Question

A polynomial equation with rational coefficients has the roots 3+ square root 6, 2- square root 5. Find the two additional roots

Answer 1

the other 2 roots will be the conjugates of the above.
In other words they are 
3 - sqrt6 and 2 + sqrt5

Answer 2

Answer:

$(3-\sqrt{6}) and (2+\sqrt{5})$

Step-by-step explanation:

If a polynomial equation with rational coefficients is $ax^2+bx+c=0$, then by quadratic formula

$x=\dfrac{b\pm\sqrt{b^2-4ac}}{2a}$

where, a,b and c are rational number.

If $\sqrt{b^2-4ac}$ is an irrational number, then we have ± sign before the irrational number.

It is given that $(3+\sqrt{6}) and (2-\sqrt{5})$ are two roots of a polynomial equation with rational coefficients. Here, $\sqrt{6} and \sqrt{5}$ are irrational numbers.

So, the root of the polynomial equation are

$(3\pm \sqrt{6}) and (2\pm \sqrt{5})$

Therefore, the remaining roots of the polynomial equation are $(3-\sqrt{6}) and (2+\sqrt{5})$.