Question

2x^2-10x+5=0 by completing the square

Answer 1

To solve the quadratic equation \(2x^2 - 10x + 5 = 0\) by completing the square, follow these steps:

1. **Divide the equation by the coefficient of \(x^2\)**: In this case, we will divide the entire equation by 2 to simplify it.

\[
x^2 - 5x + \frac{5}{2} = 0
\]

2. **Move the constant term to the right side**:

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

3. **Complete the square**: To complete the square, take half of the coefficient of \(x\) (which is \(-5\)), square it, and add it to both sides. Half of \(-5\) is \(-\frac{5}{2}\), and squaring it gives:

\[
\left(-\frac{5}{2}\right)^2 = \frac{25}{4}
\]

Add \(\frac{25}{4}\) to both sides:

\[
x^2 - 5x + \frac{25}{4} = -\frac{5}{2} + \frac{25}{4}
\]

4. **Simplify the right side**: To combine the terms on the right side, convert \(-\frac{5}{2}\) to a fraction with a denominator of 4:

\[
-\frac{5}{2} = -\frac{10}{4}
\]

Therefore, the right side becomes:

\[
-\frac{10}{4} + \frac{25}{4} = \frac{15}{4}
\]

Now we have:

\[
x^2 - 5x + \frac{25}{4} = \frac{15}{4}
\]

5. **Rewrite the left side as a square**:

\[
\left(x - \frac{5}{2}\right)^2 = \frac{15}{4}
\]

6. **Take the square root of both sides**:

\[
x - \frac{5}{2} = \pm \sqrt{\frac{15}{4}}
\]

\[
x - \frac{5}{2} = \pm \frac{\sqrt{15}}{2}
\]

7. **Solve for \(x\)**:

\[
x = \frac{5}{2} \pm \frac{\sqrt{15}}{2}
\]

This gives us two solutions:

\[
x = \frac{5 + \sqrt{15}}{2} \quad \text{and} \quad x = \frac{5 - \sqrt{15}}{2}
\]

So the solutions to the equation \(2x^2 - 10x + 5 = 0\) are:

\[
x = \frac{5 + \sqrt{15}}{2}, \quad x = \frac{5 - \sqrt{15}}{2}
\]