What is the general solution for sec^2(x)+2sec(x)+4=0 ?

The general solution for sec^2(x)+2sec(x)+4=0 is No Solution for x\in\mathbb{R}

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Popular >

sec^2(x)+2sec(x)+4=0

Solution

sec^{2} (x) + 2 sec (x) + 4 = 0

Solution

No Solution for x \in R

Solution steps

sec^{2} (x) + 2 sec (x) + 4 = 0

Solve by substitution

sec^{2} (x) + 2 sec (x) + 4 = 0

Let:

sec (x) = u

u^{2} + 2 u + 4 = 0

u^{2} + 2 u + 4 = 0 : u = - 1 + 3 i, u = - 1 - 3 i

u^{2} + 2 u + 4 = 0

Solve with the quadratic formula

u^{2} + 2 u + 4 = 0

Quadratic Equation Formula:

For

a = 1, b = 2, c = 4

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 4}{2 \cdot 1}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 4}{2 \cdot 1}

Simplify

2^{2} - 4 \cdot 1 \cdot 4 : 23 i

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

Multiply the numbers:

4 \cdot 1 \cdot 4 = 16

= 2^{2} - 16

Apply imaginary number rule:

- a = i a

= i 16 - 2^{2}

- 2^{2} + 16 = 23

- 2^{2} + 16

2^{2} = 4

= - 4 + 16

Add/Subtract the numbers:

- 4 + 16 = 12

= 12

Prime factorization of

12 : 2^{2} \cdot 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

Apply radical rule:

= 3 2^{2}

Apply radical rule:

2^{2} = 2

= 23

= 23 i

u_{1, 2} = \frac{- 2 \pm 2 3 i}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 2 + 2 3 i}{2 \cdot 1}, u_{2} = \frac{- 2 - 2 3 i}{2 \cdot 1}

u = \frac{- 2 + 2 3 i}{2 \cdot 1} : - 1 + 3 i

\frac{- 2 + 2 3 i}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2 + 2 3 i}{2}

Factor

- 2 + 23 i : 2 (- 1 + 3 i)

- 2 + 23 i

Rewrite as

= - 2 \cdot 1 + 23 i

Factor out common term

2

= 2 (- 1 + 3 i)

= \frac{2 ( - 1 + 3 i )}{2}

Divide the numbers:

\frac{2}{2} = 1

= - 1 + 3 i

u = \frac{- 2 - 2 3 i}{2 \cdot 1} : - 1 - 3 i

\frac{- 2 - 2 3 i}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2 - 2 3 i}{2}

Factor

- 2 - 23 i : - 2 (1 + 3 i)

- 2 - 23 i

Rewrite as

= - 2 \cdot 1 - 23 i

Factor out common term

2

= - 2 (1 + 3 i)

= - \frac{2 ( 1 + 3 i )}{2}

Divide the numbers:

\frac{2}{2} = 1

= - (1 + 3 i)

Negate

- (1 + 3 i) = - 1 - 3 i

= - 1 - 3 i

The solutions to the quadratic equation are:

u = - 1 + 3 i, u = - 1 - 3 i

Substitute back

u = sec (x)

sec (x) = - 1 + 3 i, sec (x) = - 1 - 3 i

sec (x) = - 1 + 3 i, sec (x) = - 1 - 3 i

sec (x) = - 1 + 3 i :

No Solution

sec (x) = - 1 + 3 i

No Solution

sec (x) = - 1 - 3 i :

No Solution

sec (x) = - 1 - 3 i

No Solution

Combine all the solutions

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for sec^2(x)+2sec(x)+4=0 ?
The general solution for sec^2(x)+2sec(x)+4=0 is No Solution for x\in\mathbb{R}