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Solve by completing the square.\newliner2−2r−5=0r^2 - 2r - 5 = 0\newlineWrite your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth.\newliner=r = _____ or r=r = _____

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Solve a quadratic equation by completing the squareright chevron icon

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Q. Solve by completing the square.\newliner2−2r−5=0r^2 - 2r - 5 = 0\newlineWrite your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth.\newliner=r = _____ or r=r = _____
  1. Rewrite and Add: r2−2r−5=0r^2 - 2r - 5 = 0\newlineRewrite the equation in the form of x2+bx=cx^2 + bx = c.\newlineAdd 55 to both sides.\newliner2−2r−5+5=0+5r^2 - 2r - 5 + 5 = 0 + 5\newliner2−2r=5r^2 - 2r = 5
  2. Complete the Square: r2−2r=5r^2 - 2r = 5\newlineChoose the equation after completing the square.\newlineSince (−2/2)2=1(-2/2)^2 = 1, add 11 on both sides.\newliner2−2r+1=5+1r^2 - 2r + 1 = 5 + 1\newliner2−2r+1=6r^2 - 2r + 1 = 6
  3. Factor and Identify: r2−2r+1=6r^2 - 2r + 1 = 6\newlineIdentify the equation after factoring the left side.\newliner2−2r+1=6r^2 - 2r + 1 = 6\newline(r−1)2=6(r - 1)^2 = 6
  4. Take Square Root: (r−1)2=6(r - 1)^2 = 6\newlineIdentify the equation after taking the square root on both sides.\newlineRound the non-terminating values to the nearest hundredth.\newlineTake the square root of both sides of the equation.\newline(r−1)2=6\sqrt{(r - 1)^2} = \sqrt{6}\newliner−1=±6r - 1 = \pm\sqrt{6}\newliner−1≈±2.45r - 1 \approx \pm 2.45
  5. Isolate Variable: We found:\newliner−1≈±2.45r - 1 \approx \pm 2.45\newlineChoose the equation after isolating the variable rr.\newlineTo isolate rr, add 11 to both sides of the equation.\newliner−1+1≈±2.45+1r - 1 + 1 \approx \pm 2.45 + 1\newliner≈1±2.45r \approx 1 \pm 2.45
  6. Isolate Variable: We found:\newliner−1≈±2.45r - 1 \approx \pm 2.45\newlineChoose the equation after isolating the variable rr.\newlineTo isolate rr, add 11 to both sides of the equation.\newliner−1+1≈±2.45+1r - 1 + 1 \approx \pm 2.45 + 1\newliner≈1±2.45r \approx 1 \pm 2.45We have:\newliner≈1±2.45r \approx 1 \pm 2.45\newlineWhat are the two values of rr?\newliner≈1+2.45r \approx 1 + 2.45 implies r≈3.45r \approx 3.45.\newlinerr00 implies rr11.\newlineValues of rr: rr33, rr44

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