Question

I NEED HELP PLS THIS IS DUE IN 3 HOURS

Answer 1

Answer:

Part 1)  $x^{2} -2x-2=(x-1-\sqrt{3})(x-1+\sqrt{3})$

Part 2)  $x^{2} -6x+4=(x-3-\sqrt{5})(x-3+\sqrt{5})$

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

Part 1)

in this problem we have

$x^{2} -2x-2=0$

so

$a=1\\b=-2\\c=-2$

substitute in the formula

$x=\frac{-(-2)(+/-)\sqrt{-2^{2}-4(1)(-2)}} {2(1)}\\\\x=\frac{2(+/-)\sqrt{12}} {2}\\\\x=\frac{2(+/-)2\sqrt{3}} {2}\\\\x_1=\frac{2(+)2\sqrt{3}} {2}=1+\sqrt{3}\\\\x_2=\frac{2(-)2\sqrt{3}} {2}=1-\sqrt{3}$

therefore

$x^{2} -2x-2=(x-(1+\sqrt{3}))(x-(1-\sqrt{3}))$

$x^{2} -2x-2=(x-1-\sqrt{3})(x-1+\sqrt{3})$

Part 2)

in this problem we have

$x^{2} -6x+4=0$

so

$a=1\\b=-6\\c=4$

substitute in the formula

$x=\frac{-(-6)(+/-)\sqrt{-6^{2}-4(1)(4)}} {2(1)}$

$x=\frac{6(+/-)\sqrt{20}} {2}$

$x=\frac{6(+/-)2\sqrt{5}} {2}$

$x_1=\frac{6(+)2\sqrt{5}}{2}=3+\sqrt{5}$

$x_2=\frac{6(-)2\sqrt{5}}{2}=3-\sqrt{5}$

therefore

$x^{2} -6x+4=(x-(3+\sqrt{5}))(x-(3-\sqrt{5}))$

$x^{2} -6x+4=(x-3-\sqrt{5})(x-3+\sqrt{5})$