Irrational root theorem: Let x and y be rational numbers and let √y be an irrational number. If x + √y is a root of the polynomial equation with rational coefficients, then x - √y is also a root.
Notice that in order for the theorem to work, √y must be an irrational number. This will ensure that x + √y is also irrational. Otherwise, the theorem cannot be used. Say for instance y is 4. √4 = 2. Since x is a rational number, then x + 2 is also a rational number. This does not work since the theorem is dealing with irrational numbers.
For example, show that 2 + √3 and 2 - √3 are roots of the polynomial equation x^{2} - 4x + 1.
First, show that 2 + √3 is a root
x^{2} - 4x + 1 = (2 + √3)^{2} - 4(2 + √3) + 1
x^{2} - 4x + 1 = 4 + 4√3 + 3 - 8 - 4√3 + 1
x^{2} - 4x + 1 = 4 + 3 + 1 - 8 + 4√3 - 4√3
x^{2} - 4x + 1 = 0
Then, show that 2 - √3 is also a root
x^{2} - 4x + 1 = (2 - √3)^{2} - 4(2 - √3) + 1
x^{2} - 4x + 1 = (2 + -√3)^{2} - 4(2 + -√3) + 1
x^{2} - 4x + 1 = 4 - 4√3 + 3 - 8 + 4√3 + 1
x^{2} - 4x + 1 = 4 + 3 + 1 - 8 - 4√3 + 4√3
x^{2} - 4x + 1 = 0
Therefore, as you can see, both 2 + √3 and 2 - √3 are roots of the polynomial equation x^{2} - 4x + 1.
Example #1
A polynomial equation with integer coefficients has the roots 8 + √7 and -√2. Find two additional roots.
By the irrational root theorem, if 8 + √7 is a root, then 8 - √7 is also a root.
By the irrational root theorem, if -√2 is a root, then √2 is also a root.
Two additional roots are 8 - √7 and √2
Example #2
A polynomial equation with integer coefficients has the roots -√16 and -√11. Find more roots using the irrational root theorem when applicable.
We cannot use the irrational root theorem for -√16 since -√16 = -4 and -4 is not an irrational number.
By the irrational root theorem, if -√11 is a root, then √11 is also a root.
An additional root is √11.
Jan 12, 22 07:48 AM
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