Raise each side of the equation to the reciprocal of the rational exponent.
x=1 and x=2
Practice makes perfect
To solve equations with a variable expression raised to a rational exponent, we raise each side of the equation to the reciprocal of the rational exponent.
x^(m n)=k ⇔ (x^(m n))^() n m=k^() n m
Remember, if m is even, then (x^(m n))^() n m=|x|. In this case m= 1, so we do not need to worry about this. In the given equation the variable term is already isolated, so we will raise each side of the equation to the power of 4.
We obtained a polynomial equation. To solve it, we will define another variable. If we let z=x^2, we can rewrite the last equation in terms of the z-variable.
x^4-5x^2+4=0 ⇔ z^2-5z+4=0
Note that the above equation in terms of z is a quadratic equation. There are many ways to solve a quadratic equation. We will use the Quadratic Formula to solve it. Let's determine a, b, and c.
z^2-5z+4=0 ⇔ 1z^2+( - 5)z+ 4=0
We see above that a= 1, b= - 5, and c= 4. Let's substitute these values into the Quadratic Formula to solve the equation.
Now we can calculate the first root using the positive sign and the second root using the negative sign.
z=5± 3/2
z=5+ 3/2
z=5- 3/2
z=4
z=1
We found that the solutions for z^2-5z+4=0 are z=4 and z=1. This means that x^2=4 and x^2=1. Let's solve them!
lcx^2=4 & (I) x^2=1 & (II)
(I): (II): sqrt(LHS)=sqrt(RHS)
lx=± 2 x=± 1
Checking the Solutions
Next, we will check the solutions by substituting 2, - 2, 1, and - 1 for x into the original equation. If the substitution produces a true statement, we know that our answer is correct. If it does not, then it is an extraneous solution. Let's first check the 2.
