Question

Como resolver f(x)= x²-2x-3 calculando x para que : f(x)=4

Answer 1

$f(x)=x^2-2x-3=4\\ \\ x^2-2x-3-4=0\\ \\ x^2-2x-7=0\\ \\ \Delta=b^2-4ac\\ \\ \Delta=(-2)^2-4.1.(-7)=4+28=32\\ \\ x=\frac{-b\pm\sqrt{\Delta}}{2a}\\ \\ x=\frac{2\pm\sqrt{32}}{2}=\frac{2\pm4\sqrt2}{2}=1\pm2\sqrt2\\ \\ S=\{1-2\sqrt2,1+2\sqrt2\}$

Answer 2

$f(x)= x^{2} -2x-3$

$f(x)=4$ ⇒ $x^{2} -2x-3 =4$

$x^{2} -2x-3 -4=0$

$x^{2} -2x-7=0$

Δ$= b^{2} -4ac= (-2)^{2} -4.1.(-7)=4+28=32$

$x= \dfrac{-(-2)+ \sqrt{32} }{2.1} = \dfrac{2+ 4\sqrt{2} }{2}=1+ 2\sqrt{2}$
ou
$x= \dfrac{-(-2)- \sqrt{32} }{2.1} = \dfrac{2- 4\sqrt{2} }{2}=1- 2\sqrt{2}$