What is the general solution for tan^2(x)-9tan(x)-10=0 ?

The general solution for tan^2(x)-9tan(x)-10=0 is x=1.47112…+pin,x=(3pi)/4+pin

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tan^2(x)-9tan(x)-10=0

Solution

tan^{2} (x) - 9 tan (x) - 10 = 0

Solution

x = 1.47112 \dots + πn, x = \frac{3 π}{4} + πn

Degrees

x = 84.28940 \dots^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (x) - 9 tan (x) - 10 = 0

Solve by substitution

tan^{2} (x) - 9 tan (x) - 10 = 0

Let:

tan (x) = u

u^{2} - 9 u - 10 = 0

u^{2} - 9 u - 10 = 0 : u = 10, u = - 1

u^{2} - 9 u - 10 = 0

Solve with the quadratic formula

u^{2} - 9 u - 10 = 0

Quadratic Equation Formula:

For

a = 1, b = - 9, c = - 10

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot ( - 10 )}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot ( - 10 )}{2 \cdot 1}

(- 9)^{2} - 4 \cdot 1 \cdot (- 10) = 11

(- 9)^{2} - 4 \cdot 1 \cdot (- 10)

Apply rule

- (- a) = a

= (- 9)^{2} + 4 \cdot 1 \cdot 10

Apply exponent rule:

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

n

is even

(- 9)^{2} = 9^{2}

= 9^{2} + 4 \cdot 1 \cdot 10

Multiply the numbers:

4 \cdot 1 \cdot 10 = 40

= 9^{2} + 40

9^{2} = 81

= 81 + 40

Add the numbers:

81 + 40 = 121

= 121

Factor the number:

121 = 1 1^{2}

= 1 1^{2}

Apply radical rule:

1 1^{2} = 11

= 11

u_{1, 2} = \frac{- ( - 9 ) \pm 11}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 9 ) + 11}{2 \cdot 1}, u_{2} = \frac{- ( - 9 ) - 11}{2 \cdot 1}

u = \frac{- ( - 9 ) + 11}{2 \cdot 1} : 10

\frac{- ( - 9 ) + 11}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{9 + 11}{2 \cdot 1}

Add the numbers:

9 + 11 = 20

= \frac{20}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{20}{2}

Divide the numbers:

\frac{20}{2} = 10

= 10

u = \frac{- ( - 9 ) - 11}{2 \cdot 1} : - 1

\frac{- ( - 9 ) - 11}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{9 - 11}{2 \cdot 1}

Subtract the numbers:

9 - 11 = - 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = 10, u = - 1

Substitute back

u = tan (x)

tan (x) = 10, tan (x) = - 1

tan (x) = 10, tan (x) = - 1

tan (x) = 10 : x = arctan (10) + πn

tan (x) = 10

Apply trig inverse properties

tan (x) = 10

General solutions for

tan (x) = 10

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (10) + πn

x = arctan (10) + πn

tan (x) = - 1 : x = \frac{3 π}{4} + πn

tan (x) = - 1

General solutions for

tan (x) = - 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

Combine all the solutions

x = arctan (10) + πn, x = \frac{3 π}{4} + πn

Show solutions in decimal form

x = 1.47112 \dots + πn, x = \frac{3 π}{4} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for tan^2(x)-9tan(x)-10=0 ?
The general solution for tan^2(x)-9tan(x)-10=0 is x=1.47112…+pin,x=(3pi)/4+pin