﻿ Solve the equation with square root coefficient a^3+2\sqrt{11}a^2+11a+\sqrt{11}+1 =0

a^3+2\sqrt{11}a^2+11a+\sqrt{11}+1 =0
(1)

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

b+a+1 = 0
(2)

or

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

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}

Steven Zheng posted 4 months ago

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