Question

se considera solutiile f, g : R-R , f(x)= 2 x la a 3 - 2x+1, g(x)= -xpatrat + 1. sa se determine solutiile reale ale ecuatiei f(x)= g(x).

Answer 1

Salut,

$2x^3-2x+1=-x^2+1,\;sau\;2x^3+x^2-2x=0,\;sau\;x(2x^2+x-2)=0,\\\\deci\;x_1=0,\;x_{2,3}=\dfrac{-1\pm\sqrt{1-4\cdot 2\cdot(-2)}}{2\cdot 2}=\dfrac{-1\pm\sqrt{17}}{4}.$

Green eyes.

Answer 2

f(x)=2x²-2x+1
g(x)=-x²+1
f(x)=g(x)
2x²-2x+1=-x²+1⇒
2x²+x²-2x+1-1=0⇒
3x²-2x=0⇒
a=3
b=2
c=0
Δ=b²-4ac⇒Δ=4-0⇒Δ=4⇒√Δ=√4⇒Δ=2.
x1x2=-b+-√Δ/2a
x1=-2+2/6⇒0/6⇒x1=0
x2=-2-2/6⇒-4/6⇒x2=-2/3
sucees