Question

radical din 3-x=2x ajutor

Answer 1

coditiide existenta

3-x≥0

si

x≥0

rezulta x∈[0;3]

rezolvare

ridicam la patrat

4x²=3-x

4x²+x-3=0

Δ=1-4*4*(-3)=1+48=49⇒√Δ=7

x1,2=(-1+/-7)/8

x1=(-1-7)/8=-8/8=-1∉[0;3]

x2=(-1+7)/8=6/8=3/4∈[0;3]

S={3/4}

verificare

√(3-3/4) =2*3/4

√(9/4)=3/2

3/2=3/2, Adevarat , bine rezolvat

extra ...

problema are si o rezolvare grafica din care se vede ca solutia este unica; in atasament este reprezentata aproximativ

Answer 2

$\sqrt{3-x} = 2x~~ \rightarrow \text{ridic\u{a}m ambele p\u{a}r\c{t}i la p\u{a}trat} \\ \\ \Big(\sqrt{3 - x}\Big)^{2} = (2x)^{2} \\ 3 - x = 4x^{2} \\3 - x - 4x^{2} = 0\Big{|}\times(-1) \\ 4x^{2} + x - 3 = 0\\ \\ \Delta = b^2- 4ac = 1 - 4\times (-3)\times 4 = 1 + 48 = 49\\ \\ x_1= \dfrac{-b-\sqrt{\Delta}}{2a} = \dfrac{-1-7}{8} = -\dfrac{8}{8} = -1\\ \\ x_2=\dfrac{-b+\sqrt{\Delta}}{2a} = \dfrac{-1 + 7}{8} = \dfrac{6}{8}= \dfrac{3}{4} \\ \\ \text{Condi\c{t}ii de existen\c{t}\u{a}:}$
$3-x\geq 0\Rightarrow -x \geq -3 \Big|\cdot(-1) \Rightarrow x \leq 3\\2x\geq 0 \Rightarrow x\geq 0\\ \\\Rightarrow 0\leq x\leq 3 \Rightarrow x\in [0,3]$


$\bullet~x_1 = -1 \notin [0,3] \\ \\ \bullet~x_2 =\dfrac{3}{4}\in [0,3]\\ \\ \Rightarrow S = \Big\{\dfrac{3}{4}\Big\}$