# If one root of the equation $x^2+ax+b=0$ is $\sqrt{3+\sqrt{8}}\;,$ If $a,b$ are rational number, Then $15ab$ is

If one root of the equation $$x^2+ax+b=0$$ is $$\sqrt{3+\sqrt{8}}\;,$$ If $$a,b$$ are rational number,

Then $$15ab$$ is

$$\bf{My\; Try::}$$ Given $$x=\sqrt{3+\sqrt{8}}$$ is a root of $$x^2+ax+b=0$$

So we get $$3+\sqrt{8}+a\cdot \left(\sqrt{3+\sqrt{8}}\right)+b=0$$

Now How can i solve after that, Thanks

$$3+\sqrt8=(\sqrt2+1)^2\implies\sqrt{3+2\sqrt2}=\sqrt2+1$$
the other root has to be $-\sqrt2+1$
$$\sqrt{3+\sqrt{8}}=\sqrt{2+1+2\sqrt{2}}=\sqrt{(\sqrt{2})^2+1^2+2\sqrt{2}}=\sqrt{2}+1$$ Since $a,b$ are rational, the other root must be $1-\sqrt{2}$. This can be understood by the fact that the coefficients of $x^2,x$, and the constant terms must be rational in the quadratic formula.
Thus, $$a=1+\sqrt{2}+1-\sqrt{2}=2$$ $$b=(1+\sqrt{2})(1-\sqrt{2})=-1$$ $$15ab=-30$$
Write $\sqrt{3+\sqrt8}$ as $\sqrt{2}+1$. Since $a$ and $b$ are rational, the other root must be $-\sqrt{2}+1$. Hence $a =-\left( \sqrt{2}+1 - \sqrt{2}+1 \right) = -2$ and similarly $b = -1$. Hence $15ab = 30$