Solve for x:
a) 4x^2 + 3x - 2 = 0
b) 4x + cuberoot x - 2 = 0. (Hint: rename cuberoot x = z)
c)[(x + 4) / (x - 2)] - [(x - 3) / (x - 4)] = 0.
d) 3x = squareroot(3 + x)
Why were you given this problem? It seems above the level that I thought you were at.
Let , then .
This has no rational factors, so the only way to do this (neglecting Cardano's method, which I'll do if you really want me to) is to use a graphing calculator to estimate it, or numerically calculate it.
I get that
First note that the solution set cannot contain 2 or 4, since that is outside the domain of the original problem.
Now add the fractions by getting a common denominator:
The only way the LHS can be 0 is for the numerator be 0. Thus set
(Note that x is neither 2 or 4, so this is a valid solution.)
Square both sides:
Again this is a quadratic that doesn't factor. So use the quadratic formula. I got:
Now, we need to check each of these in the original equation because squaring both sides tends to add spurious solutions. In this case we may simply note that if we use the "-" sign for x we don't have a solution. (I used the decimal approximation of the two x values to check in the original equation.)
So the solution is: