Question

Select the equation below that has solutions of 2- 3i and 2 + 3i. A. 0 = x^2 - 4x – 13B. 0 = x^2 - 4x + 5C.0 = x^2 - 4x - 5 D.0 = x^2 - 4x + 13

Answer 1

To know the equation with the solutions 2-3i and 2+3i

We will solve each using the quadratic formula;

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

x^2 - 4x – 13=0

a=1 b=-4 c=-13

substitute the values above into the formula and then evaluate

`x=\frac{4\pm\sqrt[]{(-4)^2-4(1)(-13)}}{2(1)}`

`=\frac{4\pm\sqrt[]{16+52}}{2}`
`=(4)/(2)\pm\frac{\sqrt[]{68}}{2}`

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

`x=\frac{4\pm\sqrt[]{(-4)^2-4(1)(5)}}{2}`
`=(4)/(2)\pm\frac{\sqrt[]{-4}}{2}`

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

`x=\frac{4\pm\sqrt[]{(-4)^2-4(1)(13)}}{2}`
`x=(4)/(2)\pm\frac{\sqrt[]{-36}}{2}`

x= 2 ± 3i

Either x= 2-3i or x= 2+3i

The correct option is D

D. 0 = x^2 - 4x + 13​