Question

Solve the following quadratic equation for all values of xx in simplest form.
25, left parenthesis, 1, minus, 2, x, right parenthesis, squared, minus, 48, equals, 1
25(1−2x)
2
−48=
1

Answer 1

To solve the equation \( 25(1 - 2x)^2 - 48 = 1 \), we can start by isolating the quadratic expression:

1. Add 48 to both sides:

\[
25(1 - 2x)^2 = 49
\]

2. Divide both sides by 25:

\[
(1 - 2x)^2 = \frac{49}{25}
\]

3. Take the square root of both sides. Remember to account for both the positive and negative square roots:

\[
1 - 2x = \pm \frac{7}{5}
\]

Now we will break this into two separate cases.

**Case 1**: \( 1 - 2x = \frac{7}{5} \)

Subtract 1 from both sides:

\[
-2x = \frac{7}{5} - 1
\]

Convert 1 to a fraction:

\[
-2x = \frac{7}{5} - \frac{5}{5} = \frac{2}{5}
\]

Now divide by -2:

\[
x = -\frac{2}{5} \div -2 = -\frac{2}{5} \cdot -\frac{1}{2} = \frac{2}{10} = \frac{1}{5}
\]

**Case 2**: \( 1 - 2x = -\frac{7}{5} \)

Subtract 1 from both sides:

\[
-2x = -\frac{7}{5} - 1
\]

Convert 1 to a fraction:

\[
-2x = -\frac{7}{5} - \frac{5}{5} = -\frac{12}{5}
\]

Now divide by -2:

\[
x = -\frac{12}{5} \div -2 = -\frac{12}{5} \cdot -\frac{1}{2} = \frac{12}{10} = \frac{6}{5}
\]

Thus, the solutions to the original equation are:

\[
x = \frac{1}{5} \quad \text{and} \quad x = \frac{6}{5}
\]