Question

X^2+18=9x solve by completing the square

Answer 1

To solve the equation \( x^2 + 18 = 9x \) by completing the square, we'll follow these steps:

1. Rearrange the equation to set it equal to zero:
\[
x^2 - 9x + 18 = 0
\]

2. Move the constant term (18) to the other side:
\[
x^2 - 9x = -18
\]

3. To complete the square, take half of the coefficient of \( x \) (which is -9), square it, and add it to both sides. Half of -9 is \(-\frac{9}{2}\), and squaring it gives:
\[
\left(-\frac{9}{2}\right)^2 = \frac{81}{4}
\]

4. Add \( \frac{81}{4} \) to both sides:
\[
x^2 - 9x + \frac{81}{4} = -18 + \frac{81}{4}
\]

5. To combine the right side, convert -18 to a fraction with a denominator of 4:
\[
-18 = -\frac{72}{4}
\]
Therefore,
\[
-18 + \frac{81}{4} = -\frac{72}{4} + \frac{81}{4} = \frac{9}{4}
\]

6. Now we have:
\[
x^2 - 9x + \frac{81}{4} = \frac{9}{4}
\]

7. The left side can now be factored as a square:
\[
\left(x - \frac{9}{2}\right)^2 = \frac{9}{4}
\]

8. Take the square root of both sides:
\[
x - \frac{9}{2} = \pm \frac{3}{2}
\]

9. Solve for \( x \) by adding \( \frac{9}{2} \) to both sides:
- For the positive root:
\[
x - \frac{9}{2} = \frac{3}{2} \implies x = \frac{9}{2} + \frac{3}{2} = \frac{12}{2} = 6
\]
- For the negative root:
\[
x - \frac{9}{2} = -\frac{3}{2} \implies x = \frac{9}{2} - \frac{3}{2} = \frac{6}{2} = 3
\]

So the solutions are:
\[
x = 6 \quad \text{and} \quad x = 3
\]