MHB What Are the Real Solutions for This Equation?

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Solve for real solution(s) for $$x^2− x + 1 = (x^2+ x + 1)(x^2+ 2x + 4)$$.
 
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anemone said:
Solve for real solution(s) for $$x^2− x + 1 = (x^2+ x + 1)(x^2+ 2x + 4)$$.

My solution:

If we expand and collect like terms, we have:

$$x^4+3x^3+6x^2+7x+3=0$$

Let's assume the LHS factors in the form:

$$x^4+3x^3+6x^2+7x+3=(x^2+ax+1)(x^2+bx+3)=x^4+(a+b)x^3+(ab+4)x^2+(3a+b)x+3$$

Equating coefficients selected to result in a linear 2X2 system, we have:

$$a+b=3$$

$$3a+b=7$$

From this we find:

$$(a,b)=(2,1)$$

Hence:

$$x^4+3x^3+6x^2+7x+3=(x^2+2x+1)(x^2+x+3)=(x+1)^2(x^2+x+3)=0$$

The discriminant of the second factor is negative, thus the only real solution is:

$$x=-1$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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