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Solve by completing the square.\newlinef214f=31f^2 - 14f = 31\newlineWrite your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth.\newlinef=f = _____ or f=f = _____

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Q. Solve by completing the square.\newlinef214f=31f^2 - 14f = 31\newlineWrite your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth.\newlinef=f = _____ or f=f = _____
  1. Write Equation Form: Write the equation in the form of f2+bf=cf^2 + bf = c. The given equation is already in this form: f214f=31f^2 - 14f = 31.
  2. Move Constant Term: Move the constant term to the right side of the equation.\newlineAdd 3131 to both sides to isolate the f2f^2 and ff terms on one side.\newlinef214f+3131=31+31f^2 - 14f + 31 - 31 = 31 + 31\newlinef214f=31f^2 - 14f = 31
  3. Find Completion Number: Find the number to complete the square.\newlineTake half of the coefficient of ff, square it, and add it to both sides of the equation.\newline(14/2)2=49(-14/2)^2 = 49\newlinef214f+49=31+49f^2 − 14f + 49 = 31 + 49\newlinef214f+49=80f^2 − 14f + 49 = 80
  4. Factor Left Side: Factor the left side of the equation.\newlineThe left side is now a perfect square trinomial.\newline(f7)2=80(f - 7)^2 = 80
  5. Take Square Root: Take the square root of both sides of the equation.\newlineRemember to consider both the positive and negative square roots.\newline(f7)2=±80\sqrt{(f - 7)^2} = \pm\sqrt{80}\newlinef7=±80f - 7 = \pm\sqrt{80}
  6. Solve for f: Solve for f by isolating the variable.\newlineAdd 77 to both sides of the equation to get ff by itself.\newlinef7+7=±80+7f - 7 + 7 = \pm\sqrt{80} + 7\newlinef=7±80f = 7 \pm\sqrt{80}
  7. Simplify Square Root: Simplify the square root and round to the nearest hundredth if necessary. 80\sqrt{80} is approximately 8.948.94 when rounded to the nearest hundredth. f=7±8.94f = 7 \pm 8.94
  8. Find Values of f: Find the two values of f.\newlinef=7+8.94f = 7 + 8.94 implies f15.94f \approx 15.94.\newlinef=78.94f = 7 - 8.94 implies f1.94f \approx -1.94.\newlineValues of f: 15.9415.94, 1.94-1.94

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