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AnnyKZ [126]
3 years ago
8

3x²+8x+4 how to solve this

Mathematics
2 answers:
Lunna [17]3 years ago
7 0
3x² + 8x + 4
First, i divide the equation into two parenthesis, so that the first parts of both multiply to make the first term, 3x². 3x * x = 3x²
(3x  +    )(x +  )
Then I find two numbers that multiply to make 4, which are 1 and 4, or 2 and 2. 
Our options are: 
(3x + 1)(x + 4) 
(3x + 4)(x + 1)
(3x + 2)(3x +2)

To figure out which one to use, I'm just going to FOIL them all. 
(3x + 1)(x + 4)  = 3x² + 1x + 12x + 4 = 3x² + 13x + 4
(3x + 4)(x + 1) = 3x² + 4x + 3x + 4 = 3x² + 7x + 4
(3x + 2)(x +2) = 3x² + 2x + 6x + 4 = 3x² + 8x + 4

The factored form is: 
(3x + 2)(x + 2) 

To solve for the roots ( where the graph crosses the x axis, where y = 0) we set the equation equal to 0:

(3x + 2)(x + 2) = 0
The zero product property says that anything times 0 is 0, so we set each individual part equal to 0 and solve for the two roots. 

3x + 2 = 0
3x = -2
x = -2/3

x + 2 = 0 
x = -2


lisov135 [29]3 years ago
3 0


In this type of equation you could use different methods. But I choose to do Quadratic Method.

a= 3, b= 8, and c= 4.

X= -(8)+_SQUARE ROOT (8)^2-4(3)(4)/2(3)

X= -8+_SQUARE ROOT16/6 Square Root of 16 is 4

X= -8+_4/6

X=  -8+4/6= -4/6= -2/3.

X= -8-4/6= -12/6= -2.

Answer is: -2/3 and -2.




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son4ous [18]

Answer:

1) w₁=4 - i w₂= -4 + i

2) w₁= 3 - i w₂= -3  + i

3) w₁= 1 + 2i w₂= - 1 - 2i

4) w₁= 2- 3i w₂= -2 + 3i

5) w₁= 5 - 2i w₂= -5 + 2i

6) w₁= 5 - 3i w₂= -5 + 3i

Step-by-step explanation:

The root of a complex number is given by:

\sqrt[n]{z}=\sqrt[n]{r}(Cos(\frac{\theta+2k\pi}{n}) + i Sin(\frac{\theta+2k\pi}{n}))

where:

r: is the module of the complex number

θ: is the angle of the complex number to the positive axis x

n: index of the root

1) z = 15 − 8i  ⇒ r=17 θ= -0.4899 rad

w₁=\sqrt{17}(Cos(\frac{-0.4899}{2}) + i Sin(\frac{-0.4899}{2}))=4-i

w₂=\sqrt{17}(Cos(\frac{-0.4899+2\pi}{2}) + i Sin(\frac{-0.4899+2\pi}{2}))=-1+i

2) z = 8 − 6i  ⇒ r=10 θ= -0.6435 rad

w₁=\sqrt{10}(Cos(\frac{ -0.6435}{2}) + i Sin(\frac{ -0.6435}{2}))= 3 - i

w₂=\sqrt{10}(Cos(\frac{ -0.6435+2\pi}{2}) + i Sin(\frac{ -0.6435+2\pi}{2}))= -3  + i

3) z = −3 + 4i  ⇒ r=5 θ= -0.9316 rad

w₁=\sqrt{5}(Cos(\frac{-0.9316}{2}) + i Sin(\frac{-0.9316}{2}))= 1 + 2i

w₂=\sqrt{5}(Cos(\frac{-0.9316+2\pi}{2}) + i Sin(\frac{-0.9316+2\pi}{2}))= -1 - 2i

4) z = −5 − 12i  ⇒ r=13 θ= 0.4426 rad

w₁=\sqrt{13}(Cos(\frac{0.4426}{2}) + i Sin(\frac{0.4426}{2}))= 2- 3i

w₂=\sqrt{13}(Cos(\frac{0.4426+2\pi}{2}) + i Sin(\frac{0.4426+2\pi}{2}))= -2 + 3i

5) z = 21 − 20i  ⇒ r=29 θ= -0.8098 rad

w₁=\sqrt{29}(Cos(\frac{-0.8098}{2}) + i Sin(\frac{-0.8098}{2}))= 5 - 2i

w₂=\sqrt{29}(Cos(\frac{-0.8098+2\pi}{2}) + i Sin(\frac{-0.8098+2\pi}{2}))= -5 + 2i

6) z = 16 − 30i ⇒ r=34 θ= -1.0808 rad

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kaheart [24]

Answer:

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Step-by-step explanation:

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The surface area of the right triangular prism is 270 sq ft

<h3>Total surface ara of the prism</h3>

The total surface area of the prism is the  sum of all the area of its faces

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A = bh

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For the two rectangles

A = 2lw

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For the third triangle;
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Taking the sum of the areas

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TSA  = 270 sq ft

Hence the surface area of the right triangular prism is 270 sq ft

Learn more on surface area of prism here; brainly.com/question/1297098

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