2x^2-8x+2=0
2(x^2 -4x +1) = 0
The roots are
x = (-1/2)(4 +/- sqrt12)
= -3.732 and -0.268
3x^2-5x+1=0
To solve the equation 2x^2-8x+2=0, you can use the quadratic formula or factoring.
1. Quadratic Formula:
The quadratic formula is given by x = (-b ± √(b^2 - 4ac)) / (2a), where the equation is in the form ax^2 + bx + c = 0.
For the given equation 2x^2 - 8x + 2 = 0, you can identify a = 2, b = -8, and c = 2.
Substituting these values into the quadratic formula, we get:
x = (-(-8) ± √((-8)^2 - 4 * 2 * 2)) / (2 * 2)
x = (8 ± √(64 - 16)) / 4
x = (8 ± √48) / 4
x = (8 ± 4√3) / 4
Simplifying further:
x = (2 ± √3)
So, the solutions to the equation 2x^2 - 8x + 2 = 0 are x = 2 + √3 and x = 2 - √3.
2. Factoring:
If the quadratic equation can be factored, it can be solved by setting each factor equal to zero.
For the equation 2x^2 - 8x + 2 = 0, we can factor it by splitting the quadratic term:
2x^2 - 4x - 4x + 2 = 0
2x(x - 2) - 2(x - 2) = 0
(x - 2)(2x - 1) = 0
Now, we can set each factor equal to zero and solve:
x - 2 = 0 or 2x - 1 = 0
Solving the first equation: x - 2 = 0, we get x = 2.
Solving the second equation: 2x - 1 = 0, we get x = 1/2.
Therefore, the solutions to the equation 2x^2 - 8x + 2 = 0 are x = 2 and x = 1/2.
You can also verify these solutions by substituting them back into the original equation to check if they satisfy it.