Answer:
The equation has 2 non real solutions.
Step-by-step explanation:
Given:
[tex]x ^2 = 4x - 5[/tex]
To Find:
The solutions of the equation = ?
Solution:
Lets find the solution using the quadratic equation formula
[tex]x ^2 = 4x - 5[/tex]
[tex]x^2-4x +5 = 0[/tex]
[tex]x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
Here
a = 1
b =-4
c = 5
Now Substituting the values,
[tex]x = \frac{-(-4) \pm \sqrt{(-4)^2-4(1)(5)}}{2(1)}[/tex]
[tex]x = \frac{4\pm \sqrt{16-20}}{2}[/tex]
[tex]x = \frac{4\pm \sqrt{-4}}{2}[/tex]
[tex]x = \frac{4\pm \sqrt{4}\times \sqrt{-1}}{2}[/tex]
[tex]x = \frac{4\pm 2 \sqrt{-1}}{2}[/tex]
[tex]x = \frac{4\pm 2i}{2}[/tex]
[tex]x= \frac{4+2i}{2}[/tex] [tex]x= \frac{4-2i}{2}[/tex]
x= 2 + i x= 2-i