(i) Solve the equation 3y^2-5y-2=0
(ii) hence, solve the equation, 3(1/x-1)^2 -5(1/x-1)-2=0
I will treat this as one question.
From the second part, let y = (1/x - 1).
The second equation thus becomes
3y^2 - 5y - 2 = 0 , which is the first one
(3y + 1)(y - 2) = 0
y = -1/3 or y = 2 , which answers the first part
then 1/x - 1 = -1/3 or 1/x - 1 = 2
1/x = 2/3 or 1/x = 3
x = 3/2 or x = 1/3
z + (-3 and 2 5ths)= -4 and 1 10ths please solve
To solve the equation 3y^2 - 5y - 2 = 0, we can use the quadratic formula. The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac))/(2a)
In our equation, a = 3, b = -5, and c = -2. Plugging these values into the quadratic formula, we get:
y = (-(-5) ± √((-5)^2 - 4*3*(-2))) / (2*3)
= (5 ± √(25 + 24)) / 6
= (5 ± √49) / 6
This simplifies to:
y = (5 ± 7) / 6
So the solutions for y are:
y = (5 + 7) / 6 = 12 / 6 = 2
y = (5 - 7) / 6 = -2 / 6 = -1/3
Now, let's solve the equation 3(1/x-1)^2 - 5(1/x-1) - 2 = 0. We need to find the values of x that satisfy this equation. To do this, we can use substitution.
Let's substitute a new variable u = 1/x-1. Rewriting the equation using this substitution, we get:
3u^2 - 5u - 2 = 0
Now, we can solve this quadratic equation for u using the quadratic formula:
u = (-(-5) ± √((-5)^2 - 4*3*(-2))) / (2*3)
= (5 ± √(25 + 24)) / 6
= (5 ± √49) / 6
This simplifies to:
u = (5 + 7) / 6 = 2
u = (5 - 7) / 6 = -1/3
Now, we need to solve for x. We can substitute back in the value of u to get:
1/x-1 = 2 or 1/x-1 = -1/3
Solving these equations for x, we get:
1/x = 3 or 1/x = 2/3
Multiplying both sides by x, we get:
x = 1/3 or x = 3
So the solutions for the equation 3(1/x-1)^2 - 5(1/x-1) - 2 = 0 are x = 1/3 and x = 3.