Bună!
[tex]x=\sqrt{4x+6} <=> \sqrt{4x+6} =x |^{2} <=> \sqrt{4x+6} ^{2} =x^{2} <=> 4x+6=x^{2} <=> -x^{2} +4x+6=0[/tex]
a=-1
b=4
c=6
Δ=b²-4ac=4²-4×(-1)×6=16+24=40 >0 ⇒
[tex]x_{1} =\frac{-b+\sqrt{delta} }{2a} =\frac{-4+\sqrt{40} }{-2} =\frac{-4+2\sqrt{10} }{-2} = -\frac{2(-2+\sqrt{10)} }{2} =-(-2+\sqrt{10} )=2-\sqrt{10}[/tex]
[tex]x_{2} = \frac{-b-\sqrt{delta} }{2a} =\frac{-4-2\sqrt{10} }{-2} =\frac{-2(2+\sqrt{10}) }{-\\2} =2+\sqrt{10}[/tex]
- pentru [tex]x=2-\sqrt{10}[/tex]
[tex]2-\sqrt{10} =\sqrt{4(2-\sqrt{10})+6 } <=> 2-\sqrt{10} =\sqrt{8-4\sqrt{10}+6 } <=> 2-\sqrt{10} =\sqrt{14-4\sqrt{10} } <=> 2-\sqrt{10} = \sqrt{(2-\sqrt{10})^{2} } <=> 2-\sqrt{10}= \sqrt{10} -2 (F)[/tex]
- pentru [tex]x=2+\sqrt{10}[/tex]
[tex]2+\sqrt{10} =\sqrt{4(2+\sqrt{10})+6 } <=> 2+\sqrt{10} =\sqrt{8+4\sqrt{10} +6} <=> 2+\sqrt{10} =\sqrt{14+4\sqrt{10} } <=> 2+\sqrt{10} =\sqrt{(2+\sqrt{10})^{2} } <=> 2+\sqrt{10} =2+\sqrt{10} (A)[/tex]
x ∈ [tex]2+\sqrt{10}[/tex]
