Answer:
[tex]3-\sqrt{11}[/tex]
Step-by-step explanation:
If you want rational coefficients then you would want the conjugate of any irrational zero given.
The question is equivalent to what is the conjugate of [tex]3+\sqrt{11}[/tex].
The conjugate of [tex]3+\sqrt{11}[/tex] is [tex]3-\sqrt{11}[/tex].
In general, the conjugate of a+b is a-b
or the conjugate of a-b is a+b.
Or maybe you like this explanation more:
Let [tex]x=3+\sqrt{11}[/tex]
Subtract 3 on both sides:
[tex]x-3=\sqrt{11}[/tex]
Square both sides:
[tex](x-3)^2=11[/tex]
Subtract 11 on both sides:
[tex](x-3)^2-11=0[/tex]
Use difference of squares to factor. I apply [tex]u^2-v^2=(u-v)(u+v)[/tex].
[tex]([x-3]-\sqrt{11})([x-3]+\sqrt{11})=0[/tex]
So you have either
[tex][x-3]-\sqrt{11}=0[/tex] or [tex][x-3}+\sqrt{11}=0[/tex]
Solve both for x-3 and then x.
Add sqrt(11) on both sides for first equation and subtract sqrt(11) on both sides for second equation:
[tex]x-3=\sqrt{11}[/tex] or [tex]x-3=-\sqrt{11}[/tex]
Add 3 on both sides:
[tex]x=3+\sqrt{11}[/tex] or [tex]x=3-\sqrt{11}[/tex]
