Respuesta :
To know the equation with the solutions 2-3i and 2+3i
We will solve each using the quadratic formula;
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]x^2 - 4x – 13=0
a=1 b=-4 c=-13
substitute the values above into the formula and then evaluate
[tex]x=\frac{4\pm\sqrt[]{(-4)^2-4(1)(-13)}}{2(1)}[/tex][tex]=\frac{4\pm\sqrt[]{16+52}}{2}[/tex][tex]=\frac{4}{2}\pm\frac{\sqrt[]{68}}{2}[/tex]The above has no such root
Let's move on and check the second option
x^2 - 4x + 5=0
a=1 b= -4 c=5
substitute into the formula and evaluate
[tex]x=\frac{4\pm\sqrt[]{(-4)^2-4(1)(5)}}{2}[/tex][tex]=\frac{4}{2}\pm\frac{\sqrt[]{-4}}{2}[/tex]x =2 ± i
The above is not the option
Let's check the next option
x^2 - 4x - 5
a=1 b= -4 c= -5
Option c is also not an option since c=-5, it will give a positive number on the square root
Let;s move on and check option D
x^2 - 4x + 13
a=1 b= -4 c=13
substitute into the formula and then evaluate
[tex]x=\frac{4\pm\sqrt[]{(-4)^2-4(1)(13)}}{2}[/tex][tex]x=\frac{4}{2}\pm\frac{\sqrt[]{-36}}{2}[/tex]x= 2 ± 3i
Either x= 2-3i or x= 2+3i
The correct option is D
D. 0 = x^2 - 4x + 13
