[tex]\implies {\blue {\boxed {\boxed {\purple {\sf { \: x = - 1 \: + \: i \sqrt{6} \:(or) \: \: x = - 1 \: -\: i \sqrt{6} }}}}}}[/tex]
[tex]\implies {\blue {\boxed {\boxed {\purple {\sf {x\:=\:2}}}}}}[/tex]
[tex]\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\red{:}}}}}[/tex]
[tex] \: {x}^{3} + 3x - 14 = 0[/tex]
➺[tex] \: {x}^{2} (x + 1) - x(x + 1) + 4(x + 1) = 18[/tex]
➺[tex] \: (x + 1)( {x}^{2} - x + 4) = 18[/tex]
➺[tex] \: {x}^{2} - x + 4 = \frac{18}{(x + 1)} [/tex]
➺[tex] \: {x}^{2} - x + 4 - 6 = \frac{18}{(x + 1)} - 6[/tex]
➺[tex] \: (x - 2)(x + 1) = \frac{18 - 6(x + 1)}{(x + 1)} [/tex]
➺[tex] \: (x - 2)(x + 1) = \frac{18 - 6x - 6}{(x + 1)} [/tex]
➺[tex] \: (x - 2)(x + 1) = \frac{12 - 6x}{(x + 1)} [/tex]
➺[tex] \: (x - 2)(x + 1) = \frac{ - 6(x - 2)}{(x + 1)} [/tex]
➺[tex] \: (x + 1 )² = \frac{ - 6(x - 2)}{(x + 1)(x - 2)} [/tex]
➺[tex] \: (x + 1)² = \frac{ - 6}{(x + 1)} [/tex]
[tex]\sf\pink{Error\:corrected\:here. }[/tex]
➺[tex] \: {x}^{2} + 2x + 1 = - 6[/tex]
➺[tex] \: {x}^{2} + 2x + 7 = 0[/tex]
By quadratic formula, we have
➺[tex] \: x = \frac{ - b± \sqrt{ {b}^{2} - 4ac} }{2a} [/tex]
➺[tex] \: x = \frac{ - 2± \sqrt{ {2}^{2} - 4.1.7} }{2 \times 1} [/tex]
➺[tex]x = \frac{ - 2± \sqrt{ - 24} }{2 } [/tex]
➺[tex] \: x = \frac{ - 2± \sqrt{ - 1 \times 4 \times 6} }{2} [/tex]
➺[tex] \: x = \frac{ - 2± \sqrt{ - 1} \times \sqrt{4} \times \sqrt{6} }{2} [/tex]
➺[tex] \: x = \frac{ - 2 \: ± \: i \times 2 \times \sqrt{6} }{2} [/tex]
➺[tex] \: x = \frac{ - 2 \: ± \:i \: 2 \sqrt{6} }{2} [/tex]
➺[tex] \: x = \frac{ 2 \: ( - 1 \: ± \: i \: \sqrt{6}) }{2} [/tex]
➺[tex] \: x = - 1 \: ± \: i \sqrt{6} [/tex]
Therefore, the two values of [tex]x[/tex] are ([tex] \: - 1 \: + \: i \sqrt{6}[/tex]) and ([tex] \: - 1 \: -\: i \sqrt{6}[/tex]).
[tex]x[/tex]³ + 3 [tex]x[/tex] - 14 = 0
➼ [tex]x[/tex]³ + 3 [tex]x[/tex] = 14
➼ [tex]x[/tex] ( [tex]x[/tex]² + 3 ) = 14
Factors of 14 = 1, 2, 7 and 14.
a) Substituting [tex]x\:=\:1[/tex], we have
➼ 1 ( 1 + 3 ) ≠ 14
➼ 1 x 4 ≠ 14
➼ [tex]\boxed{ 4\: ≠ \:14 }[/tex]
b) Substituting [tex]x\:=\:2[/tex], we have
➼ 2 ( 2² + 3 ) = 14
➼ 2 ( 4 + 3 ) = 14
➼ 2 x 7 = 14
➼ [tex]\boxed{ 14 \:= \:14 }[/tex]
c) Substituting [tex]x\:=\:7[/tex], we have
➼ 7 ( 7² + 3 ) ≠ 14
➼ 7 ( 49 + 3 ) ≠ 14
➼ 7 x 52 ≠ 14
➼ [tex]\boxed{ 364\: ≠ \:14 }[/tex]
d) Substituting [tex]x\:=\:14[/tex], we have
➼ 14 ( 14² + 3 ) ≠ 14
➼ 14 x 199 ≠ 14
➼ [tex]\boxed{ 2786\: ≠ \:14 }[/tex]
Hence, our only real solution is 2.
[tex]\large\mathfrak{{\pmb{\underline{\orange{Mystique35 }}{\orange{❦}}}}}[/tex]
