Question
Solve the equation with square root coefficient
a^3+2\sqrt{11}a^2+11a+\sqrt{11}+1 =0
Solve the equation with square root coefficient
a^3+2\sqrt{11}a^2+11a+\sqrt{11}+1 =0
Let
b = \sqrt{11}
then
b^2 = 11
The equation is transformed to
a^3+2ba^2+b^2a+b+1 = 0
Rearrange the terms to make the equation a quadratic equation in terms of b
ab^2+(2a^2+1)b+a^3+1 = 0
Factorizing the terms
ab^2+(2a^2+1)b+(a+1)(a^2-a+1) = 0
Cross-multiply the coefficients of terms for factorization
\begin{array}{cc} a & a^2-a+1 & a^2-a+1 \\ 1 & a+1 & a^2+a\\ \hline & & 2a^2+1 \end{array}
Then the equation is factored to
(ab+a^2-a+1)(b+a+1) = 0
Then
or
From (2), we get
a = -(b+1) = -(\sqrt{11}+1 )
Now we evaluate equation (3). Substitute b=\sqrt{5} into the equation
a^2+(\sqrt{11}-1 )a+1=0
The discriminant
\Delta = (\sqrt{11}-1)^2-4
=(\sqrt{11}-3 )(\sqrt{11}+1 )
=(\sqrt{11}-\sqrt{9} )(\sqrt{11}+1 )>0
Then
\Delta = 8-2\sqrt{11}
Therefore
The quadratic equation has two distinct roots
a = \dfrac{-\sqrt{11}+1\pm\sqrt{8-2\sqrt{11} } }{2}
In summary,
the equation a^3+2\sqrt{11}a^2+11a+\sqrt{11}+1 =0 has three real roots, which are -\sqrt{11}-1, \dfrac{-\sqrt{11}+1+\sqrt{8-2\sqrt{11} } }{2} , \dfrac{-\sqrt{11}+1-\sqrt{8-2\sqrt{11} } }{2}