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lozanna [386]
4 years ago
5

I NEED HELP PLS THIS IS DUE IN 3 HOURS

Mathematics
1 answer:
Mariulka [41]4 years ago
5 0

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})

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