solve
6x^2 - 5x +1 = 0
is this correct?
6x^2 - 2x - 3x + 1=0
-2x(-3x+1) 1(-3x +1)
(2x-1)=0 (3x-1)=0
x=1/2 x=1/3
The final answer is correct but there is an intermediate step that is wrong. You should have written
(2x-1)(3x-1) = 0
as the factoring step.
You did break up the quadratic correctly into
-2x(3x-1) + (-3x+ 1)
which can be written
(-3x+1)(-2x+1) = (3x-1)(2x-1)
To solve the equation 6x^2 - 5x + 1 = 0, the correct steps are as follows:
1. First, check if the equation is factorable. In this case, it is not factorable, so we need to use the quadratic formula or complete the square method.
2. Let's use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a), where a, b, and c are the coefficients of the equation.
In this case, a = 6, b = -5, and c = 1. Plugging these values into the quadratic formula, we get:
x = (-(-5) ± √((-5)^2 - 4 * 6 * 1)) / (2 * 6)
x = (5 ± √(25 - 24)) / 12
x = (5 ± √1) / 12
x = (5 ± 1) / 12
3. Simplifying further, we get two possible values for x:
x1 = (5 + 1) / 12 = 6 / 12 = 1/2
x2 = (5 - 1) / 12 = 4 / 12 = 1/3
So, the solutions to the equation 6x^2 - 5x + 1 = 0 are x = 1/2 and x = 1/3.