191k views
1 vote
(20x^3-7x^2+3x-7)/-13x^2-5

User Myluco
by
7.1k points

1 Answer

5 votes

Answer:

x = ((30 sqrt(1462809) + 36253)^(2/3) - 131)/(60 (30 sqrt(1462809) + 36253)^(1/3)) + 7/60 or x = 7/60 - 1/60 ((-1)/(30 sqrt(1462809) + 36253))^(1/3) (131 (-1)^(1/3) + (30 sqrt(1462809) + 36253)^(2/3)) or x = 1/60 (131 (-1/(36253 + 30 sqrt(1462809)))^(1/3) + (-1)^(2/3) (36253 + 30 sqrt(1462809))^(1/3)) + 7/60

see atachement it's more legible.

Explanation:

Solve for x:

(20 x^3 - 7 x^2 + 3 x - 7)/(-13 x^2 - 5) = 0

Hint: | Multiply both sides by a polynomial to clear fractions.

Multiply both sides by -13 x^2 - 5:

20 x^3 - 7 x^2 + 3 x - 7 = 0

Hint: | Look for a simple substitution that eliminates the quadratic term of 20 x^3 - 7 x^2 + 3 x - 7.

Eliminate the quadratic term by substituting y = x - 7/60:

-7 + 3 (y + 7/60) - 7 (y + 7/60)^2 + 20 (y + 7/60)^3 = 0

Hint: | Write the cubic polynomial on the left hand side in standard form.

Expand out terms of the left hand side:

20 y^3 + (131 y)/60 - 36253/5400 = 0

Hint: | Write the cubic equation in standard form.

Divide both sides by 20:

y^3 + (131 y)/1200 - 36253/108000 = 0

Hint: | Perform the substitution y = z + λ/z.

Change coordinates by substituting y = z + λ/z, where λ is a constant value that will be determined later:

-36253/108000 + (131 (z + λ/z))/1200 + (z + λ/z)^3 = 0

Hint: | Transform the rational equation into a polynomial equation.

Multiply both sides by z^3 and collect in terms of z:

z^6 + z^4 (3 λ + 131/1200) - (36253 z^3)/108000 + z^2 (3 λ^2 + (131 λ)/1200) + λ^3 = 0

Hint: | Find an appropriate value for λ in order to make the coefficients of z^2 and z^4 both zero.

Substitute λ = -131/3600 and then u = z^3, yielding a quadratic equation in the variable u:

u^2 - (36253 u)/108000 - 2248091/46656000000 = 0

Hint: | Solve for u.

Find the positive solution to the quadratic equation:

u = (36253 + 30 sqrt(1462809))/216000

Hint: | Perform back substitution on u = (36253 + 30 sqrt(1462809))/216000.

Substitute back for u = z^3:

z^3 = (36253 + 30 sqrt(1462809))/216000

Hint: | Take the cube root of both sides.

Taking cube roots gives 1/60 (36253 + 30 sqrt(1462809))^(1/3) times the third roots of unity:

z = 1/60 (36253 + 30 sqrt(1462809))^(1/3) or z = -1/60 (-36253 - 30 sqrt(1462809))^(1/3) or z = 1/60 (-1)^(2/3) (36253 + 30 sqrt(1462809))^(1/3)

Hint: | Perform back substitution with y = z - 131/(3600 z).

Substitute each value of z into y = z - 131/(3600 z):

y = 1/60 (30 sqrt(1462809) + 36253)^(1/3) - 131/(60 (30 sqrt(1462809) + 36253)^(1/3)) or y = -1/60 (-30 sqrt(1462809) - 36253)^(1/3) - (131 (-1)^(2/3))/(60 (30 sqrt(1462809) + 36253)^(1/3)) or y = 131/60 ((-1)/(30 sqrt(1462809) + 36253))^(1/3) + 1/60 (-1)^(2/3) (30 sqrt(1462809) + 36253)^(1/3)

Hint: | Simplify each solution.

Bring each solution to a common denominator and simplify:

y = ((30 sqrt(1462809) + 36253)^(2/3) - 131)/(60 (36253 + 30 sqrt(1462809))^(1/3)) or y = -1/60 (-1/(36253 + 30 sqrt(1462809)))^(1/3) ((30 sqrt(1462809) + 36253)^(2/3) + 131 (-1)^(1/3)) or y = 1/60 (131 ((-1)/(30 sqrt(1462809) + 36253))^(1/3) + (-1)^(2/3) (30 sqrt(1462809) + 36253)^(1/3))

Hint: | Perform back substitution on the three roots.

Substitute back for x = y + 7/60:

Answer: x = ((30 sqrt(1462809) + 36253)^(2/3) - 131)/(60 (30 sqrt(1462809) + 36253)^(1/3)) + 7/60 or x = 7/60 - 1/60 ((-1)/(30 sqrt(1462809) + 36253))^(1/3) (131 (-1)^(1/3) + (30 sqrt(1462809) + 36253)^(2/3)) or x = 1/60 (131 (-1/(36253 + 30 sqrt(1462809)))^(1/3) + (-1)^(2/3) (36253 + 30 sqrt(1462809))^(1/3)) + 7/60

(20x^3-7x^2+3x-7)/-13x^2-5-example-1
User SaoPauloooo
by
7.8k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories