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3 years ago

2|3x - 2|= 14 how would you solve this absolute value equation?

Mathematics

2 answers:

IRINA_888 [86]3 years ago

7 0

Answer:

x = 3
x = - 5/3

Step-by-step explanation:

2|3x - 2|= 14
|3x - 2|= 7

1) 3x - 2 ≥ 0 ⇒ 3x ≥ 2 ⇒ x ≥ 2/3

3x - 2 = 7
3x = 9
x = 3 is valid root as 3 ≥ 2/3

2) 3x - 2 < 0 ⇒ 3x < 2 ⇒ x < 2/3

- (3x - 2) = 7
3x - 2 = -7
3x = -5
x = -5/3 is valid root as -5/3 < 2/3

Feliz [49]3 years ago

6 0

Answer: 2. |3x-2|= 14

|3x-2|=2.7/2

|3x-2|=7

x∈ {3,-5/3

Step-by-step explanation: \frac{5}{3}

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for In the given expression, calculate as indicated. Express your answer as a single polynomial in standard form. 4x^(3)-2x^(2)

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The polynomial in standard form is $y = 14 + 6\cdot x - 8\cdot x^{2}+4\cdot x^{3}$ .

Step-by-step explanation:

A polynomial in standard fulfills the following condition:

$y = \Sigma_{i=0}^{m}\,c_{i}\cdot x^{i}$ , $\forall\,i\in \mathbb{N}$ , $0 \leq i \leq m$

Let be $y = (4\cdot x^{3}-2\cdot x^{2}+9)-(6\cdot x^{2}-6\cdot x + 5)$ , which is now handled algebraically:

1) $(4\cdot x^{3}-2\cdot x^{2}+9)-(6\cdot x^{2}-6\cdot x + 5)$ Given

2) $(4\cdot x^{3}-2\cdot x^{2}+9) +(-1)\cdot (6\cdot x^{2}-6\cdot x + 5)$ $(-1)\cdot a = -a$

3) $(4\cdot x^{3}-2\cdot x^{2}+9) +[(-1)\cdot (6\cdot x^{2})+(-6\cdot x)\cdot (-1)+5]$ Definition of substraction/Distributive property

4) $(4\cdot x^{3}-2\cdot x^{2}+9)+(-6\cdot x^{2}+6\cdot x + 5)$ $(-1)\cdot a = -a$ / $(-a)\cdot (-b) = a\cdot b$

5) $4\cdot x^{3}+ (-2\cdot x^{2}-6\cdot x^{2})+6\cdot x + (9+5)$ Associative and commutative properties

6) $4\cdot x^{3}-8\cdot x^{2}+6\cdot x + 14$ Distributive property/Definition of addition

7) $14 + 6\cdot x -8\cdot x^{2}+4\cdot x^{3}$ Commutative property/Result

The polynomial in standard form is $y = 14 + 6\cdot x - 8\cdot x^{2}+4\cdot x^{3}$ .

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