To solve the equation \( x^2 - x - 7 = 0 \) by completing the square, follow these steps:
1. **Rewrite the equation:**
\[
x^2 - x - 7 = 0
\]
2. **Move the constant term to the other side:**
\[
x^2 - x = 7
\]
3. **Complete the square:**
- The coefficient of \( x \) is \(-1\).
- Half of \(-1\) is \(-\frac{1}{2}\), and squaring it gives \(\left(-\frac{1}{2}\right)^2 = \frac{1}{4}\).
- Add \(\frac{1}{4}\) to both sides to complete the square.
\[
x^2 - x + \frac{1}{4} = 7 + \frac{1}{4}
\]
4. **Simplify the equation:**
\[
x^2 - x + \frac{1}{4} = \frac{28}{4} + \frac{1}{4}
\]
\[
x^2 - x + \frac{1}{4} = \frac{29}{4}
\]
Now we notice that the left side is a perfect square:
\[
\left(x - \frac{1}{2}\right)^2 = \frac{29}{4}
\]
5. **Take the square root of both sides:**
\[
x - \frac{1}{2} = \pm \sqrt{\frac{29}{4}}
\]
\[
x - \frac{1}{2} = \pm \frac{\sqrt{29}}{2}
\]
6. **Solve for \( x \):**
\[
x = \frac{1}{2} \pm \frac{\sqrt{29}}{2}
\]
Split into two cases:
\[
x = \frac{1 + \sqrt{29}}{2}
\]
\[
x = \frac{1 - \sqrt{29}}{2}
\]
7. **Calculate the approximate values:**
\[
\sqrt{29} \approx 5.39
\]
So:
\[
x = \frac{1 + 5.39}{2} = \frac{6.39}{2} \approx 3.195 \approx 3.19 \text{ (rounded to nearest hundredth)}
\]
\[
x = \frac{1 - 5.39}{2} = \frac{-4.39}{2} \approx -2.195 \approx -2.19 \text{ (rounded to nearest hundredth)}
\]
Therefore, the solutions are approximately \( x \approx 3.19 \) and \( x \approx -2.19 \).
The correct answer is:
\[
3.19, -2.19
\]
1.
Solve the equation by completing the square. Round to the nearest hundredth if necessary.
x2 – x – 7 = 0
–6.75, 7.75
3.19, –2.19
3.05, –2.05
2.6, 2.69
1 answer
1 answer