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tigry1 [53]
2 years ago
15

Find all possible values of b such that 4x² + bx + 3 can be factored. Check all that apply.

Mathematics
1 answer:
aleksklad [387]2 years ago
3 0

The possible values of b are 7 and 8

<h3>How to determine the possible values of x?</h3>

The expression is given as:

4x² + bx + 3

Next, we test the options to determine the values of b

<u>Option 1: b = 13</u>

So, we have:

4x² + 13x + 3 ---- this cannot be factorized

<u>Option 2: b = 7</u>

So, we have:

4x² + 7x + 3  

Expand

4x² + 4x + 3x + 3

Factorize

4x(x +1) + 3(x + 1)

Factor out x + 1

(4x + 3)(x + 1) ------ this can be factorized

<u>Option 3: b = 8</u>

So, we have:

4x² + 8x + 3

Expand

4x² + 6x + 2x + 3

Factorize

2x(2x +3) + 1(2x + 3)

Factor out 2x + 3

(2x + 3)(2x + 1) ------ this can be factorized

<u>Option 4: b = 1</u>

So, we have:

4x² + x + 3 ---- this cannot be factorized

Hence, the possible values of b are 7 and 8

Read more about factorized expressions at:

brainly.com/question/723406

#SPJ1

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Write an equation of a line that is parallel to the line whose equation is 3y=x+6 and that passes through the point (-3,4).
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Step-by-step explanation:

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(\frac{1}{81})^{\frac{x}{243}}=(\frac{1}{9})^{-3}-1

To solve this equation you use the following properties:

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Thne, by using this propwerty in the equation (1) you obtain for x

log_{(\frac{1}{81})}(\frac{1}{81})^{\frac{x}{243}}=log_{\frac{1}{81}}[(\frac{1}{81})-1]\\\\\frac{x}{243}=log_{\frac{1}{81}}[(\frac{1}{81})-1]\\\\x=(243)log_{\frac{1}{81}}[(\frac{1}{81})-1]

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