Question

Find the root of irrational equation. Guys i really need you help!

Answer 1

   
[tex]\displaystyle\\ x^2+3( \sqrt{3}-1)x-3 \sqrt{27}=0 \\ x^2+3( \sqrt{3}-1)x-3 \sqrt{9\cdot 3}=0 \\ x^2+3( \sqrt{3}-1)x-3\cdot 3 \sqrt{3}=0 \\ x^2+3( \sqrt{3}-1)x-9 \sqrt{3}=0 \\ \\ x_{12}= \frac{-b\pm \sqrt{b^2 -4ac} }{2a}= \\ \\ =\frac{-3( \sqrt{3}-1)\pm \sqrt{[3( \sqrt{3}-1)]^2 -4\cdot1 \cdot ( -9\sqrt{3})} }{2\cdot 1}= \\ \\ =\frac{-3( \sqrt{3}-1)\pm \sqrt{9(3-2\sqrt{3}+1) +36\sqrt{3}} }{2}= \\ \\ =\frac{-3( \sqrt{3}-1)\pm \sqrt{9(4-2\sqrt{3}) +36\sqrt{3}} }{2}= [/tex]


[tex]\displaystyle \\ =\frac{-3( \sqrt{3}-1)\pm \sqrt{36-18\sqrt{3} +36\sqrt{3}} }{2}= \\ \\ =\frac{-3( \sqrt{3}-1)\pm \sqrt{36+18\sqrt{3} } }{2}= \\ \\ =\frac{-3( \sqrt{3}-1)\pm \sqrt{9(4+2\sqrt{3}) } }{2}= \\ \\ =\frac{-3( \sqrt{3}-1)\pm 3\sqrt{4+2\sqrt{3} } }{2}= \\ \\ =\frac{-3( \sqrt{3}-1)\pm 3\sqrt{3+1+2\sqrt{3} } }{2}= \\ \\ =\frac{-3( \sqrt{3}-1)\pm 3\sqrt{3+2\sqrt{3}+1 } }{2}= \\ \\ =\frac{-3( \sqrt{3}-1)\pm 3\sqrt{(\sqrt{3}+1 )^2} }{2}= [/tex]


[tex]\displaystyle\\ =\frac{-3( \sqrt{3}-1)\pm 3(\sqrt{3}+1 ) }{2}=\boxed{\frac{-3\sqrt{3}+3\pm (3\sqrt{3}+3) }{2}} \\ \\ \\ x_1 = \frac{-3\sqrt{3}+3 + 3\sqrt{3}+3 }{2}= \frac{6}{2} =\boxed{3} \\ \\ \\ x_2= \frac{-3\sqrt{3}+3 -3\sqrt{3}-3 }{2}= \frac{-6\sqrt{3}}{2}=\boxed{-3\sqrt{3}}[/tex]

Answer 2

x²+3(√3-1)x-3√27=0, sau: x²+3(√3-1)x-3*3*√3=0
Avem relatiile lui Vieta: [tex]S= x_{1} + x_{2}=- \frac{b}{a}=-[3( \sqrt{3} -1)]=-3 \sqrt{3}+3 [/tex]
 $P= x_{1} x_{2} = \frac{c}{a}=-3 \sqrt{27}=(-3 \sqrt{9*3})=(-3 \sqrt{3})*3$
Deci: $x_{1}=-3 \sqrt{3},si, x_{2}=3$