Question

Testul 2, subiectul 1, exercițiul 3. Va rog ❤️

Answer 1

Bună!

$x=\sqrt{4x+6} <=> \sqrt{4x+6} =x |^{2} <=> \sqrt{4x+6} ^{2} =x^{2} <=> 4x+6=x^{2} <=> -x^{2} +4x+6=0$

a=-1

b=4

c=6

Δ=b²-4ac=4²-4×(-1)×6=16+24=40 >0 ⇒

    $x_{1} =\frac{-b+\sqrt{delta} }{2a} =\frac{-4+\sqrt{40} }{-2} =\frac{-4+2\sqrt{10} }{-2} = -\frac{2(-2+\sqrt{10)} }{2} =-(-2+\sqrt{10} )=2-\sqrt{10}$

     $x_{2} = \frac{-b-\sqrt{delta} }{2a} =\frac{-4-2\sqrt{10} }{-2} =\frac{-2(2+\sqrt{10}) }{-\\2} =2+\sqrt{10}$

• pentru $x=2-\sqrt{10}$

$2-\sqrt{10} =\sqrt{4(2-\sqrt{10})+6 } <=> 2-\sqrt{10} =\sqrt{8-4\sqrt{10}+6 } <=> 2-\sqrt{10} =\sqrt{14-4\sqrt{10} } <=> 2-\sqrt{10} = \sqrt{(2-\sqrt{10})^{2} } <=> 2-\sqrt{10}= \sqrt{10} -2 (F)$

• pentru $x=2+\sqrt{10}$

$2+\sqrt{10} =\sqrt{4(2+\sqrt{10})+6 } <=> 2+\sqrt{10} =\sqrt{8+4\sqrt{10} +6} <=> 2+\sqrt{10} =\sqrt{14+4\sqrt{10} } <=> 2+\sqrt{10} =\sqrt{(2+\sqrt{10})^{2} } <=> 2+\sqrt{10} =2+\sqrt{10} (A)$

                       x ∈ $2+\sqrt{10}$