Question

Considere as funções f e g, tais que f(x) = x^2 + 3x - 5 e g(x) = 2x^2 + 7x - 2. o valores de x para que f(x) = g(x) são *

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo-a-passo:

$\sf f(x) = x^2 + 3x - 5$

$\sf g(x) = 2x^2 + 7x - 2$

$\sf f(x) = g(x)$

$\sf x^2 + 3x - 5 = 2x^2 + 7x - 2$

$\sf 2x^2 - x^2 + 7x - 3x -2 + 5 = 0$

$\sf x^2 + 4x + 3 = 0$

$\sf \Delta = b^2 - 4.a.c$

$\sf \Delta = 4^2 - 4.1.3$

$\sf \Delta = 16 - 12$

$\sf \Delta = 4$

$\sf x = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{-4 \pm \sqrt{4}}{2.1} \Rightarrow \begin{cases}\sf x' = \dfrac{-4 + 2}{2} = \dfrac{-2}{2} = -1\\\sf \\\sf x'' = \dfrac{-4 - 2}{2} = \dfrac{-6}{2} = -3\end{cases}$

$\boxed{\boxed{\sf S = \{-1;-3\}}}$

Answer 2

Explicação passo-a-passo:

$\sf x^2+3x-5=2x^2+7x-2$

$\sf 2x^2-x^2+7x-3x-2+5=0$

$\sf x^2+4x+3=0$

$\sf \Delta=4^2-4\cdot1\cdot4$

$\sf \Delta=16-12$

$\sf \Delta=4$

$\sf x=\dfrac{-4\pm\sqrt{4}}{2\cdot1}=\dfrac{-4\pm2}{2}$

$\sf x'=\dfrac{-4+2}{2}~\Rightarrow~x'=\dfrac{-2}{2}~\Rightarrow~\red{x'=-1}$

$\sf x"=\dfrac{-4-2}{2}~\Rightarrow~x"=\dfrac{-6}{2}~\Rightarrow~\red{x"=-3}$