To solve the quadratic equation [tex]\(10x = 5x^2 - 1\)[/tex] by completing the square, follow these steps:
1. Rewrite the equation in standard quadratic form [tex]\(ax^2 + bx + c = 0\)[/tex]:
[tex]\[
5x^2 - 10x - 1 = 0
\][/tex]
2. Divide the entire equation by the coefficient of [tex]\(x^2\)[/tex] (which is 5), to make the coefficient of [tex]\(x^2\)[/tex] equal to 1:
[tex]\[
x^2 - 2x - \frac{1}{5} = 0
\][/tex]
3. Isolate the constant term on the right side of the equation:
[tex]\[
x^2 - 2x = \frac{1}{5}
\][/tex]
4. Complete the square on the left side of the equation:
- Take the coefficient of [tex]\(x\)[/tex] (which is -2), divide it by 2, and square it:
[tex]\[
\left(\frac{-2}{2}\right)^2 = 1
\][/tex]
- Add and subtract this square inside the equation:
[tex]\[
x^2 - 2x + 1 - 1 = \frac{1}{5}
\][/tex]
[tex]\[
(x - 1)^2 - 1 = \frac{1}{5}
\][/tex]
- Rearrange it to balance the equation:
[tex]\[
(x - 1)^2 = \frac{1}{5} + 1
\][/tex]
[tex]\[
(x - 1)^2 = \frac{1}{5} + \frac{5}{5}
\][/tex]
[tex]\[
(x - 1)^2 = \frac{6}{5}
\][/tex]
5. Solve for [tex]\(x - 1\)[/tex] by taking the square root of both sides:
[tex]\[
x - 1 = \pm \sqrt{\frac{6}{5}}
\][/tex]
- This can be simplified to:
[tex]\[
x - 1 = \pm \frac{\sqrt{6}}{\sqrt{5}}
\][/tex]
[tex]\[
x - 1 = \pm \frac{\sqrt{30}}{5}
\][/tex]
- Thus, [tex]\(x\)[/tex] is:
[tex]\[
x = 1 \pm \frac{\sqrt{30}}{5}
\][/tex]
6. Write the solutions in the standard form:
[tex]\[
x = \frac{5 \pm \sqrt{30}}{5}
\][/tex]
So, the correct answer is:
B. [tex]\( x = \frac{5 \pm \sqrt{30}}{5} \)[/tex]
