Answer:
Part 1) Â [tex]x^{2} -2x-2=(x-1-\sqrt{3})(x-1+\sqrt{3})[/tex]
Part 2) Â [tex]x^{2} -6x+4=(x-3-\sqrt{5})(x-3+\sqrt{5})[/tex]
Step-by-step explanation:
we know that
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
Part 1)
in this problem we have
[tex]x^{2} -2x-2=0[/tex]
so
[tex]a=1\\b=-2\\c=-2[/tex]
substitute in the formula
[tex]x=\frac{-(-2)(+/-)\sqrt{-2^{2}-4(1)(-2)}} {2(1)}\\\\x=\frac{2(+/-)\sqrt{12}} {2}\\\\x=\frac{2(+/-)2\sqrt{3}} {2}\\\\x_1=\frac{2(+)2\sqrt{3}} {2}=1+\sqrt{3}\\\\x_2=\frac{2(-)2\sqrt{3}} {2}=1-\sqrt{3}[/tex]
therefore
[tex]x^{2} -2x-2=(x-(1+\sqrt{3}))(x-(1-\sqrt{3}))[/tex]
[tex]x^{2} -2x-2=(x-1-\sqrt{3})(x-1+\sqrt{3})[/tex]
Part 2)
in this problem we have
[tex]x^{2} -6x+4=0[/tex]
so
[tex]a=1\\b=-6\\c=4[/tex]
substitute in the formula
[tex]x=\frac{-(-6)(+/-)\sqrt{-6^{2}-4(1)(4)}} {2(1)}[/tex]
[tex]x=\frac{6(+/-)\sqrt{20}} {2}[/tex]
[tex]x=\frac{6(+/-)2\sqrt{5}} {2}[/tex]
[tex]x_1=\frac{6(+)2\sqrt{5}}{2}=3+\sqrt{5}[/tex]
[tex]x_2=\frac{6(-)2\sqrt{5}}{2}=3-\sqrt{5}[/tex]
therefore
[tex]x^{2} -6x+4=(x-(3+\sqrt{5}))(x-(3-\sqrt{5}))[/tex]
[tex]x^{2} -6x+4=(x-3-\sqrt{5})(x-3+\sqrt{5})[/tex]
