Questions: n(v)=-x^2+5x+1

Transcript text: (c) $n(v)=-x^{2}+5 x+1$

Solution

Solution Steps

To find the roots of the quadratic function $ n(v) = -x^2 + 5x + 1 $, we can use the quadratic formula, which is given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

where $ a $, $ b $, and $ c $ are the coefficients of the quadratic equation $ ax^2 + bx + c = 0 $. In this case, $ a = -1 $, $ b = 5 $, and $ c = 1 $.

Step 1: Identify the Coefficients

For the quadratic function $ n(v) = -x^2 + 5x + 1 $, we identify the coefficients as follows:

$ a = -1 $
$ b = 5 $
$ c = 1 $

Step 2: Calculate the Discriminant

The discriminant $ D $ is calculated using the formula: \[ D = b^2 - 4ac \] Substituting the values: \[ D = 5^2 - 4(-1)(1) = 25 + 4 = 29 \] Taking the square root gives: \[ \sqrt{D} = \sqrt{29} \approx 5.3852 \]

Step 3: Find the Roots

Using the quadratic formula: \[ x = \frac{-b \pm \sqrt{D}}{2a} \] we find the two roots: \[ x_1 = \frac{-5 + \sqrt{29}}{2(-1)} = \frac{-5 + 5.3852}{-2} \approx -0.1926 \] \[ x_2 = \frac{-5 - \sqrt{29}}{2(-1)} = \frac{-5 - 5.3852}{-2} \approx 5.1926 \]

Final Answer

The roots of the quadratic function are: \[ \boxed{x_1 \approx -0.1926} \] \[ \boxed{x_2 \approx 5.1926} \]

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