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Question
Solve the equation x4 + 2x3 - 13x2 + 2x + 1 = 0.
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Solution
GIven equation
x4 + 2x3 - 13x2 + 2x + 1 = 0
Dividing both sides by x2, we get
x2 + 2x - 13 +`(2)/x + (1)/x^2` = 0
⇒ `(x^2 + 1/x^2) +2(x + 1/x) -13` = 0
Put `x + (1)/x = y, "squaring" x^2 + (1)/x^2 + 2 = y^2`
⇒ `x^2 + (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 `x + (1)/x = -5`
Then `x + (1)/x = -5`
⇒ x2 + 1 = -5
⇒ x2 + 5x + 1 = 0
⇒ x = `(-b ± sqrt(b^2 - 4ac))/(2a)`
x = `(-5 ± sqrt(25 - 4))/(2 xx 1)`
x = `(-5 ± sqrt(25 - 4))/(2)`
x = `(-5 ± sqrt(21))/(2)`
or `x + (1)/x = 3`
⇒ x2 + 1 = 3
⇒ x2 - 3x + 1 = 0
⇒ x = `(-b ± sqrt(b^2 - 4ac))/(2a)`
x = `(-(-3) ± sqrt(9 - 4))/(2)`
x = `(3 ± sqrt(5))/(2)`
Hence x = `(-5 ± sqrt(21))/(2), (3 ± sqrt(5))/(2)`
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