Question

Rezolvati exercitiul.. x la a doua +x+1=3 va rog sa  ma ajutati

Answer 1

x^2+x+1-3=0

x^2+x-2=0

a=1, b=1, c=-2

Δ=b^2-4ac

Δ=1^2-4*1*(-2)

Δ=1+8

Δ=9

x1=(-b+√Δ)/2a

x1=(-1+√9)/2*1

x1=(-1+3)/2

x1=2/2

x1=1

x2=(-b-√Δ)/2a

x2=(-1-√9)/2*1

x2=(-1-3)/2

x2=-4/2

x2=-2

S={-2, 1}

Answer 2

Răspuns:

solutiile sunt 1 si -2

Explicație pas cu pas:

$\displaystyle x^2+x+1=3 \\ \text{Trecem pe 3 in stanga cu semn schimbat} \\ x^2+x+1-3 =0 \\x^2+x-2 =0 \\ \\ a=1 \\ b=1 \\ c=-2 \\ \\ \Delta=b^2-4ac \\ \Delta=1^2-4\cdot1\cdot(-2) \\ \Delta=1+8 \\ \Delta=9 \\ \\ X_{1,2}=\frac{-b\pm\sqrt{\Delta} }{2a} \\ \\ X_1=\frac{-b-\sqrt{\Delta} }{2a} =\frac{-1-3}{2} =\frac{-4}{2} =-2 \\ \\ X_2=\frac{-b+\sqrt{\Delta} }{2a} =\frac{-1+3}{2} = \frac{2}{2} =1$