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Otrada [13]
3 years ago
8

Quadratic 6X2 + 5X = -7

Mathematics
1 answer:
poizon [28]3 years ago
7 0
Ok, so here your being asked to solve 6x2<span> + 5x = -7 

The procedure that I did was using this formula it led me to get the following:
</span>Using the formula:

x = -(-5) ± √(-5)² - 4(6)(-6)/ 2(6)
x = 5 ± √ 25 + 144 / 12
x = 5 ± √ 169 / 12
x = 5 ± 13/12

x1 = 5 + 13/12
x1 = 18/12
x1 = 3/2

x2 = 5 - 13/12
x2 = -8/12
<span> x2 = - 2/3 

Hope this helped :)</span>
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Set up, but do not evaluate, the integral that represents the length of the curve given by x = 1 + 3t^2, y = 4 + 2t^3 over the i
kherson [118]

L

=

∫

t

f

t

i

√

(

d

x

d

t

)

2

+

(

d

y

d

t

)

2

d

t

. Since  

x

and  

y

are perpendicular, it's not difficult to see why this computes the arclength.

It isn't very different from the arclength of a regular function:  

L

=

∫

b

a

√

1

+

(

d

y

d

x

)

2

d

x

. If you need the derivation of the parametric formula, please ask it as a separate question.

We find the 2 derivatives:

d

x

d

t

=

3

−

3

t

2

d

y

d

t

=

6

t

And we substitute these into the integral:

L

=

∫

√

3

0

√

(

3

−

3

t

2

)

2

+

(

6

t

)

2

d

t

And solve:

=

∫

√

3

0

√

9

−

18

t

2

+

9

t

4

+

36

t

2

d

t

=

∫

√

3

0

√

9

+

18

t

2

+

9

t

4

d

t

=

∫

√

3

0

√

(

3

+

3

t

2

)

2

d

t

=

∫

√

3

0

(

3

+

3

t

2

)

d

t

=

3

t

+

t

3

∣

∣

√

3

0

=

3

√

3

+

3

√

3

=6The arclength of a parametric curve can be found using the formula:  

L

=

∫

t

f

t

i

√

(

d

x

d

t

)

2

+

(

d

y

d

t

)

2

d

t

. Since  

x

and  

y

are perpendicular, it's not difficult to see why this computes the arclength.

It isn't very different from the arclength of a regular function:  

L

=

∫

b

a

√

1

+

(

d

y

d

x

)

2

d

x

. If you need the derivation of the parametric formula, please ask it as a separate question.

We find the 2 derivatives:

d

x

d

t

=

3

−

3

t

2

d

y

d

t

=

6

t

And we substitute these into the integral:

L

=

∫

√

3

0

√

(

3

−

3

t

2

)

2

+

(

6

t

)

2

d

t

And solve:

=

∫

√

3

0

√

9

−

18

t

2

+

9

t

4

+

36

t

2

d

t

=

∫

√

3

0

√

9

+

18

t

2

+

9

t

4

d

t

=

∫

√

3

0

√

(

3

+

3

t

2

)

2

d

t

=

∫

√

3

0

(

3

+

3

t

2

)

d

t

=

3

t

+

t

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∣

∣

√

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0

=

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√

3

+

3

√

3

=

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√

3

Be aware that arclength usually has a difficult function to integrate. Most integrable functions look like the above where a binomial is squared and adding the two terms will flip the sign of the binomial.    

Be aware that arclength usually has a difficult function to integrate. Most integrable functions look like the above where a binomial is squared and adding the two terms will flip the sign of the binomial.

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