Answer :
To solve the equation [tex]\((x-1)^2 = 50\)[/tex], we can follow these steps:
1. Take the square root of both sides:
[tex]\[ (x-1)^2 = 50 \][/tex]
[tex]\[ x-1 = \pm \sqrt{50} \][/tex]
2. Simplify [tex]\(\sqrt{50}\)[/tex]:
[tex]\[ \sqrt{50} = \sqrt{25 \cdot 2} = 5 \sqrt{2} \][/tex]
3. So, we have two equations:
[tex]\[ x - 1 = 5 \sqrt{2} \quad \text{or} \quad x - 1 = -5 \sqrt{2} \][/tex]
4. Solve for [tex]\(x\)[/tex] in both cases:
[tex]\[ x - 1 = 5 \sqrt{2} \quad \Rightarrow \quad x = 1 + 5 \sqrt{2} \][/tex]
[tex]\[ x - 1 = -5 \sqrt{2} \quad \Rightarrow \quad x = 1 - 5 \sqrt{2} \][/tex]
Thus, the solutions to the equation [tex]\((x-1)^2 = 50\)[/tex] are:
[tex]\[ x = 1 + 5 \sqrt{2} \quad \text{and} \quad x = 1 - 5 \sqrt{2} \][/tex]
From the given choices:
- [tex]\( x = -49 \)[/tex]
- [tex]\( x = 51 \)[/tex]
- [tex]\( x = 1 + 5 \sqrt{2} \)[/tex]
- [tex]\( x = 1 - 5 \sqrt{2} \)[/tex]
The correct values of [tex]\( x \)[/tex] are [tex]\( x = 1 + 5 \sqrt{2} \)[/tex] and [tex]\( x = 1 - 5 \sqrt{2} \)[/tex].
1. Take the square root of both sides:
[tex]\[ (x-1)^2 = 50 \][/tex]
[tex]\[ x-1 = \pm \sqrt{50} \][/tex]
2. Simplify [tex]\(\sqrt{50}\)[/tex]:
[tex]\[ \sqrt{50} = \sqrt{25 \cdot 2} = 5 \sqrt{2} \][/tex]
3. So, we have two equations:
[tex]\[ x - 1 = 5 \sqrt{2} \quad \text{or} \quad x - 1 = -5 \sqrt{2} \][/tex]
4. Solve for [tex]\(x\)[/tex] in both cases:
[tex]\[ x - 1 = 5 \sqrt{2} \quad \Rightarrow \quad x = 1 + 5 \sqrt{2} \][/tex]
[tex]\[ x - 1 = -5 \sqrt{2} \quad \Rightarrow \quad x = 1 - 5 \sqrt{2} \][/tex]
Thus, the solutions to the equation [tex]\((x-1)^2 = 50\)[/tex] are:
[tex]\[ x = 1 + 5 \sqrt{2} \quad \text{and} \quad x = 1 - 5 \sqrt{2} \][/tex]
From the given choices:
- [tex]\( x = -49 \)[/tex]
- [tex]\( x = 51 \)[/tex]
- [tex]\( x = 1 + 5 \sqrt{2} \)[/tex]
- [tex]\( x = 1 - 5 \sqrt{2} \)[/tex]
The correct values of [tex]\( x \)[/tex] are [tex]\( x = 1 + 5 \sqrt{2} \)[/tex] and [tex]\( x = 1 - 5 \sqrt{2} \)[/tex].
