Answer

a^3+2\sqrt{7}a^2+7a+\sqrt{7}+1 =0
(1)

Let

b = \sqrt{7}

then

b^2 = 7

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{7}+1 )

Now we evaluate equation (3). Substitute b=\sqrt{7} into the equation

a^2+(\sqrt{7}-1 )a+1=0

The discriminant

\Delta = (\sqrt{7}-1)^2-4

=(\sqrt{7}-3 )(\sqrt{7}+1 )

=(\sqrt{7}-\sqrt{9} )(\sqrt{7}+1 )<0

Therefore

There is no real root for the quadratic equation

In summary,

the equation a^3+2\sqrt{7}a^2+7a+\sqrt{7}+1 =0 has one real root -\sqrt{7}-1

Steven Zheng posted 1 hour ago

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