Solve A Fourth Degree Equation


Leo Giugiuc has kindly posted a problem at the CutTheKnotMath facebook page with a solution (Solution 1) by Leo Giugiuc and Dan Sitaru. Solution 3 is by Kunihiko Chikaya.

Solve the equation


Solution 1

Let $f(x)=x^4+8x^3+23x^2+28x+9.\;$ We observe that $f(-1)=f(-2)=f(-3)=-3.\;$ From that we deduce

$\displaystyle\begin{align} f(x)&=(x+1)(x+2)^2(x+3)-3\\ &=(x^2+4x+3)(x^2+4x+4)-3\\ &=(x^2+4x+3)^2+(x^2+4x+3)-3\\ &=\left[(x^2+4x+3)-\frac{-1+\sqrt{13}}{2}\right]\left[(x^2+4x+3)+\frac{1+\sqrt{13}}{2}\right] \end{align}$

which reduces solving $f(x)=0\;$ to solving two quadratic equations.

Solution 2

Let's find a change of variables $x=y+a\;$ that eliminates the third power of the unknown variable. Upon the substitution, the coefficient by the third power equals $4\cdot a + 8\;$ which suggests the substitution $x=y-2.\;$ Luckily, the linear term also disappears, reducing the equation to $y^4-y^2-3=0.\;$ This is a biquadratic equation which can be solved in two steps.

Solution 3

Observe that

$(x^2+ax+b)^2=x^4+2a\cdot x^3+(a^2+2b)\cdot x^2+\ldots$

Solving $2a=8$ with $a^2+2b=23\;$ gives $a=4\;$ and $b=\displaystyle\frac{7}{2}.\;$ Furthermore, other coefficients appear to fit in:

$\displaystyle x^4+8x^3+23x^2+28x+9=\left(x^2+4x+\frac{7}{2}\right)^2-\frac{13}{4},$

showing that the original equation is equivalent to $\displaystyle\left(x^2+4x+\frac{7}{2}\right)^2=\frac{13}{4}.$


There is an interesting followup problem.

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