`4x+root(4)(x)(x-2)=-(2-x)^2` Solve. Please explain how you arrive at the answer.Thanks
Solve `4x+root(4)(x)(x-2)=-(2-x)^2` :
There is no solution.
The left hand side is not defined in the reals for x<0 while the right hand side is continuous for all x, so we need not look at x<0.
The right hand side is nonpositive for all x, while the left hand side is positive for all x>.5 (by inspection) so we need not consider any x>.5
I. Looking at a graph we find no intersection:
(The parabola is in red)
II. We can use calculus to find the minimum of the left hand side on the interval (0,.5):
If `f(x)=4x+x^(1/4)(x-2)` then `f'(x)=4+1/4x^(-3/4)(x-2)+x^(1/4)` Setting `f'(x)=0` we get:
`4+1/4x^(-3/4)(x-2)+x^(1/4)=0` Using technology the zero is at `x~~.05192395` with a y-value of `y~~-.722229` The values at the endpoints are nonnegative so this is the minimum.
For the right hand side the derivative is 4-2x so the critical point is at 2, which is not in the interval. Then g(0)=-4 and g(.5)=-2.25. The maximum value for the parabola is 2.25 on the interval.
Since the minimum value for the left side is greater than the maximal value of the right side, the graphs will not intersect.
A bigger picture of the graphs: