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Solve using the quadratic formula.\newline4f2−6f+1=04f^2 - 6f + 1 = 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 using the quadratic formula.\newline4f2−6f+1=04f^2 - 6f + 1 = 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. Identify values: Identify the values of aa, bb, and cc in the quadratic equation 4f2−6f+1=04f^2 − 6f + 1 = 0. The quadratic equation is in the form af2+bf+c=0af^2 + bf + c = 0, so by comparison: a=4a = 4 b=−6b = -6 c=1c = 1
  2. Substitute values: Substitute the values of aa, bb, and cc into the quadratic formula.\newlineThe quadratic formula is f=−b±b2−4ac2af = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\newlineSubstituting the values we get:\newlinef=−(−6)±(−6)2−4⋅4⋅12⋅4f = \frac{-(-6) \pm \sqrt{(-6)^2 - 4\cdot4\cdot1}}{2\cdot4}
  3. Simplify equation: Simplify the equation.\newlinef=6±36−168f = \frac{6 \pm \sqrt{36 - 16}}{8}\newlinef=6±208f = \frac{6 \pm \sqrt{20}}{8}\newlineSince 20\sqrt{20} simplifies to 252\sqrt{5}, we can write:\newlinef=6±258f = \frac{6 \pm 2\sqrt{5}}{8}
  4. Simplify fractions: Simplify the fractions.\newlineWe can divide the numerator and the denominator by 22 to simplify the expression:\newlinef=3±54f = \frac{3 \pm \sqrt{5}}{4}
  5. Calculate values: Calculate the two possible values for ff.f=3+54f = \frac{3 + \sqrt{5}}{4} or f=3−54f = \frac{3 - \sqrt{5}}{4}Now we can approximate 5\sqrt{5} to the nearest hundredth:5≈2.24\sqrt{5} \approx 2.24f≈3+2.244f \approx \frac{3 + 2.24}{4} or f≈3−2.244f \approx \frac{3 - 2.24}{4}f≈5.244f \approx \frac{5.24}{4} or f≈0.764f \approx \frac{0.76}{4}f≈1.31f \approx 1.31 or f=3+54f = \frac{3 + \sqrt{5}}{4}00

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