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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

1 answer

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. \(x = 13, -5\)