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Given [tex](x-7)^2 = 36[/tex], select the values of [tex]x[/tex].

A. [tex]x = 13[/tex]

B. [tex]x = 1[/tex]

C. [tex]x = -29[/tex]

D. [tex]x = 42[/tex]


Sagot :

Let's solve the equation [tex]\((x-7)^2 = 36\)[/tex].

Step 1: Start with the given equation:
[tex]\[ (x - 7)^2 = 36 \][/tex]

Step 2: To eliminate the square, take the square root of both sides. Remember, taking the square root gives us both positive and negative solutions:
[tex]\[ x - 7 = \pm \sqrt{36} \][/tex]

Step 3: Since [tex]\(\sqrt{36} = 6\)[/tex], we have:
[tex]\[ x - 7 = \pm 6 \][/tex]

This gives us two equations to solve:
1. [tex]\( x - 7 = 6 \)[/tex]
2. [tex]\( x - 7 = -6 \)[/tex]

Step 4: Solve the first equation:
[tex]\[ x - 7 = 6 \][/tex]
Add 7 to both sides:
[tex]\[ x = 6 + 7 \][/tex]
[tex]\[ x = 13 \][/tex]

Step 5: Solve the second equation:
[tex]\[ x - 7 = -6 \][/tex]
Add 7 to both sides:
[tex]\[ x = -6 + 7 \][/tex]
[tex]\[ x = 1 \][/tex]

So, the solutions to the equation [tex]\((x-7)^2 = 36\)[/tex] are [tex]\(x = 13\)[/tex] and [tex]\(x = 1\)[/tex].

Step 6: Now, we need to check which of the given values [tex]\(x = 13\)[/tex], [tex]\(x = 1\)[/tex], [tex]\(x = -29\)[/tex], and [tex]\(x = 42\)[/tex] are solutions to the equation.

From our solutions, [tex]\( x = 13 \)[/tex] and [tex]\( x = 1 \)[/tex] are the solutions. However, [tex]\( x = -29 \)[/tex] and [tex]\( x = 42 \)[/tex] are not solutions because they do not satisfy the equation.

Therefore, the values of [tex]\( x \)[/tex] that satisfy the equation [tex]\((x-7)^2 = 36\)[/tex] are:
[tex]\[ x = 13 \][/tex]
[tex]\[ x = 1 \][/tex]