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Original Question
Solve the equation for all values of x by completing the square. Express your answer in simplest form. x, squared, minus, 8, x,...Question
Solve the equation for all values of x by completing the square. Express your answer in simplest form.
x, squared, minus, 2, x, equals, 48
x
2
−2x=
48
x, squared, minus, 2, x, equals, 48
x
2
−2x=
48
Answers
Answered by
GPT-4o mini
To solve the equation \( x^2 - 2x = 48 \) by completing the square, follow these steps:
1. **Move the constant to the other side**:
\[
x^2 - 2x - 48 = 0
\]
We can actually keep it as \( x^2 - 2x = 48 \) for completing the square.
2. **Complete the square**:
Take the coefficient of \( x \) (which is -2), halve it, and square it:
\[
\left(-\frac{2}{2}\right)^2 = (-1)^2 = 1
\]
3. **Add and subtract this square inside the equation**:
\[
x^2 - 2x + 1 - 1 = 48
\]
This can be rewritten as:
\[
(x - 1)^2 - 1 = 48
\]
4. **Move the constant back to the right side**:
\[
(x - 1)^2 = 48 + 1
\]
\[
(x - 1)^2 = 49
\]
5. **Take the square root of both sides**:
\[
x - 1 = \pm 7
\]
6. **Solve for \( x \)**:
- If \( x - 1 = 7 \):
\[
x = 7 + 1 = 8
\]
- If \( x - 1 = -7 \):
\[
x = -7 + 1 = -6
\]
Thus, the solutions for the equation \( x^2 - 2x = 48 \) are:
\[
\boxed{8} \text{ and } \boxed{-6}
\]
1. **Move the constant to the other side**:
\[
x^2 - 2x - 48 = 0
\]
We can actually keep it as \( x^2 - 2x = 48 \) for completing the square.
2. **Complete the square**:
Take the coefficient of \( x \) (which is -2), halve it, and square it:
\[
\left(-\frac{2}{2}\right)^2 = (-1)^2 = 1
\]
3. **Add and subtract this square inside the equation**:
\[
x^2 - 2x + 1 - 1 = 48
\]
This can be rewritten as:
\[
(x - 1)^2 - 1 = 48
\]
4. **Move the constant back to the right side**:
\[
(x - 1)^2 = 48 + 1
\]
\[
(x - 1)^2 = 49
\]
5. **Take the square root of both sides**:
\[
x - 1 = \pm 7
\]
6. **Solve for \( x \)**:
- If \( x - 1 = 7 \):
\[
x = 7 + 1 = 8
\]
- If \( x - 1 = -7 \):
\[
x = -7 + 1 = -6
\]
Thus, the solutions for the equation \( x^2 - 2x = 48 \) are:
\[
\boxed{8} \text{ and } \boxed{-6}
\]
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