Question

If an angle a of a abc satisfies 5 cos a + 3 = 0, then the roots of the quadratic equation 9x2 + 27x + 20 = 0 are

Answer 1

Answer:

sec A, tan A

Step-by-step explanation:

Given If an angle a of abc satisfies 5 cos a + 3 = 0, then the roots of the quadratic equation 9x2 + 27x + 20 = 0 are

Eqn is 5 cos a + 3 = 0

     So 5 cos a = - 3

          cos a = -3/5

So sec A = 1 / cos A

sec A = 1 / - 3/5

sec A = - 5 / 3

sec ^2 A - tan ^2 A = 1

tan^2 A = sec ^2 A - 1

tan A = √sec ^A - 1

         = √25/9 - 1

        = √16/9

tan A = 4/3

So roots of the equation are  sec A, tan A

Answer 2

➡️Answer:

Roots of Quadratic equation are tan a and sec a

➡️Solution:

Roots of
$9 {x}^{2} + 27x + 20 = 0 \\ \\ quadratic \: formula \\ \\ x_{1,2} = \frac{ - b ± \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ x_{1,2} = \frac{ - 27 ± \sqrt{ {(27)}^{2} - 4 \times 9 \times 20} }{2 \times 9} \\ \\ x_{1,2}= \frac{ - 27 ± \sqrt{729 - 720} }{18} \\ \\ x_{1,2} = \frac{ - 27 ±3}{18} \\ \\ x_{1} = \frac{ - 27 + 3}{18} = - \frac{24}{18} = \frac{ - 4}{3} \\ \\ x_{2} = \frac{ - 27 - 3}{18} = - \frac{30}{18} = \frac{ - 5}{3}$
As it is given that

$5 \: cos \: a + 3 = 0 \\ \\ cos \: a = \frac{ - 3}{5} \\ \\ so \\ sec \: a = \frac{ - 5}{3} = x_{2}$

By this way find the other trigonometric ratio for another root.

As we know that sinA =√(1-cos^2 a)

sin a= 4/5

tan a = 4/5/-3/5=-4/3= x1

So,roots of given Quadratic equation are tan a and sec a.

Hope it helps you.