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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

User Manuchehr
by
7.2k points

2 Answers

2 votes

Answer:

d. x2 – 2x – 4 = 0

Explanation:

User Mndrix
by
8.1k points
1 vote

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

User Rakesh Tembhurne
by
8.1k points

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