Question

Rezolvati ecuatia irationala si stabiliti si domeniul: radical din 2x^2+x+4=radical din 7.

Answer 1

$\sqrt{2x {}^{2} + x + 4 } = \sqrt{7} = > 2 {x}^{2} + x + 4 = 7 = > 2x {}^{2} + x + 4 - 7 = 0 = > 2x {}^{2} + x - 3 = 0 = > x = \frac{ - 1 - \sqrt{ {1}^{2} - 4 \times 2 \times ( - 3)} }{2 \times 2} . \: \: x = \frac{ - 1 + \sqrt{ {1}^{2} - 4 \times 2 \times ( - 3)} }{2 \times 2} . = > x = \frac{ - 1 + \sqrt{1 + 24} }{4} . \: \: x = \frac{ - 1 - \sqrt{1 + 24} }{4} . \: \: x = \frac{ - 1 + \sqrt{25} }{4} . \: \: x = \frac{ - 1 - \sqrt{25} }{4} . \: x = \frac{ - 1 - 5}{4} . \: x = \frac{ - 1 + 5}{4} = > x = 1. \: x = - \frac{3}{2}$

Answer 2

Răspuns:

x∈{-3/2;1}

Explicație pas cu pas:

Domeniu

2x²+x+4≥0

Δ=1-482*4<0, x∈R=D

rezolvare

√(2x^2+x+4)=√7

ridicam la patrat

2x^2+x+4=7

2x^2+x-3=0

Δ=1-4*2*(-3)=1+24=25

x1,2=(-1+-√25)/2*2=(-1+-5)/4

x1=-6/4=-3/2

x2=4/4=1

S={-3/2;1}