Solve the equation x^4 + 2x^3 - 13x^2 + 2x + 1 = 0.

Question

Solve the equation $x^4 + 2x^3 - 13x^2 + 2x + 1 = 0.$

✓

Answer

Given equation
$x^4+2 x^3-13 x^2+2 x+1=0$
Dividing both sides by x2, we get
$x ^2+2 x -13+\frac{2}{x}+\frac{1}{x^2}=0$
$\Rightarrow\left(x^2+\frac{1}{x^2}\right)+2\left(x+\frac{1}{x}\right)-13=0$
Put $x+\frac{1}{x}=y$, squaring $x^2+\frac{1}{x^2}+2=y^2$
$\Rightarrow x^2+\frac{1}{x^2}=y^2-2$
Then $y ^2-2+2 y -13=0$
$\Rightarrow y^2+2 y-15=0$
$\Rightarrow y^2+5 y-3 y-15=0 $
$\Rightarrow y(y+5)-3(y+5)=0$
$\Rightarrow(y+5)(y-3)=0$
$\Rightarrow y+5=0 \text { or } y=-5$
or $y-3=0$ or $y=3$
But $x+\frac{1}{x}=-5$
Then $x+\frac{1}{x}=-5$
$\Rightarrow x^2+1=-5 $
$\Rightarrow x^2+5 x+1=0 $
$\Rightarrow x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} $
$x=\frac{-5 \pm \sqrt{25-4}}{2 \times 1} $
$x=\frac{-5 \pm \sqrt{25-4}}{2} $
$x=\frac{-5 \pm \sqrt{21}}{2}$
$\text { or } x+\frac{1}{x}=3 $
$\Rightarrow x^2+1=3 $
$\Rightarrow x^2-3 x +1=0 $
$\Rightarrow x =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} $
$x =\frac{-(-3) \pm \sqrt{9-4}}{2} $
$x =\frac{3 \pm \sqrt{5}}{2}$
Hence $x=\frac{-5 \pm \sqrt{21}}{2}, \frac{3 \pm \sqrt{5}}{2}$

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