Question

Solve the equation x⁴ + 2x³ - 13x² + 2x + 1 = 0.

Answer 1

GIven equation
x⁴ + 2x³ - 13x² + 2x + 1 = 0
Dividing both sides by x2, we  get
x² + 2x - 13 +${\displaystyle \frac{2}{x}+\frac{1}{x^2}}$ = 0
⇒ ${\displaystyle (x^2+\frac{1}{x^2})+2(x+\frac{1}{x})-13}$ = 0
Put ${\displaystyle x+\frac{1}{x}=y,\text{squaring} x^2+\frac{1}{x^2}+2=y^2}$
⇒ ${\displaystyle x^2+\frac{1}{x^2}=y^2-2}$
Then y² - 2 + 2y - 13 = 0
⇒ y² + 2y - 15 = 0
⇒ y² + 5y - 3y - 15 = 0
⇒ y(y + 5) -3(y + 5) = 0
⇒ (y + 5)(y - 3) = 0
⇒ y + 5 = 0 or y = -5
or y - 3 = 0 or y =3
But ${\displaystyle x+\frac{1}{x}=-5}$
Then ${\displaystyle x+\frac{1}{x}=-5}$
⇒ x² + 1 = -5
⇒ x² + 5x + 1 = 0
⇒ x = ${\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$
x = ${\displaystyle \frac{-5\pm \sqrt{25-4}}{2\times 1}}$
x = ${\displaystyle \frac{-5\pm \sqrt{25-4}}{2}}$
x = ${\displaystyle \frac{-5\pm \sqrt{21}}{2}}$
or ${\displaystyle x+\frac{1}{x}=3}$
⇒ x² + 1 = 3
⇒ x² - 3x + 1 = 0
⇒ x = ${\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$
x = ${\displaystyle \frac{-(-3)\pm \sqrt{9-4}}{2}}$
x = ${\displaystyle \frac{3\pm \sqrt{5}}{2}}$
Hence x = ${\displaystyle \frac{-5\pm \sqrt{21}}{2},\frac{3\pm \sqrt{5}}{2}}$