What's the conjugate root?

$\sqrt{3+\sqrt8}=1+\sqrt2$ .

As $a$ and $b$ are rational and one root is irrational, the other root must also be irrational and it must also be the conjugate of that root.

Thus the second root is $1-\sqrt2$ .

Now we get the required quadratic by multiplying $(x-(1+\sqrt2))(x-(1-\sqrt2))$ , but an easier method is by using Vieta's Formula.

By Vieta's Formula, sum of the roots is $\dfrac {-a} 1=1+\sqrt2+1-\sqrt2$ or $a=-2$ .

and product of the roots is $\dfrac {b} 1=(1+\sqrt2)(1-\sqrt2)=-1$ or $b=-1$ .

Thus $15ab=15 \times -2 \times -1=\boxed{30}$ .

Actually, by finding that (3 + 8^0.5)^0.5 = 1+ 2^0.5, allows you to subtitute that as the root of the quadratic equation, you know that the coefficient of x squared is 1, from there you subtitute 1 in A (in the original quadratic equation), from there, you know that -B(in the original quadratic equation) divided by 2 has to be 1(to be equal to 1 from 1 + 2^0.5)so you solve the linear equation -B÷2 = 1, where you find that B is equal to -2. you subtitute that in (B^2 - 4AC)^0.5 (from the original quadratic equation) where you are left with ((-2)^2 - 4 1 C)÷2 = 2^0.5, after you solved the linear equation you are left with C = -1, where in your question A is the coefficient of x^2, B is a and C is b. After that you do (-1) (-2) 15 = 30. sorry for my english, and if somebody doesn't get the solution, write it in the comments. *EDIT Some of the letters in my comment are messed up(- before the B in -B÷2 = 1. and it is 4 times 1 times C instead of 41C.)