Ответ:

[tex]1)\;x^2+2\sqrt{3}-23=0\\\\2)\; x^2+2\sqrt{2}x-23=0\\\\3)\; x^2-5=0[/tex]

Объяснение:

[tex]a(x-x_1)(x-x_2)=0[/tex]

Пусть а=1, тогда квадратное уравнение будет таким:

[tex](x-x_1)(x-x_2)=0[/tex]

[tex]1)x_{1,2}=2\pm\sqrt{3}\\\\(x+(2+\sqrt{3}))(x-(2-\sqrt{3}))=0\\x^2+(2+\sqrt{3})x-(2-\sqrt{3})x-(2+\sqrt{3})(2-\sqrt{3})=0\\x^2+2x+\sqrt{3}x-2x+\sqrt{3}x-(4-3)=0\\x^2+2\sqrt{3}-1=0[/tex]

[tex]2)x_{1,2}=5\pm\sqrt{2}\\\\(x+(5+\sqrt{2}))(x-(5-\sqrt{2}))=0\\x^2+(5+\sqrt{2})x-(5-\sqrt{2})x-(5+\sqrt{2})(5-\sqrt{2}) =0\\x^2+5x+\sqrt{2}x-5x+\sqrt{2}x-(25-2)=0\\x^2+2\sqrt{2}x-23=0[/tex]

[tex]3)x_{1,2}=\pm\sqrt{5}\\\\(x-\sqrt{5})(x-(-\sqrt{5}))=0\\(x-\sqrt{5})(x+\sqrt{5})=0\\x^2-5=0[/tex]