Which equation has the solutions x = 1 plus-or-minus StartRoot 5 EndRoot? x2 + 2x + 4 = 0 x2 – 2x + 4 = 0 x2 + 2x – 4 = 0 x2 – 2x – 4 = 0
Answers
Answer:
We will proceed to solve each case to determine the solution of the problem.
case a) x^{2}+2x+4=0x
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=-4x
2
+2x=−4
Complete the square. Remember to balance the equation by adding the same constants to each side.
x^{2}+2x+1=-4+1x
2
+2x+1=−4+1
x^{2}+2x+1=-3x
2
+2x+1=−3
Rewrite as perfect squares
(x+1)^{2}=-3(x+1)
2
=−3
\begin{gathered}(x+1)=(+/-)\sqrt{-3}\\(x+1)=(+/-)\sqrt{3}i\\x=-1(+/-)\sqrt{3}i\end{gathered}
(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=0x
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=-4x
2
−2x=−4
Complete the square. Remember to balance the equation by adding the same constants to each side.
x^{2}-2x+1=-4+1x
2
−2x+1=−4+1
x^{2}-2x+1=-3x
2
−2x+1=−3
Rewrite as perfect squares
(x-1)^{2}=-3(x−1)
2
=−3
\begin{gathered}(x-1)=(+/-)\sqrt{-3}\\(x-1)=(+/-)\sqrt{3}i\\x=1(+/-)\sqrt{3}i\end{gathered}
(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=0x
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=4x
2
+2x=4
Complete the square. Remember to balance the equation by adding the same constants to each side.
x^{2}+2x+1=4+1x
2
+2x+1=4+1
x^{2}+2x+1=5x
2
+2x+1=5
Rewrite as perfect squares
(x+1)^{2}=5(x+1)
2
=5
\begin{gathered}(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}\end{gathered}
(x+1)=(+/−)
5
x=−1(+/−)
5
therefore
case c) is not the solution of the problem
case d) x^{2}-2x-4=0x
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=4x
2
−2x=4
Complete the square. Remember to balance the equation by adding the same constants to each side.
x^{2}-2x+1=4+1x
2
−2x+1=4+1
x^{2}-2x+1=5x
2
−2x+1=5
Rewrite as perfect squares
(x-1)^{2}=5(x−1)
2
=5
\begin{gathered}(x-1)=(+/-)\sqrt{5}\\x=1(+/-)\sqrt{5}\end{gathered}
(x−1)=(+/−)
5
x=1(+/−)
5
therefore
case d) is the solution of the problem
therefore
the answer is
x^{2}-2x-4=0x
2
−2x−4=0