If an angle A of a [ ]ABC satiesfies 5 cosA + 3 = 0, then the roots of the quadratic equat…

Question

If an angle A of a $$\Delta $$ABC satiesfies 5 cosA + 3 = 0,  then the roots of the quadratic equation, 9x2 + 27x + 20 = 0 are :

Accepted Answer

Here, 9x2 + 27x + 20 = 0

$$\therefore\,\,\,$$ x = $${{ - b \pm \sqrt {{b^2} - 4ac} } \over {2a}}$$

$$ \Rightarrow $$ $\,\,\,$x =${{ - 27 \pm \sqrt {{{27}^2} - 4 \times 9 \times 20} } \over {2 \times 9}}$$ \Rightarrow $$ $\,\,\,$$ x = - $${4 \over 3}$$, - $${5 \over 3}$$

Given, cosA = - $${3 \over 5}$$

$$\therefore\,\,\,$$ sec A = $${1 \over {\cos A}}$$ = - $${5 \over 3}$$

Here, A is an obtuse angle.

$$\therefore\,\,\,$$ tan A = - $$\sqrt {{{\sec }^2}A - 1}  =  - {4 \over 3}.$$

Hence, roots of the equation are sec A and tan A.

Solutions · 1