What is the general solution for (sech^4(x))=tanh^4(x) ?

The general solution for (sech^4(x))=tanh^4(x) is x=ln(sqrt(2)+1),x=ln(sqrt(2)-1)

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(sech^4(x))=tanh^4(x)

Solution

(sech^{4} (x)) = tanh^{4} (x)

Solution

x = ln (2 + 1), x = ln (2 - 1)

Degrees

x = 50.49898 \dots^{\circ}, x = - 50.49898 \dots^{\circ}

Solution steps

(sech^{4} (x)) = tanh^{4} (x)

Rewrite using trig identities

sech^{4} (x) = tanh^{4} (x)

Use the Hyperbolic identity:

tanh (x) = \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}

sech^{4} (x) = (\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}})^{4}

Use the Hyperbolic identity:

sech (x) = \frac{2}{e ^{x} + e ^{- x}}

(\frac{2}{e ^{x} + e ^{- x}})^{4} = (\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}})^{4}

(\frac{2}{e ^{x} + e ^{- x}})^{4} = (\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}})^{4}

(\frac{2}{e ^{x} + e ^{- x}})^{4} = (\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}})^{4} : x = ln (2 + 1), x = ln (2 - 1)

(\frac{2}{e ^{x} + e ^{- x}})^{4} = (\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}})^{4}

Apply exponent rules

(\frac{2}{e ^{x} + e ^{- x}})^{4} = (\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}})^{4}

Apply exponent rule:

a^{b c} = (a^{b})^{c}

e^{- x} = (e^{x})^{- 1}

(\frac{2}{e ^{x} + ( e ^{x} ) ^{- 1}})^{4} = (\frac{e ^{x} - ( e ^{x} ) ^{- 1}}{e ^{x} + ( e ^{x} ) ^{- 1}})^{4}

(\frac{2}{e ^{x} + ( e ^{x} ) ^{- 1}})^{4} = (\frac{e ^{x} - ( e ^{x} ) ^{- 1}}{e ^{x} + ( e ^{x} ) ^{- 1}})^{4}

Rewrite the equation with

e^{x} = u

(\frac{2}{u + ( u ) ^{- 1}})^{4} = (\frac{u - ( u ) ^{- 1}}{u + ( u ) ^{- 1}})^{4}

Solve

(\frac{2}{u + u ^{- 1}})^{4} = (\frac{u - u ^{- 1}}{u + u ^{- 1}})^{4} : u = 2 + 1, u = 2 - 1

(\frac{2}{u + u ^{- 1}})^{4} = (\frac{u - u ^{- 1}}{u + u ^{- 1}})^{4}

Refine

\frac{16 u ^{4}}{( u ^{2} + 1 ) ^{4}} = \frac{( u ^{2} - 1 ) ^{4}}{( u ^{2} + 1 ) ^{4}}

Multiply both sides by

(u^{2} + 1)^{4}

\frac{16 u ^{4}}{( u ^{2} + 1 ) ^{4}} = \frac{( u ^{2} - 1 ) ^{4}}{( u ^{2} + 1 ) ^{4}}

Multiply both sides by

(u^{2} + 1)^{4}

\frac{16 u ^{4}}{( u ^{2} + 1 ) ^{4}} (u^{2} + 1)^{4} = \frac{( u ^{2} - 1 ) ^{4}}{( u ^{2} + 1 ) ^{4}} (u^{2} + 1)^{4}

Simplify

\frac{16 u ^{4}}{( u ^{2} + 1 ) ^{4}} (u^{2} + 1)^{4} = \frac{( u ^{2} - 1 ) ^{4}}{( u ^{2} + 1 ) ^{4}} (u^{2} + 1)^{4}

Simplify

\frac{16 u ^{4}}{( u ^{2} + 1 ) ^{4}} (u^{2} + 1)^{4} : 16 u^{4}

\frac{16 u ^{4}}{( u ^{2} + 1 ) ^{4}} (u^{2} + 1)^{4}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{16 u ^{4} ( u ^{2} + 1 ) ^{4}}{( u ^{2} + 1 ) ^{4}}

Cancel the common factor:

(u^{2} + 1)^{4}

= 16 u^{4}

Simplify

\frac{( u ^{2} - 1 ) ^{4}}{( u ^{2} + 1 ) ^{4}} (u^{2} + 1)^{4} : (u^{2} - 1)^{4}

\frac{( u ^{2} - 1 ) ^{4}}{( u ^{2} + 1 ) ^{4}} (u^{2} + 1)^{4}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( u ^{2} - 1 ) ^{4} ( u ^{2} + 1 ) ^{4}}{( u ^{2} + 1 ) ^{4}}

Cancel the common factor:

(u^{2} + 1)^{4}

= (u^{2} - 1)^{4}

16 u^{4} = (u^{2} - 1)^{4}

16 u^{4} = (u^{2} - 1)^{4}

16 u^{4} = (u^{2} - 1)^{4}

Solve

16 u^{4} = (u^{2} - 1)^{4} : u = 2 + 1, u = 2 - 1

16 u^{4} = (u^{2} - 1)^{4}

Expand

(u^{2} - 1)^{4} : u^{8} - 4 u^{6} + 6 u^{4} - 4 u^{2} + 1

(u^{2} - 1)^{4}

Apply binomial theorem:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

a = u^{2}, b = - 1

= i = 0 \sum 4 (i 4) (u^{2})^{(4 - i)} (- 1)^{i}

Expand summation

(i n) = \frac{n !}{i ! ( n - i ) !}

i = 0 : \frac{4 !}{0 ! ( 4 - 0 ) !} (u^{2})^{4} (- 1)^{0}

i = 1 : \frac{4 !}{1 ! ( 4 - 1 ) !} (u^{2})^{3} (- 1)^{1}

i = 2 : \frac{4 !}{2 ! ( 4 - 2 ) !} (u^{2})^{2} (- 1)^{2}

i = 3 : \frac{4 !}{3 ! ( 4 - 3 ) !} (u^{2})^{1} (- 1)^{3}

i = 4 : \frac{4 !}{4 ! ( 4 - 4 ) !} (u^{2})^{0} (- 1)^{4}

= \frac{4 !}{0 ! ( 4 - 0 ) !} (u^{2})^{4} (- 1)^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} (u^{2})^{3} (- 1)^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} (u^{2})^{2} (- 1)^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} (u^{2})^{1} (- 1)^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} (u^{2})^{0} (- 1)^{4}

= \frac{4 !}{0 ! ( 4 - 0 ) !} (u^{2})^{4} (- 1)^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} (u^{2})^{3} (- 1)^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} (u^{2})^{2} (- 1)^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} (u^{2})^{1} (- 1)^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} (u^{2})^{0} (- 1)^{4}

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (u^{2})^{4} (- 1)^{0} : u^{8}

\frac{4 !}{0 ! ( 4 - 0 ) !} (u^{2})^{4} (- 1)^{0}

Apply rule

a^{0} = 1, a \neq = 0

(- 1)^{0} = 1

= 1 \cdot \frac{4 !}{0 ! ( 4 - 0 ) !} (u^{2})^{4}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= 1 \cdot \frac{4 ! ( u ^{2} ) ^{4}}{0 ! ( 4 - 0 ) !}

Simplify

\frac{4 ! ( u ^{2} ) ^{4}}{0 ! ( 4 - 0 ) !} : u^{8}

\frac{4 ! ( u ^{2} ) ^{4}}{0 ! ( 4 - 0 ) !}

0! (4 - 0)! = 4!

0! (4 - 0)!

Subtract the numbers:

4 - 0 = 4

= 0! \cdot 4!

Apply factorial rule:

0! = 1

= 1 \cdot 4!

Multiply:

1 \cdot 4! = 4!

= 4!

= \frac{4 ! ( u ^{2} ) ^{4}}{4 !}

Cancel the common factor:

4!

= (u^{2})^{4}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= u^{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= u^{8}

= 1 \cdot u^{8}

Multiply:

1 \cdot u^{8} = u^{8}

= u^{8}

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (u^{2})^{3} (- 1)^{1} : - 4 u^{6}

\frac{4 !}{1 ! ( 4 - 1 ) !} (u^{2})^{3} (- 1)^{1}

Apply rule

a^{1} = a

(- 1)^{1} = - 1

= (- 1) \frac{4 !}{1 ! ( 4 - 1 ) !} (u^{2})^{3}

Remove parentheses:

(- a) = - a

= - \frac{4 !}{1 ! ( 4 - 1 ) !} (u^{2})^{3} \cdot 1

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= - 1 \cdot \frac{4 ! ( u ^{2} ) ^{3}}{1 ! ( 4 - 1 ) !}

Simplify

\frac{4 ! ( u ^{2} ) ^{3}}{1 ! ( 4 - 1 ) !} : 4 u^{6}

\frac{4 ! ( u ^{2} ) ^{3}}{1 ! ( 4 - 1 ) !}

Subtract the numbers:

4 - 1 = 3

= \frac{4 ! ( u ^{2} ) ^{3}}{1 ! \cdot 3 !}

Cancel the factorials:

\frac{n !}{( n - m ) !} = n \cdot (n - 1) \dots (n - m + 1), n > m

\frac{4 !}{3 !} = 4

= \frac{4 ( u ^{2} ) ^{3}}{1 !}

(u^{2})^{3} = u^{6}

(u^{2})^{3}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= u^{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= u^{6}

= \frac{4 u ^{6}}{1 !}

Apply factorial rule:

n! = 1 \cdot 2 \cdot 3 \cdot \dots \cdot n

1! = 1

= \frac{4 u ^{6}}{1}

Apply rule

\frac{a}{1} = a

= 4 u^{6}

= - 1 \cdot 4 u^{6}

Multiply the numbers:

1 \cdot 4 = 4

= - 4 u^{6}

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (u^{2})^{2} (- 1)^{2} : 6 u^{4}

\frac{4 !}{2 ! ( 4 - 2 ) !} (u^{2})^{2} (- 1)^{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{4 ! ( u ^{2} ) ^{2} ( - 1 ) ^{2}}{2 ! ( 4 - 2 ) !}

4! (u^{2})^{2} (- 1)^{2} = 4! (u^{2})^{2}

4! (u^{2})^{2} (- 1)^{2}

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

= 1 \cdot 4! (u^{2})^{2}

Multiply:

(u^{2})^{2} \cdot 1 = (u^{2})^{2}

= 4! (u^{2})^{2}

= \frac{4 ! ( u ^{2} ) ^{2}}{2 ! ( 4 - 2 ) !}

Subtract the numbers:

4 - 2 = 2

= \frac{4 ! ( u ^{2} ) ^{2}}{2 ! \cdot 2 !}

Cancel the factorials:

\frac{n !}{( n - m ) !} = n \cdot (n - 1) \dots (n - m + 1), n > m

\frac{4 !}{2 !} = 4 \cdot 3

= \frac{4 \cdot 3 ( u ^{2} ) ^{2}}{2 !}

Refine

= \frac{12 ( u ^{2} ) ^{2}}{2 !}

(u^{2})^{2} = u^{4}

(u^{2})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= u^{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= u^{4}

= \frac{12 u ^{4}}{2 !}

2! = 2

2!

Apply factorial rule:

n! = 1 \cdot 2 \cdot 3 \cdot \dots \cdot n

2! = 1 \cdot 2

= 1 \cdot 2

Multiply the numbers:

1 \cdot 2 = 2

= 2

= \frac{12 u ^{4}}{2}

Divide the numbers:

\frac{12}{2} = 6

= 6 u^{4}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (u^{2})^{1} (- 1)^{3} : - 4 u^{2}

\frac{4 !}{3 ! ( 4 - 3 ) !} (u^{2})^{1} (- 1)^{3}

Apply rule

a^{1} = a

(u^{2})^{1} = u^{2}

= (- 1)^{3} \frac{4 !}{3 ! ( 4 - 3 ) !} u^{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{4 ! u ^{2} ( - 1 ) ^{3}}{3 ! ( 4 - 3 ) !}

4! u^{2} (- 1)^{3} = - 4! u^{2}

4! u^{2} (- 1)^{3}

(- 1)^{3} = - 1

(- 1)^{3}

Apply exponent rule:

(- a)^{n} = - a^{n},

n

is odd

(- 1)^{3} = - 1^{3}

= - 1^{3}

Apply rule

1^{a} = 1

= - 1

= (- 1) \cdot 4! u^{2}

Refine

= - 4! u^{2}

= \frac{- 4 ! u ^{2}}{3 ! ( 4 - 3 ) !}

Subtract the numbers:

4 - 3 = 1

= \frac{- 4 ! u ^{2}}{3 ! \cdot 1 !}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{4 ! u ^{2}}{3 ! \cdot 1 !}

Cancel the factorials:

\frac{n !}{( n - m ) !} = n \cdot (n - 1) \dots (n - m + 1), n > m

\frac{4 !}{3 !} = 4

= - \frac{4 u ^{2}}{1 !}

Apply factorial rule:

n! = 1 \cdot 2 \cdot 3 \cdot \dots \cdot n

1! = 1

= - \frac{4 u ^{2}}{1}

Apply rule

\frac{a}{1} = a

= - 4 u^{2}

\frac{4 !}{4 ! ( 4 - 4 ) !} (u^{2})^{0} (- 1)^{4} = 1

\frac{4 !}{4 ! ( 4 - 4 ) !} (u^{2})^{0} (- 1)^{4}

Apply rule

a^{0} = 1, a \neq = 0

(u^{2})^{0} = 1

= (- 1)^{4} \cdot 1 \cdot \frac{4 !}{4 ! ( 4 - 4 ) !}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= 1 \cdot \frac{( - 1 ) ^{4} \cdot 4 !}{4 ! ( 4 - 4 ) !}

Cancel the common factor:

4!

= 1 \cdot \frac{( - 1 ) ^{4}}{( 4 - 4 ) !}

\frac{( - 1 ) ^{4}}{( 4 - 4 ) !} = 1

\frac{( - 1 ) ^{4}}{( 4 - 4 ) !}

(- 1)^{4} = 1

(- 1)^{4}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 1)^{4} = 1^{4}

= 1^{4}

Apply rule

1^{a} = 1

= 1

= \frac{1}{( 4 - 4 ) !}

(4 - 4)! = 1

(4 - 4)!

Subtract the numbers:

4 - 4 = 0

= 0!

Apply factorial rule:

0! = 1

= 1

= \frac{1}{1}

Apply rule

\frac{a}{1} = a

= 1

= 1 \cdot 1

Multiply the numbers:

1 \cdot 1 = 1

= 1

= u^{8} - 4 u^{6} + 6 u^{4} - 4 u^{2} + 1

16 u^{4} = u^{8} - 4 u^{6} + 6 u^{4} - 4 u^{2} + 1

Switch sides

u^{8} - 4 u^{6} + 6 u^{4} - 4 u^{2} + 1 = 16 u^{4}

Move

16 u^{4}

to the left side

u^{8} - 4 u^{6} + 6 u^{4} - 4 u^{2} + 1 = 16 u^{4}

Subtract

16 u^{4}

from both sides

u^{8} - 4 u^{6} + 6 u^{4} - 4 u^{2} + 1 - 16 u^{4} = 16 u^{4} - 16 u^{4}

Simplify

u^{8} - 4 u^{6} - 10 u^{4} - 4 u^{2} + 1 = 0

u^{8} - 4 u^{6} - 10 u^{4} - 4 u^{2} + 1 = 0

Rewrite the equation with

v = u^{2}, v^{2} = u^{4}, v^{3} = u^{6}

and

v^{4} = u^{8}

v^{4} - 4 v^{3} - 10 v^{2} - 4 v + 1 = 0

Solve

v^{4} - 4 v^{3} - 10 v^{2} - 4 v + 1 = 0 : v = - 1, v = 3 + 22, v = 3 - 22

v^{4} - 4 v^{3} - 10 v^{2} - 4 v + 1 = 0

Factor

v^{4} - 4 v^{3} - 10 v^{2} - 4 v + 1 : (v + 1)^{2} (v^{2} - 6 v + 1)

v^{4} - 4 v^{3} - 10 v^{2} - 4 v + 1

Use the rational root theorem

a_{0} = 1, a_{n} = 1

The dividers of

a_{0} : 1,

The dividers of

a_{n} : 1

Therefore, check the following rational numbers:

\pm \frac{1}{1}

- \frac{1}{1}

is a root of the expression, so factor out

v + 1

= (v + 1) \frac{v ^{4} - 4 v ^{3} - 10 v ^{2} - 4 v + 1}{v + 1}

\frac{v ^{4} - 4 v ^{3} - 10 v ^{2} - 4 v + 1}{v + 1} = v^{3} - 5 v^{2} - 5 v + 1

\frac{v ^{4} - 4 v ^{3} - 10 v ^{2} - 4 v + 1}{v + 1}

Divide

\frac{v ^{4} - 4 v ^{3} - 10 v ^{2} - 4 v + 1}{v + 1} : \frac{v ^{4} - 4 v ^{3} - 10 v ^{2} - 4 v + 1}{v + 1} = v^{3} + \frac{- 5 v ^{3} - 10 v ^{2} - 4 v + 1}{v + 1}

Divide the leading coefficients of the numerator

v^{4} - 4 v^{3} - 10 v^{2} - 4 v + 1

and the divisor

v + 1 : \frac{v ^{4}}{v} = v^{3}

Quotient = v^{3}

Multiply

v + 1

v^{3} : v^{4} + v^{3}

Subtract

v^{4} + v^{3}

from

v^{4} - 4 v^{3} - 10 v^{2} - 4 v + 1

to get new remainder

Remainder = - 5 v^{3} - 10 v^{2} - 4 v + 1

Therefore

\frac{v ^{4} - 4 v ^{3} - 10 v ^{2} - 4 v + 1}{v + 1} = v^{3} + \frac{- 5 v ^{3} - 10 v ^{2} - 4 v + 1}{v + 1}

= v^{3} + \frac{- 5 v ^{3} - 10 v ^{2} - 4 v + 1}{v + 1}

Divide

\frac{- 5 v ^{3} - 10 v ^{2} - 4 v + 1}{v + 1} : \frac{- 5 v ^{3} - 10 v ^{2} - 4 v + 1}{v + 1} = - 5 v^{2} + \frac{- 5 v ^{2} - 4 v + 1}{v + 1}

Divide the leading coefficients of the numerator

- 5 v^{3} - 10 v^{2} - 4 v + 1

and the divisor

v + 1 : \frac{- 5 v ^{3}}{v} = - 5 v^{2}

Quotient = - 5 v^{2}

Multiply

v + 1

- 5 v^{2} : - 5 v^{3} - 5 v^{2}

Subtract

- 5 v^{3} - 5 v^{2}

from

- 5 v^{3} - 10 v^{2} - 4 v + 1

to get new remainder

Remainder = - 5 v^{2} - 4 v + 1

Therefore

\frac{- 5 v ^{3} - 10 v ^{2} - 4 v + 1}{v + 1} = - 5 v^{2} + \frac{- 5 v ^{2} - 4 v + 1}{v + 1}

= v^{3} - 5 v^{2} + \frac{- 5 v ^{2} - 4 v + 1}{v + 1}

Divide

\frac{- 5 v ^{2} - 4 v + 1}{v + 1} : \frac{- 5 v ^{2} - 4 v + 1}{v + 1} = - 5 v + \frac{v + 1}{v + 1}

Divide the leading coefficients of the numerator

- 5 v^{2} - 4 v + 1

and the divisor

v + 1 : \frac{- 5 v ^{2}}{v} = - 5 v

Quotient = - 5 v

Multiply

v + 1

- 5 v : - 5 v^{2} - 5 v

Subtract

- 5 v^{2} - 5 v

from

- 5 v^{2} - 4 v + 1

to get new remainder

Remainder = v + 1

Therefore

\frac{- 5 v ^{2} - 4 v + 1}{v + 1} = - 5 v + \frac{v + 1}{v + 1}

= v^{3} - 5 v^{2} - 5 v + \frac{v + 1}{v + 1}

Divide

\frac{v + 1}{v + 1} : \frac{v + 1}{v + 1} = 1

Divide the leading coefficients of the numerator

v + 1

and the divisor

v + 1 : \frac{v}{v} = 1

Quotient = 1

Multiply

v + 1

1 : v + 1

Subtract

v + 1

from

v + 1

to get new remainder

Remainder = 0

Therefore

\frac{v + 1}{v + 1} = 1

= v^{3} - 5 v^{2} - 5 v + 1

= v^{3} - 5 v^{2} - 5 v + 1

Factor

v^{3} - 5 v^{2} - 5 v + 1 : (v + 1) (v^{2} - 6 v + 1)

v^{3} - 5 v^{2} - 5 v + 1

Use the rational root theorem

a_{0} = 1, a_{n} = 1

The dividers of

a_{0} : 1,

The dividers of

a_{n} : 1

Therefore, check the following rational numbers:

\pm \frac{1}{1}

- \frac{1}{1}

is a root of the expression, so factor out

v + 1

= (v + 1) \frac{v ^{3} - 5 v ^{2} - 5 v + 1}{v + 1}

\frac{v ^{3} - 5 v ^{2} - 5 v + 1}{v + 1} = v^{2} - 6 v + 1

\frac{v ^{3} - 5 v ^{2} - 5 v + 1}{v + 1}

Divide

\frac{v ^{3} - 5 v ^{2} - 5 v + 1}{v + 1} : \frac{v ^{3} - 5 v ^{2} - 5 v + 1}{v + 1} = v^{2} + \frac{- 6 v ^{2} - 5 v + 1}{v + 1}

Divide the leading coefficients of the numerator

v^{3} - 5 v^{2} - 5 v + 1

and the divisor

v + 1 : \frac{v ^{3}}{v} = v^{2}

Quotient = v^{2}

Multiply

v + 1

v^{2} : v^{3} + v^{2}

Subtract

v^{3} + v^{2}

from

v^{3} - 5 v^{2} - 5 v + 1

to get new remainder

Remainder = - 6 v^{2} - 5 v + 1

Therefore

\frac{v ^{3} - 5 v ^{2} - 5 v + 1}{v + 1} = v^{2} + \frac{- 6 v ^{2} - 5 v + 1}{v + 1}

= v^{2} + \frac{- 6 v ^{2} - 5 v + 1}{v + 1}

Divide

\frac{- 6 v ^{2} - 5 v + 1}{v + 1} : \frac{- 6 v ^{2} - 5 v + 1}{v + 1} = - 6 v + \frac{v + 1}{v + 1}

Divide the leading coefficients of the numerator

- 6 v^{2} - 5 v + 1

and the divisor

v + 1 : \frac{- 6 v ^{2}}{v} = - 6 v

Quotient = - 6 v

Multiply

v + 1

- 6 v : - 6 v^{2} - 6 v

Subtract

- 6 v^{2} - 6 v

from

- 6 v^{2} - 5 v + 1

to get new remainder

Remainder = v + 1

Therefore

\frac{- 6 v ^{2} - 5 v + 1}{v + 1} = - 6 v + \frac{v + 1}{v + 1}

= v^{2} - 6 v + \frac{v + 1}{v + 1}

Divide

\frac{v + 1}{v + 1} : \frac{v + 1}{v + 1} = 1

Divide the leading coefficients of the numerator

v + 1

and the divisor

v + 1 : \frac{v}{v} = 1

Quotient = 1

Multiply

v + 1

1 : v + 1

Subtract

v + 1

from

v + 1

to get new remainder

Remainder = 0

Therefore

\frac{v + 1}{v + 1} = 1

= v^{2} - 6 v + 1

= v^{2} - 6 v + 1

= (v + 1) (v^{2} - 6 v + 1)

= (v + 1) (v + 1) (v^{2} - 6 v + 1)

Refine

= (v + 1)^{2} (v^{2} - 6 v + 1)

(v + 1)^{2} (v^{2} - 6 v + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

v + 1 = 0 or v^{2} - 6 v + 1 = 0

Solve

v + 1 = 0 : v = - 1

v + 1 = 0

Move

1

to the right side

v + 1 = 0

Subtract

1

from both sides

v + 1 - 1 = 0 - 1

Simplify

v = - 1

v = - 1

Solve

v^{2} - 6 v + 1 = 0 : v = 3 + 22, v = 3 - 22

v^{2} - 6 v + 1 = 0

Solve with the quadratic formula

v^{2} - 6 v + 1 = 0

Quadratic Equation Formula:

For

a = 1, b = - 6, c = 1

v_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

v_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot 1 = 42

(- 6)^{2} - 4 \cdot 1 \cdot 1

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 6)^{2} = 6^{2}

= 6^{2} - 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 6^{2} - 4

6^{2} = 36

= 36 - 4

Subtract the numbers:

36 - 4 = 32

= 32

Prime factorization of

32 : 2^{5}

32

32

divides by

232 = 16 \cdot 2

= 2 \cdot 16

16

divides by

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

= 2^{5}

= 2^{5}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{4} \cdot 2

Apply radical rule:

= 2 2^{4}

Apply radical rule:

2^{4} = 2^{\frac{4}{2}} = 2^{2}

= 2^{2} 2

Refine

= 42

v_{1, 2} = \frac{- ( - 6 ) \pm 4 2}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- ( - 6 ) + 4 2}{2 \cdot 1}, v_{2} = \frac{- ( - 6 ) - 4 2}{2 \cdot 1}

v = \frac{- ( - 6 ) + 4 2}{2 \cdot 1} : 3 + 22

\frac{- ( - 6 ) + 4 2}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{6 + 4 2}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{6 + 4 2}{2}

Factor

6 + 42 : 2 (3 + 22)

6 + 42

Rewrite as

= 2 \cdot 3 + 2 \cdot 22

Factor out common term

2

= 2 (3 + 22)

= \frac{2 ( 3 + 2 2 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 3 + 22

v = \frac{- ( - 6 ) - 4 2}{2 \cdot 1} : 3 - 22

\frac{- ( - 6 ) - 4 2}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{6 - 4 2}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{6 - 4 2}{2}

Factor

6 - 42 : 2 (3 - 22)

6 - 42

Rewrite as

= 2 \cdot 3 - 2 \cdot 22

Factor out common term

2

= 2 (3 - 22)

= \frac{2 ( 3 - 2 2 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 3 - 22

The solutions to the quadratic equation are:

v = 3 + 22, v = 3 - 22

The solutions are

v = - 1, v = 3 + 22, v = 3 - 22

v = - 1, v = 3 + 22, v = 3 - 22

Substitute back

v = u^{2},

solve for

u

Solve

u^{2} = - 1 :

No Solution for

u \in R

u^{2} = - 1

x^{2}

cannot be negative for

x \in R

No Solution for u \in R

Solve

u^{2} = 3 + 22 : u = 2 + 1

u^{2} = 3 + 22

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 3 + 22, u = - 3 + 22

3 + 22 = 2 + 1

3 + 22

= 2 + 22 + 1

= (2)^{2} + 22 + (1)^{2}

1 = 1

1

Apply rule

1 = 1

= 1

= (2)^{2} + 22 + 1^{2}

22 \cdot 1 = 22

22 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 22

= (2)^{2} + 22 \cdot 1 + 1^{2}

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

(2)^{2} + 22 \cdot 1 + 1^{2} = (2 + 1)^{2}

= (2 + 1)^{2}

Apply radical rule:

(2 + 1)^{2} = 2 + 1

= 2 + 1

- 3 + 22 = 2 + 1

- 3 + 22

3 + 22 = 2 + 1

3 + 22

= 2 + 22 + 1

= (2)^{2} + 22 + (1)^{2}

1 = 1

1

Apply rule

1 = 1

= 1

= (2)^{2} + 22 + 1^{2}

22 \cdot 1 = 22

22 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 22

= (2)^{2} + 22 \cdot 1 + 1^{2}

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

(2)^{2} + 22 \cdot 1 + 1^{2} = (2 + 1)^{2}

= (2 + 1)^{2}

Apply radical rule:

(2 + 1)^{2} = 2 + 1

= 2 + 1

= 2 + 1

u = 2 + 1

Solve

u^{2} = 3 - 22 : u = 2 - 1

u^{2} = 3 - 22

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 3 - 22, u = - 3 - 22

3 - 22 = 2 - 1

3 - 22

= 2 - 22 + 1

= (2)^{2} - 22 + (1)^{2}

1 = 1

1

Apply rule

1 = 1

= 1

= (2)^{2} - 22 + 1^{2}

22 \cdot 1 = 22

22 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 22

= (2)^{2} - 22 \cdot 1 + 1^{2}

Apply Perfect Square Formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

(2)^{2} - 22 \cdot 1 + 1^{2} = (2 - 1)^{2}

= (2 - 1)^{2}

Apply radical rule:

(2 - 1)^{2} = 2 - 1

= 2 - 1

- 3 - 22 = 2 - 1

- 3 - 22

3 - 22 = 2 - 1

3 - 22

= 2 - 22 + 1

= (2)^{2} - 22 + (1)^{2}

1 = 1

1

Apply rule

1 = 1

= 1

= (2)^{2} - 22 + 1^{2}

22 \cdot 1 = 22

22 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 22

= (2)^{2} - 22 \cdot 1 + 1^{2}

Apply Perfect Square Formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

(2)^{2} - 22 \cdot 1 + 1^{2} = (2 - 1)^{2}

= (2 - 1)^{2}

Apply radical rule:

(2 - 1)^{2} = 2 - 1

= 2 - 1

= 2 - 1

u = 2 - 1

The solutions are

u = 2 + 1, u = 2 - 1

u = 2 + 1, u = 2 - 1

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

(\frac{2}{u + u ^{- 1}})^{4}

and compare to zero

u = 0

Take the denominator(s) of

(\frac{u - u ^{- 1}}{u + u ^{- 1}})^{4}

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = 2 + 1, u = 2 - 1

u = 2 + 1, u = 2 - 1

Substitute back

u = e^{x},

solve for

x

Solve

e^{x} = 2 + 1 : x = ln (2 + 1)

e^{x} = 2 + 1

Apply exponent rules

e^{x} = 2 + 1

f (x) = g (x)

, then

ln (f (x)) = ln (g (x))

ln (e^{x}) = ln (2 + 1)

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (2 + 1)

x = ln (2 + 1)

Solve

e^{x} = 2 - 1 : x = ln (2 - 1)

e^{x} = 2 - 1

Apply exponent rules

e^{x} = 2 - 1

f (x) = g (x)

, then

ln (f (x)) = ln (g (x))

ln (e^{x}) = ln (2 - 1)

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (2 - 1)

x = ln (2 - 1)

x = ln (2 + 1), x = ln (2 - 1)

x = ln (2 + 1), x = ln (2 - 1)

Popular Examples

(sin(x))/(2+cos(x)-1)=0

\frac{sin ( x )}{2 + cos ( x ) - 1} = 0

solvefor y,arcsin(y)=ln(x)solve for

y, arcsin (y) = ln (x)

sin(9x+6)=cos(3x-4)

sin (9 x + 6) = cos (3 x - 4)

csc(x)= 1/(sec(x))

csc (x) = \frac{1}{sec ( x )}

sin(x+30)=cos(2x)

sin (x + 3 0^{\circ}) = cos (2 x)

Frequently Asked Questions (FAQ)

What is the general solution for (sech^4(x))=tanh^4(x) ?
The general solution for (sech^4(x))=tanh^4(x) is x=ln(sqrt(2)+1),x=ln(sqrt(2)-1)