Irrational equation is equation, that contains variable under radical or variable is a base of power with fractional exponent. For example, `sqrt(x-2)=2x-1` and `x^(1/4)-5=0` are irrational equations.
There are two methods that are used to solve irrational equation:
- Method of raising both sides of equation to a same power;
- Method of introducing new variable.
Example 1. Solve equation `sqrt(x-1)+sqrt(2x+6)=6`.
Let's rewrite it a bit: `sqrt(2x+6)=6-sqrt(x-1)`.
Now, square both sides: `(sqrt(2x+6))^2=(6-sqrt(x-1))^2` or `2x+6=36-12sqrt(x-1)+x-1`.
From this we have that `12sqrt(x-1)=29-x`. Again square both sides: `144(x-1)=(29-x)^2`, i.e. `x^2-202x+895=0`. This equation has two roots: `x_1=5` and `x_2=197`.
After squaring, there could appear extraneous roots, so we need to check all roots.
- When x=5 we have that `sqrt(5-1)+sqrt(2*5+6)=6`, i.e. x=5 is root of initial equation.
- When x=197 we have that `sqrt(197-1)+sqrt(2*197+6)!=6`, i.e. x=197 is extraneous root.
Therefore, initial eqaution has only one root: x=5.
Example 2. Solve equation `x^2+3-sqrt(2x^2-3x+2)=3/2(x+4)`.
If we begin to square both sides of the equation, then we will face difficult and tedious calculations. However, note that this equation can be transformed into quadratic.
To do this multiply both sides of equation by 2: `2x^2+6-2sqrt(2x^2-3x+2)=3x+12` or `2x^2-3x+2-2sqrt(2x^2-3x+2)-8=0`.
Let `y=sqrt(2x^2-3x+2)` then equation can be rewritten as `y^2-2y-8=0`. This equation has two solutions: `y_1=4,y_2=-2`.
Thus, we obtained set of equations: `sqrt(2x^2-3x+2)=4,sqrt(2x^2-3x+2)=-2`.
First equation has two roots: `x_1=7/2,x_2=-2`; Second equation doesn't have roots, because square root can't be negative.
- If `x=7/2` then `(7/2)^2+3-sqrt(2*(7/2)^2-3*(7/2)+2)=3/2(7/2+4)`, so `x=7/2` is root of the initial equation.
- If `x=-2` then `(-2)^2+3-sqrt(2*(-2)^2-3*(-2)+2)=3/2(-2+4)`, so `x=3` is root of the initial equation.
Therefore, initial equation has two roots: `7/2` and `-2`.