Question

Find x

x³ + 3x - 14 = 0

x³ + x² - x² - x + 4x + 4 = 18
x²(x + 1) - x(x + 1) + 4(x + 1) = 18
(x + 1)(x² - x + 4) = 18
x² - x + 4 = 18/(x + 1)
x² - x + 4 - 6 = 18/(x + 1) - 6
x² - x - 2 = 18/(x + 1) - 6
(x - 2)(x + 1) = (18 - 6(x + 1))/(x + 1)
(x - 2)(x + 1) = (18 - 6x - 6)/(x + 1)
(x - 2)(x + 1) = (12 - 6x)/(x + 1)
(x - 2)(x + 1) = (-6(x - 2))/(x + 1)
x + 1 = (-6(x - 2))/(x + 1)(x - 2)
x + 1 = -6/(x + 1)
(x + 1)² = -6
x² + 2x + 8 = 0
x = (-b +- √(b² - 4ac))/2a
x = (-2 +- √(4 - 32))/2
x = (-2 +- √(-28)/2
x = (-2 +- i√28)/2​

Something's wrong.​

Answer 1

`\implies {\blue {\boxed {\boxed {\purple {\sf { \: x = - 1 \: + \: i √(6) \:(or) \: \: x = - 1 \: -\: i √(6) }}}}}}`

And

`\implies {\blue {\boxed {\boxed {\purple {\sf {x\:=\:2}}}}}}`

`\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\red{:}}}}}`

`\: {x}^(3) + 3x - 14 = 0`

➺
`\: {x}^(2) (x + 1) - x(x + 1) + 4(x + 1) = 18`

➺
`\: (x + 1)( {x}^(2) - x + 4) = 18`

➺
`\: {x}^(2) - x + 4 = (18)/((x + 1))`

➺
`\: {x}^(2) - x + 4 - 6 = (18)/((x + 1)) - 6`

➺
`\: (x - 2)(x + 1) = (18 - 6(x + 1))/((x + 1))`

➺
`\: (x - 2)(x + 1) = (18 - 6x - 6)/((x + 1))`

➺
`\: (x - 2)(x + 1) = (12 - 6x)/((x + 1))`

➺
`\: (x - 2)(x + 1) = ( - 6(x - 2))/((x + 1))`

➺
`\: (x + 1 )² = ( - 6(x - 2))/((x + 1)(x - 2))`

➺
`\: (x + 1)² = ( - 6)/((x + 1))`

`\sf\pink{Error\:corrected\:here. }`

➺
`\: {x}^(2) + 2x + 1 = - 6`

➺
`\: {x}^(2) + 2x + 7 = 0`

By quadratic formula, we have

➺
`\: x = \frac{ - b± \sqrt{ {b}^(2) - 4ac} }{2a}`

➺
`\: x = \frac{ - 2± \sqrt{ {2}^(2) - 4.1.7} }{2 * 1}`

➺
`x = ( - 2± √( - 24) )/(2 )`

➺
`\: x = ( - 2± √( - 1 * 4 * 6) )/(2)`

➺
`\: x = ( - 2± √( - 1) * √(4) * √(6) )/(2)`

➺
`\: x = ( - 2 \: ± \: i * 2 * √(6) )/(2)`

➺
`\: x = ( - 2 \: ± \:i \: 2 √(6) )/(2)`

➺
`\: x = ( 2 \: ( - 1 \: ± \: i \: √(6)) )/(2)`

➺
`\: x = - 1 \: ± \: i √(6)`

Therefore, the two values of
`x` are (
`\: - 1 \: + \: i √(6)`) and (
`\: - 1 \: -\: i √(6)`).

Let us look at another method.

`x`³ + 3
`x` - 14 = 0

➼
`x`³ + 3
`x` = 14

➼
`x` (
`x`² + 3 ) = 14

Factors of 14 = 1, 2, 7 and 14.

a) Substituting
`x\:=\:1`, we have

➼ 1 ( 1 + 3 ) ≠ 14

➼ 1 x 4 ≠ 14

➼
`\boxed{ 4\: ≠ \:14 }`

b) Substituting
`x\:=\:2`, we have

➼ 2 ( 2² + 3 ) = 14

➼ 2 ( 4 + 3 ) = 14

➼ 2 x 7 = 14

➼
`\boxed{ 14 \:= \:14 }`

c) Substituting
`x\:=\:7`, we have

➼ 7 ( 7² + 3 ) ≠ 14

➼ 7 ( 49 + 3 ) ≠ 14

➼ 7 x 52 ≠ 14

➼
`\boxed{ 364\: ≠ \:14 }`

d) Substituting
`x\:=\:14`, we have

➼ 14 ( 14² + 3 ) ≠ 14

➼ 14 x 199 ≠ 14

➼
`\boxed{ 2786\: ≠ \:14 }`

Hence, our only real solution is 2.

`\large\mathfrak{{\pmb{\underline{\orange{Mystique35 }}{\orange{❦}}}}}`