a^3+2\sqrt{5}a^2+5a+\sqrt{5}-1 =0
(1)

Let

b = \sqrt{5}

then

b^2 = 5

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

b+a-1 = 0
(2)

or

ab+a^2+a+1 = 0
(3)

From (2), we get

a = -b+1 = -\sqrt{5}+1

Now we evaluate equation (3). Substitute b=\sqrt{5} into the equation

a^2+(\sqrt{5}+1 )a+1=0

The discriminant

\Delta = (\sqrt{5}+1)^2-4

=2+2\sqrt{5} >0

Then

The quadratic equation has two distinct roots

a = \dfrac{-\sqrt{5}-1\pm\sqrt{2+2\sqrt{5} } }{2}

In summary,

the equation a^3+2\sqrt{5}a^2+5a+\sqrt{5}+1 =0 has three real roots, which are -\sqrt{5}+1, \dfrac{-\sqrt{5}-1+\sqrt{2+2\sqrt{5} } }{2} , \dfrac{-\sqrt{5}-1-\sqrt{2+2\sqrt{5} } }{2}

Steven Zheng posted 2 days ago

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