Answer:
[tex]x = 1\pm \sqrt{5}[/tex]
Step-by-step explanation:
Equation: [tex]ax^2+bx+c=0[/tex] --1
Quadratic formula : [tex]x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Now Solve the given option using this formula
A)[tex]x^2 + 2x + 4 = 0[/tex]
On comparing with 1
a = 1 , b = 2 , c= 4
So, [tex]x = \frac{-2\pm \sqrt{2^2-4(1)(4)}}{2(1)}[/tex]
[tex]x = \frac{-2\pm \sqrt{-12}}{2}[/tex]
B)[tex]x^2 - 2x + 4 = 0[/tex]
On comparing with 1
a = 1 , b = -2 , c= 4
So, [tex]x = \frac{2\pm \sqrt{(-2)^2-4(1)(4)}}{2(1)}[/tex]
[tex]x = \frac{2\pm \sqrt{-12}}{2}[/tex]
C)[tex]x^2 + 2x - 4 = 0[/tex]
On comparing with 1
a = 1 , b = 2 , c=- 4
So, [tex]x = \frac{-2\pm \sqrt{2^2-4(1)(-4)}}{2(1)}[/tex]
[tex]x = \frac{-2\pm \sqrt{20}}{2}[/tex]
[tex]x = \frac{-2\pm 2\sqrt{5}}{2}[/tex]
[tex]x =-1\pm \sqrt{5}[/tex]
D)[tex]x^2 - 2x - 4 = 0[/tex]
On comparing with 1
a = 1 , b = -2 , c= -4
So, [tex]x = \frac{2\pm \sqrt{(-2)^2-4(1)(-4)}}{2(1)}[/tex]
[tex]x = \frac{2\pm \sqrt{20}}{2}[/tex]
[tex]x = \frac{2\pm 2\sqrt{5}}{2}[/tex]
[tex]x = 1\pm \sqrt{5}[/tex]
Hence Option D is correct.
[tex]x^2 - 2x - 4 = 0[/tex] has a solution [tex]x = 1\pm \sqrt{5}[/tex]
