da equacao: X elevador a 2 menos 4X mais 3 igual a zero.
a) 3
b) 6
c) -4
d) 0
e) 2
f) 1

Answer 1

Olá, bom dia!

$x^2-4x+3=0\\\\a=1,\;b=-4\;e\;c=3\\\\\Delta=b^2-4ac\\\Delta=(-4)^2-4\cdot1\cdot3\\\Delta=16-12\\\Delta=4\\\\x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\x=\dfrac{-(-4)\pm\sqrt{4}}{2\cdot1}\\\\x=\dfrac{4\pm2}{2}\\\\x_1=\dfrac{4+2}{2}=\dfrac{6}{2}=\boxed{3}\\\\x_2=\dfrac{4-2}{2}=\dfrac{2}{2}=\boxed{1}$

Podemos concluir que apenas 1 e 3 são raízes desta equação, portanto, alternativas (a) e (f).



Espero ter ajudado.
Bons estudos! :)

Answer 2

${x}^{2} - 4x + 3 = 0 \\ x = 3 \\ {3}^{2} - 4 \times 3 + 3 = 0 \\ 9 - 12 + 3 = 0 \\ - 3 + 3 = 0 \\ 0 = 0 \\ a)3 \: e \: raiz \\ x = 6 \\ {6}^{2} - 4 \times 6 + 3 = 0 \\ 36 - 24 + 3 = 0 \\ 12 + 3 = 0 \\ 15 = 0 \\ b)6 \: nao \: e \: raiz \\ x = - 4 \\ { ( - 4)}^{2} - 4( - 4) + 3 = 0 \\ 16 + 16 + 3 = 0 \\ 35 = 0 \\ c) - 4 \: nao \: e \: raiz \\ x = 0 \\ {0}^{2} - 4 \times 0 + 3 = 0 \\ 3 = 0 \\ d)0 \: nao \: e \: raiz \\ x = 2 \\ {2}^{2} - 4 \times 2 + 3 = 0 \\ 4 - 8 + 3 = 0 \\ 1 = 0 \\ e)2 \: nao \: e \: raiz \\ x = 1 \\ {1}^{2} - 4 \times 1 + 3 = 0 \\ 1 - 4 + 3 = 0 \\ - 3 + 3 = 0 \\ 0 = 0 \\ f)1e \: raiz$