Question

Solve this problem out

Answer 1

Answer:

The equation D has (1-i) as a solution

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

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

Option A

in this problem we have

$x^{2} +2x-2=0$  

so

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

substitute in the formula

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

$x=\frac{-2(+/-)\sqrt{12}} {2}$

$x=\frac{-2(+/-)2\sqrt{3}} {2}$

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

Has two real solutions

Option B

in this problem we have

$x^{2} +2x+2=0$  

so

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

substitute in the formula

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

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

Remember that

$i=\sqrt{-1}$

$x=\frac{-2(+/-)2i} {2}$

$x=-1(+/-)i$

$x=-1+i$

$x=-1-i$

Option C

in this problem we have

$x^{2} -2x-2=0$  

so

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

substitute in the formula

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

$x=\frac{2(+/-)\sqrt{12}} {2}$

$x=\frac{2(+/-)2\sqrt{3}} {2}$

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

Has two real solutions

Option D

in this problem we have

$x^{2} -2x+2=0$  

so

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

substitute in the formula

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

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

Remember that

$i=\sqrt{-1}$

$x=\frac{2(+/-)2i} {2}$

$x=1(+/-)i$

$x=1+i$

$x=1-i$

therefore

The equation D has (1-i) as a solution