Question

solve each equation by completing the square
y^2-4y-5=0
2x^2-3x+1=0
x^2+4x+1=0
x^2-8x-65=0

Answer 1

Let's solve each equation by completing the square.

### 1. \(y^2 - 4y - 5 = 0\)

1. Rearrange the equation:
\(y^2 - 4y = 5\)

2. Complete the square:
- Take half of the coefficient of \(y\) (which is \(-4\)), square it:
\(\left(-\frac{4}{2}\right)^2 = 4\)
- Add and subtract this square on the left side:
\(y^2 - 4y + 4 - 4 = 5\) ⇒ \(y^2 - 4y + 4 = 9\)

3. Rewrite it as a perfect square:
\((y - 2)^2 = 9\)

4. Solve for \(y\):
\(y - 2 = \pm 3\)

Thus,
\(y = 5\) or \(y = -1\).

### 2. \(2x^2 - 3x + 1 = 0\)

1. Rearrange the equation:
\(2x^2 - 3x = -1\)

2. Divide the whole equation by 2:
\(x^2 - \frac{3}{2}x = -\frac{1}{2}\)

3. Complete the square:
- Take half of the coefficient of \(x\) (which is \(-\frac{3}{2}\)), square it:
\(\left(-\frac{3}{4}\right)^2 = \frac{9}{16}\)
- Add and subtract this square on the left side:
\(x^2 - \frac{3}{2}x + \frac{9}{16} - \frac{9}{16} = -\frac{1}{2}\)
⇒
\(x^2 - \frac{3}{2}x + \frac{9}{16} = -\frac{1}{2} + \frac{9}{16}\)

4. Simplify the right side:
Convert \(-\frac{1}{2}\) to sixteenths:
\(-\frac{1}{2} = -\frac{8}{16}\)
\(-\frac{8}{16} + \frac{9}{16} = \frac{1}{16}\)

5. Rewrite as a perfect square:
\(\left(x - \frac{3}{4}\right)^2 = \frac{1}{16}\)

6. Solve for \(x\):
\(x - \frac{3}{4} = \pm \frac{1}{4}\)
Thus, \(x = 1\) or \(x = \frac{1}{2}\).

### 3. \(x^2 + 4x + 1 = 0\)

1. Rearrange the equation:
\(x^2 + 4x = -1\)

2. Complete the square:
- Take half of the coefficient of \(x\) (which is \(4\)), square it:
\(\left(\frac{4}{2}\right)^2 = 4\)
- Add and subtract this square on the left side:
\(x^2 + 4x + 4 - 4 = -1\)
⇒
\(x^2 + 4x + 4 = 3\)

3. Rewrite as a perfect square:
\((x + 2)^2 = 3\)

4. Solve for \(x\):
\(x + 2 = \pm \sqrt{3}\)
Thus, \(x = -2 + \sqrt{3}\) or \(x = -2 - \sqrt{3}\).

### 4. \(x^2 - 8x - 65 = 0\)

1. Rearrange the equation:
\(x^2 - 8x = 65\)

2. Complete the square:
- Take half of the coefficient of \(x\) (which is \(-8\)), square it:
\(\left(-\frac{8}{2}\right)^2 = 16\)
- Add and subtract this square on the left side:
\(x^2 - 8x + 16 - 16 = 65\)
⇒
\(x^2 - 8x + 16 = 81\)

3. Rewrite as a perfect square:
\((x - 4)^2 = 81\)

4. Solve for \(x\):
\(x - 4 = \pm 9\)
Thus, \(x = 13\) or \(x = -5\).

### Summary of Solutions:
1. \(y = 5, -1\)
2. \(x = 1, \frac{1}{2}\)
3. \(x = -2 + \sqrt{3}, -2 - \sqrt{3}\)
4. \(x = 13, -5\)