Question

зх² = 2(х – 1)² + 3 =​

Answer 1

Explicação passo-a-passo:

confiaaaa na call plis ...............

Answer 2

$3x^{2} =2\cdot(x-1)^{2} +3\\ \\ 3x^{2} =2\cdot[(x^{2} -2\cdotx\cdot1+(-1)^{2}]+3\\ \\ 3x^{2} =2\cdot(x^{2} -2x+1)+3\\ \\ 3x^{2} =2x^{2} -4x+2+3\\ \\ 3x^{2} =2x^{2} -4x+5\\ \\ 3x^{2} -2x^{2} +4x-5=0\\ \\ x^{2} +4x-5=0~~~~~\to ~~~~~Bhaskara$

Bháskara:

$x^{2}+4x-5=0~~~\to ~~~a=1~~,~~b=4~~,~~c=-5\\ \\\\ \dfrac{-b\pm\sqrt{b^{2}-4\cdot a\cdot c } }{2\cdot a} ~~=~~\dfrac{-4\pm\sqrt{4^{2}-4\cdot 1\cdot (-5) } }{2\cdot 1} ~~=~~\dfrac{-4\pm\sqrt{36 } }{2}$

$x'=\dfrac{-4+6}{2} =\dfrac{2}{2} ~~~\to~~~\boxed{x'=1} \\ \\ \\ x''=\dfrac{-4-6}{2} =-\dfrac{10}{2} ~~~\to~~~\boxed{x''=-5 }$

Resposta:

Os valores de x que satisfazem a questão são  x = 1   e   x = -5 .

:)