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​

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By quadratic formula, we have

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Therefore, the two values of are () and ().

³ + 3 - 14 = 0

➼ ³ + 3 = 14

➼ ( ² + 3 ) = 14

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

a) Substituting , we have

➼ 1 ( 1 + 3 ) ≠ 14

➼ 1 x 4 ≠ 14

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b) Substituting , we have

➼ 2 ( 2² + 3 ) = 14

➼ 2 ( 4 + 3 ) = 14

➼ 2 x 7 = 14

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c) Substituting , we have

➼ 7 ( 7² + 3 ) ≠ 14

➼ 7 ( 49 + 3 ) ≠ 14

➼ 7 x 52 ≠ 14

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d) Substituting , we have

➼ 14 ( 14² + 3 ) ≠ 14

➼ 14 x 199 ≠ 14

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Hence, our only real solution is 2.

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step-by-step explanation:

step-by-step explanation:

the true statements are:

r = set of real numbers

n = set of natural numbers, 1,2,

w = n u{0} = {0,1,}