Question

If $ \sec \alpha $ and $ \cos ec\alpha $ are the roots of the equation $ {x^2} + px + q = 0 $ , then:
(A) $ {p^2} = p + 2q $
(B) $ {q^2} = p + 2q $
(C) $ {p^2} = q\left( {q + 2} \right) $
(D) $ {q^2} = p\left( {p + 2} \right) $
(E) $ {p^2} = q\left( {q - 2} \right) $

Answer 1

Hint: In the given problem, we are given that the roots of the equation $ {x^2} + px + q = 0 $ are the trigonometric functions $ \sec \alpha $ and $ \cos ec\alpha $ . Hence, we will use the relationships between the roots of the equation and the coefficients of the terms. The trigonometric formulae and algebraic identity will come into significant use once we start solving the problem.

Complete step-by-step answer:
So, we are given the quadratic equation $ {x^2} + px + q = 0 $ .
Then, we know the sum of the roots of a quadratic equation $ a{x^2} + bx + c = 0 $ is given by $ \left( { - \dfrac{b}{a}} \right) $ . Also, the product of roots is given by $ \left( {\dfrac{c}{a}} \right) $ .
So, the sum of roots of the equation $ {x^2} + px + q = 0 $ is $ - p $ and the product of roots is $ q $ .
We also know that the roots of the quadratic equation are: $ \sec \alpha $ and $ \cos ec\alpha $ .
Hence, we get, $ \sec \alpha + \cos ec\alpha = - p $ and $ \sec \alpha \cos ec\alpha = q $ .
Converting the trigonometric functions secant and cosecant into sine and cosine, we get,
 $ \dfrac{1}{{\cos \alpha }} + \dfrac{1}{{\sin \alpha }} = - p $ and $ \dfrac{1}{{\sin \alpha \cos \alpha }} = q $ .
Now, we take reciprocals on both sides of the equation $ \dfrac{1}{{\sin \alpha \cos \alpha }} = q $ . So, we get,
 $ \Rightarrow \sin \alpha \cos \alpha = \dfrac{1}{q} $
Now, taking the LCM of denominators in equation $ \dfrac{1}{{\cos \alpha }} + \dfrac{1}{{\sin \alpha }} = - p $ .
So, we get, $ \dfrac{{\sin \alpha + \cos \alpha }}{{\sin \alpha \cos \alpha }} = - p $
Substituting the value of $ \sin \alpha \cos \alpha $ in the equation, we get,
 $ \Rightarrow \sin \alpha + \cos \alpha = \left( { - p} \right) \times \sin \alpha \cos \alpha $
 $ \Rightarrow \sin \alpha + \cos \alpha = - \dfrac{p}{q} $
Squaring both sides of the equation, we get,
 $ \Rightarrow {\left( {\sin \alpha + \cos \alpha } \right)^2} = {\left( { - \dfrac{p}{q}} \right)^2} $
Now, using the algebraic identity $ {\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} $ , we get,
 $ \Rightarrow {\sin ^2}\alpha + 2\sin \alpha \cos \alpha + {\cos ^2}\alpha = {\left( { - \dfrac{p}{q}} \right)^2} $
We know that trigonometric identity $ {\sin ^2}x + {\cos ^2}x = 1 $ . So, we get,
 $ \Rightarrow 1 + 2\sin \alpha \cos \alpha = \dfrac{{{p^2}}}{{{q^2}}} $
Substituting the value of $ \sin \alpha \cos \alpha $ , we get,
 $ \Rightarrow 1 + \dfrac{2}{q} = \dfrac{{{p^2}}}{{{q^2}}} $
Multiplying both sides of the equation by $ {q^2} $ . So, we get,
 $ \Rightarrow {q^2} + 2q = {p^2} $
Factoring out the common terms, we get,
 $ \Rightarrow {p^2} = q\left( {q + 2} \right) $
Hence, $ {p^2} = q\left( {q + 2} \right) $ .
So, the correct answer is “Option C”.

Note: Quadratic equations are the polynomial equations with degree of the variable or unknown as $ 2 $ . Relationships between the coefficients of the terms and the roots of the equation must be known so as to be able to solve the questions. Care should be taken while doing the calculations as it can change our final answer.