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

Question

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

Answer 1

$$3+\sqrt8=(\sqrt2+1)^2\implies\sqrt{3+2\sqrt2}=\sqrt2+1$$

Like The Conjugate Roots Theorem for Irrational Roots,

the other root has to be $-\sqrt2+1$

Can you take it home from here?

Answer 2

$$\sqrt{3+\sqrt{8}}=\sqrt{2+1+2\sqrt{2}}=\sqrt{(\sqrt{2})^2+1^2+2\sqrt{2}}=\sqrt{2}+1$$ 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, $$a=1+\sqrt{2}+1-\sqrt{2}=2$$ $$b=(1+\sqrt{2})(1-\sqrt{2})=-1$$ $$15ab=-30$$

Answer 3

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 $15ab = 30$