Question

6) 18x' – 6x² + 3x – 1=0 - -​

Answer 1

6) 18x' – 6x² + 3x – 1=0 - -
6x ²-3=0
3x(2x-1)=0
x=0
x= 1/2

Answer 2

To solve the equation 18x' – 6x² + 3x – 1 = 0, you can follow these steps:

1. Rearrange the terms so that the equation is in standard quadratic form (ax² + bx + c = 0):

   -6x² + 18x' + 3x - 1 = 0

2. Combine like terms:

   -6x² + (18x' + 3x) - 1 = 0

3. Factor out the common factors from the terms:

   -3x(2x² - 6x' - 1) = 0

4. Now, you have a quadratic equation in the form ax² + bx + c = 0:

   2x² - 6x' - 1 = 0

5. To solve this quadratic equation, you can use the quadratic formula:

   x = (-b ± √(b² - 4ac)) / (2a)

   In this case, a = 2, b = -6, and c = -1. Plug these values into the quadratic formula:

   x = (-(-6) ± √((-6)² - 4 * 2 * (-1))) / (2 * 2)

   x = (6 ± √(36 + 8)) / 4

   x = (6 ± √44) / 4

6. Simplify the square root:

   x = (6 ± 2√11) / 4

7. Divide both the numerator and denominator by 2 to simplify further:

   x = (3 ± √11) / 2

So, the solutions to the equation 18x' – 6x² + 3x – 1 = 0 are:

x = (3 + √11) / 2 x = (3 - √11) / 2