What are the solutions to two decimals place where 0 degrees is less or equal to x or less or equal to2 pi. This is the equation 5x^2+4x-1=0

Answer 1

To find the solutions to the equation \(5x^2 + 4x - 1 = 0\) where \(0° \leq x \leq 2\pi\), we can use the quadratic formula:

The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In our equation, \(a = 5\), \(b = 4\), and \(c = -1\).

Substitute these values into the formula:

\[ x = \frac{-4 \pm \sqrt{4^2 - 4(5)(-1)}}{2(5)} \]
\[ x = \frac{-4 \pm \sqrt{16 + 20}}{10} \]
\[ x = \frac{-4 \pm \sqrt{36}}{10} \]
\[ x = \frac{-4 \pm 6}{10} \]

This gives two possible solutions:

\( x_1 = \frac{-4 + 6}{10} = \frac{2}{10} = 0.2 \)

and

\( x_2 = \frac{-4 - 6}{10} = \frac{-10}{10} = -1 \)

However, we need to check if these solutions fall within the given interval \(0 \leq x \leq 2\pi\).

Since \(x_1 = 0.2\) falls within the interval, it is a valid solution.

However, \(x_2 = -1\) is not a valid solution as it is outside the given interval.

Therefore, the solution to the equation \(5x^2 + 4x - 1 = 0\) within the interval \(0° \leq x \leq 2\pi\) is \(x = 0.2\).