anonymous
  • anonymous
Solve by completing the square: x2-8x+7=0 That is an x squared. Thanks.
HippoCampus Algebra & Geometry
katieb
  • katieb
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anonymous
  • anonymous
Let's use \(x^2 - 4x - 8 = 0\) to demonstrate Completing the Square. 1. Put the constant term on the other side of the equation by adding 8 to both sides. \(x^2 - 4x = 8\) 2. Look at the coefficient of the x-term (-4) -> take half this term (-2) and square it (+4). Add this to both sides. \(x^2 - 4x + 4 = 12\) This is a perfect square and can be factored. 3. Factor: \((x-2)^2\) (hint: The constant term is the same as half of the x term that was squared in step 2.) 4. Solve for x: \(\sqrt{(x-2)^2} = \sqrt{12}\) \(x - 2 = \pm 2 \sqrt{3}\) \(x = 2 \pm 2 \sqrt{3}\) Please ask if anything is unclear.

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