Question

Which equation has the solutions x=1+- square root of 5

A. x2 + 2x + 4 = 0
B. x2 – 2x + 4 = 0
C. x2 + 2x – 4 = 0
D. x2 – 2x – 4 = 0

Answer 1

x2 – 2x – 4 = 0 is the right option.

Answer 2

`x = 1\pm √(5)`

Explanation:

Equation:
`ax^2+bx+c=0` --1

Quadratic formula :
`x = (-b\pm √(b^2-4ac))/(2a)`

Now Solve the given option using this formula

A)
`x^2 + 2x + 4 = 0`

On comparing with 1

a = 1 , b = 2 , c= 4

So,
`x = (-2\pm √(2^2-4(1)(4)))/(2(1))`

`x = (-2\pm √(-12))/(2)`

B)
`x^2 - 2x + 4 = 0`

On comparing with 1

a = 1 , b = -2 , c= 4

So,
`x = (2\pm √((-2)^2-4(1)(4)))/(2(1))`

`x = (2\pm √(-12))/(2)`

C)
`x^2 + 2x - 4 = 0`

On comparing with 1

a = 1 , b = 2 , c=- 4

So,
`x = (-2\pm √(2^2-4(1)(-4)))/(2(1))`

`x = (-2\pm √(20))/(2)`

`x = (-2\pm 2√(5))/(2)`

`x =-1\pm √(5)`

D)
`x^2 - 2x - 4 = 0`

On comparing with 1

a = 1 , b = -2 , c= -4

So,
`x = (2\pm √((-2)^2-4(1)(-4)))/(2(1))`

`x = (2\pm √(20))/(2)`

`x = (2\pm 2√(5))/(2)`

`x = 1\pm √(5)`

Hence Option D is correct.

`x^2 - 2x - 4 = 0` has a solution
`x = 1\pm √(5)`