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Solve by completing the square.\newlinef2+6f−31=0f^2 + 6f - 31 = 0\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.\newlinef2+6f−31=0f^2 + 6f - 31 = 0\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. Rewrite Equation: Rewrite the equation in the form of f2+bf=cf^2 + bf = c. Add 3131 to both sides to isolate the f2f^2 and ff terms. f2+6f−31+31=0+31f^2 + 6f - 31 + 31 = 0 + 31 f2+6f=31f^2 + 6f = 31
  2. Complete the Square: Choose the number to add to both sides to complete the square.\newlineSince (62)2=9(\frac{6}{2})^2 = 9, add 99 to both sides.\newlinef2+6f+9=31+9f^2 + 6f + 9 = 31 + 9\newlinef2+6f+9=40f^2 + 6f + 9 = 40
  3. Factor Trinomial: Factor the left side as a perfect square trinomial.\newlinef2+6f+9=40f^2 + 6f + 9 = 40\newline(f+3)2=40(f + 3)^2 = 40
  4. Take Square Root: Take the square root of both sides.\newline(f+3)2=±40\sqrt{(f + 3)^2} = \pm\sqrt{40}\newlinef+3=±40f + 3 = \pm\sqrt{40}
  5. Isolate Variable: Isolate the variable ff.\newlineSubtract 33 from both sides.\newlinef+3−3=±40−3f + 3 - 3 = \pm\sqrt{40} - 3\newlinef=−3±40f = -3 \pm \sqrt{40}
  6. Simplify Square Root: Simplify the square root and round to the nearest hundredth if necessary. 40\sqrt{40} is approximately 6.326.32 (to two decimal places). So, f=−3±6.32f = -3 \pm 6.32
  7. Find Values: Find the two values of ff.f=−3+6.32f = -3 + 6.32 implies f≈3.32f \approx 3.32.f=−3−6.32f = -3 - 6.32 implies f≈−9.32f \approx -9.32.Values of ff: 3.323.32, −9.32-9.32

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