Solve.<br> 3 - 2 | 2-5

anyanavicka [17]

Latus rectum is the line segment that passes through the focus, is perpendicular to the axis, and has both endpoints on the curve.
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8 0

3 years ago

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(x+2/x-7) - (x^2+4x+13/x^2-4x-21)

olya-2409 [2.1K]

Answer:

x = -2.98079 or x = -1.15272 or x = 0.892002 or x = 4.24151

Step-by-step explanation:

Solve for x:

-x^2 + x + 14 + 2/x - 13/x^2 = 0

Bring -x^2 + x + 14 + 2/x - 13/x^2 together using the common denominator x^2:

(-x^4 + x^3 + 14 x^2 + 2 x - 13)/x^2 = 0

Multiply both sides by x^2:

-x^4 + x^3 + 14 x^2 + 2 x - 13 = 0

Multiply both sides by -1:

x^4 - x^3 - 14 x^2 - 2 x + 13 = 0

Eliminate the cubic term by substituting y = x - 1/4:

13 - 2 (y + 1/4) - 14 (y + 1/4)^2 - (y + 1/4)^3 + (y + 1/4)^4 = 0

Expand out terms of the left hand side:

y^4 - (115 y^2)/8 - (73 y)/8 + 2973/256 = 0

Add (sqrt(2973) y^2)/8 + (115 y^2)/8 + (73 y)/8 to both sides:

y^4 + (sqrt(2973) y^2)/8 + 2973/256 = (sqrt(2973) y^2)/8 + (115 y^2)/8 + (73 y)/8

y^4 + (sqrt(2973) y^2)/8 + 2973/256 = (y^2 + sqrt(2973)/16)^2:

(y^2 + sqrt(2973)/16)^2 = (sqrt(2973) y^2)/8 + (115 y^2)/8 + (73 y)/8

Add 2 (y^2 + sqrt(2973)/16) λ + λ^2 to both sides:

(y^2 + sqrt(2973)/16)^2 + 2 λ (y^2 + sqrt(2973)/16) + λ^2 = (73 y)/8 + (sqrt(2973) y^2)/8 + (115 y^2)/8 + 2 λ (y^2 + sqrt(2973)/16) + λ^2

(y^2 + sqrt(2973)/16)^2 + 2 λ (y^2 + sqrt(2973)/16) + λ^2 = (y^2 + sqrt(2973)/16 + λ)^2:

(y^2 + sqrt(2973)/16 + λ)^2 = (73 y)/8 + (sqrt(2973) y^2)/8 + (115 y^2)/8 + 2 λ (y^2 + sqrt(2973)/16) + λ^2

(73 y)/8 + (sqrt(2973) y^2)/8 + (115 y^2)/8 + 2 λ (y^2 + sqrt(2973)/16) + λ^2 = (2 λ + 115/8 + sqrt(2973)/8) y^2 + (73 y)/8 + (sqrt(2973) λ)/8 + λ^2:

(y^2 + sqrt(2973)/16 + λ)^2 = y^2 (2 λ + 115/8 + sqrt(2973)/8) + (73 y)/8 + (sqrt(2973) λ)/8 + λ^2

Complete the square on the right hand side:

(y^2 + sqrt(2973)/16 + λ)^2 = (y sqrt(2 λ + 115/8 + sqrt(2973)/8) + 73/(16 sqrt(2 λ + 115/8 + sqrt(2973)/8)))^2 + (4 (2 λ + 115/8 + sqrt(2973)/8) (λ^2 + (sqrt(2973) λ)/8) - 5329/64)/(4 (2 λ + 115/8 + sqrt(2973)/8))

To express the right hand side as a square, find a value of λ such that the last term is 0.

This means 4 (2 λ + 115/8 + sqrt(2973)/8) (λ^2 + (sqrt(2973) λ)/8) - 5329/64 = 1/64 (512 λ^3 + 96 sqrt(2973) λ^2 + 3680 λ^2 + 460 sqrt(2973) λ + 11892 λ - 5329) = 0.

Thus the root λ = 1/48 (-3 sqrt(2973) - 115) + 1/12 (-i sqrt(3) + 1) ((3 i sqrt(10705335) - 8327)/2)^(1/3) + (173 (i sqrt(3) + 1))/(3 2^(2/3) (3 i sqrt(10705335) - 8327)^(1/3)) allows the right hand side to be expressed as a square.

(This value will be substituted later):

(y^2 + sqrt(2973)/16 + λ)^2 = (y sqrt(2 λ + 115/8 + sqrt(2973)/8) + 73/(16 sqrt(2 λ + 115/8 + sqrt(2973)/8)))^2

Take the square root of both sides:

y^2 + sqrt(2973)/16 + λ = y sqrt(2 λ + 115/8 + sqrt(2973)/8) + 73/(16 sqrt(2 λ + 115/8 + sqrt(2973)/8)) or y^2 + sqrt(2973)/16 + λ = -y sqrt(2 λ + 115/8 + sqrt(2973)/8) - 73/(16 sqrt(2 λ + 115/8 + sqrt(2973)/8))

Solve using the quadratic formula:

y = 1/8 (sqrt(2) sqrt(16 λ + 115 + sqrt(2973)) + sqrt(2) sqrt((10252 - 32 sqrt(2973) λ - 256 λ^2 + 292 sqrt(2) sqrt(16 λ + 115 + sqrt(2973)))/(16 λ + 115 + sqrt(2973)))) or y = 1/8 (sqrt(2) sqrt(16 λ + 115 + sqrt(2973)) - sqrt(2) sqrt((10252 - 32 sqrt(2973) λ - 256 λ^2 + 292 sqrt(2) sqrt(16 λ + 115 + sqrt(2973)))/(16 λ + 115 + sqrt(2973)))) or y = 1/8 (sqrt(2) sqrt((10252 - 32 sqrt(2973) λ - 256 λ^2 - 292 sqrt(2) sqrt(16 λ + 115 + sqrt(2973)))/(16 λ + 115 + sqrt(2973))) - sqrt(2) sqrt(16 λ + 115 + sqrt(2973))) or y = 1/8 (-sqrt(2) sqrt(16 λ + 115 + sqrt(2973)) - sqrt(2) sqrt((10252 - 32 sqrt(2973) λ - 256 λ^2 - 292 sqrt(2) sqrt(16 λ + 115 + sqrt(2973)))/(16 λ + 115 + sqrt(2973)))) where λ = 1/48 (-3 sqrt(2973) - 115) + 1/12 (-i sqrt(3) + 1) ((3 i sqrt(10705335) - 8327)/2)^(1/3) + (173 (i sqrt(3) + 1))/(3 2^(2/3) (3 i sqrt(10705335) - 8327)^(1/3))

Substitute λ = 1/48 (-3 sqrt(2973) - 115) + 1/12 (-i sqrt(3) + 1) ((3 i sqrt(10705335) - 8327)/2)^(1/3) + (173 (i sqrt(3) + 1))/(3 2^(2/3) (3 i sqrt(10705335) - 8327)^(1/3)) and approximate:

y = -3.23079 or y = -1.40272 or y = 0.642002 or y = 3.99151

Substitute back for y = x - 1/4:

x - 1/4 = -3.23079 or y = -1.40272 or y = 0.642002 or y = 3.99151

Add 1/4 to both sides:

x = -2.98079 or y = -1.40272 or y = 0.642002 or y = 3.99151

Substitute back for y = x - 1/4:

x = -2.98079 or x - 1/4 = -1.40272 or y = 0.642002 or y = 3.99151

Add 1/4 to both sides:

x = -2.98079 or x = -1.15272 or y = 0.642002 or y = 3.99151

Substitute back for y = x - 1/4:

x = -2.98079 or x = -1.15272 or x - 1/4 = 0.642002 or y = 3.99151

Add 1/4 to both sides:

x = -2.98079 or x = -1.15272 or x = 0.892002 or y = 3.99151

Substitute back for y = x - 1/4:

x = -2.98079 or x = -1.15272 or x = 0.892002 or x - 1/4 = 3.99151

Add 1/4 to both sides:

Answer: x = -2.98079 or x = -1.15272 or x = 0.892002 or x = 4.24151

7 0

3 years ago

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Write an equation of a line that contains the point (-2,3) with a slope of 0.

Aleksandr [31]

Answer:

y =3

Step-by-step explanation:

Slope of zero means it is a horizontal line

Horizontal lines are of the form

y =

The y coordinate is 3 so

y =3

4 0

3 years ago

A diver descends to a location of -50 1 over 4 meters relative to the surface of the water. He then descends another 10 1 over 2

Yanka [14]

Answer:

$-60\frac{3}{4}$ meters.

Step-by-step explanation:

To find diver's final location relative to the surface of water we will subtract $10\frac{1}{2}$ from $-50\frac{1}{4}$ because diver dives away another $10\frac{1}{2}$ from his previous location of $-50\frac{1}{4}$ .

$-50\frac{1}{4} -10\frac{1}{2}$

Now we will convert our mixed fraction into improper fraction and then we will find a common denominator.

$\frac{-201}{4} -\frac{21}{2} =\frac{-201-42}{4}$

$=\frac{-243}{4}$

Now let us convert our answer in mixed fraction.

$=\frac{-243}{4}=[tex]-60\frac{3}{4}$ [/tex]

Therefore, diver's final location relative to the surface of the water is $-60\frac{3}{4}$ meters.

3 0

3 years ago

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What is the domain and the range of these functions?? quick please!!!!!!!!!!

Sever21 [200]

The domain of a function f(x) is the set of all values for which the function is defined, and the range of the function is the set of all values that f takes. A rational function is a function of the form f(x)=p(x)q(x) , where p(x) and q(x) are polynomials and q(x)≠0 .

6 0

3 years ago

Solve. 3 - 2 | 2-5 | + 3