Question

Which equation has the solutions x=1+or-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

d. x2 – 2x – 4 = 0

Explanation:

Answer 2

We will proceed to solve each case to determine the solution of the problem.

case a)
`x^(2)+2x+4=0`

Group terms that contain the same variable, and move the constant to the opposite side of the equation

`x^(2)+2x=-4`

Complete the square. Remember to balance the equation by adding the same constants to each side.

`x^(2)+2x+1=-4+1`

`x^(2)+2x+1=-3`

Rewrite as perfect squares

`(x+1)^(2)=-3`

`(x+1)=(+/-)√(-3)\\(x+1)=(+/-)√(3)i\\x=-1(+/-)√(3)i`

therefore

case a) is not the solution of the problem

case b)
`x^(2)-2x+4=0`

Group terms that contain the same variable, and move the constant to the opposite side of the equation

`x^(2)-2x=-4`

Complete the square. Remember to balance the equation by adding the same constants to each side.

`x^(2)-2x+1=-4+1`

`x^(2)-2x+1=-3`

Rewrite as perfect squares

`(x-1)^(2)=-3`

`(x-1)=(+/-)√(-3)\\(x-1)=(+/-)√(3)i\\x=1(+/-)√(3)i`

therefore

case b) is not the solution of the problem

case c)
`x^(2)+2x-4=0`

Group terms that contain the same variable, and move the constant to the opposite side of the equation

`x^(2)+2x=4`

Complete the square. Remember to balance the equation by adding the same constants to each side.

`x^(2)+2x+1=4+1`

`x^(2)+2x+1=5`

Rewrite as perfect squares

`(x+1)^(2)=5`

`(x+1)=(+/-)√(5)\\x=-1(+/-)√(5)`

therefore

case c) is not the solution of the problem

case d)
`x^(2)-2x-4=0`

Group terms that contain the same variable, and move the constant to the opposite side of the equation

`x^(2)-2x=4`

Complete the square. Remember to balance the equation by adding the same constants to each side.

`x^(2)-2x+1=4+1`

`x^(2)-2x+1=5`

Rewrite as perfect squares

`(x-1)^(2)=5`

`(x-1)=(+/-)√(5)\\x=1(+/-)√(5)`

therefore

case d) is the solution of the problem

therefore

the answer is

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