Solveeit Logo
Login

Question

Question: If \( \sec \alpha \) and \( \cos ec\alpha \) are the roots of the equation \( {x^2} + px + q = 0 \) ...

If sec⁡α\sec \alpha and cos⁡ecα\cos ec\alpha are the roots of the equation x2+px+q=0{x^2} + px + q = 0 , then:
(A) p2=p+2q{p^2} = p + 2q
(B) q2=p+2q{q^2} = p + 2q
(C) p2=q(q+2){p^2} = q\left( {q + 2} \right)
(D) q2=p(p+2){q^2} = p\left( {p + 2} \right)
(E) p2=q(q−2){p^2} = q\left( {q - 2} \right)

Explanation

Solution

Hint : In the given problem, we are given that the roots of the equation x2+px+q=0{x^2} + px + q = 0 are the trigonometric functions sec⁡α\sec \alpha and cos⁡ecα\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 x2+px+q=0{x^2} + px + q = 0 .
Then, we know the sum of the roots of a quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0 is given by (−ba)\left( { - \dfrac{b}{a}} \right) . Also, the product of roots is given by (ca)\left( {\dfrac{c}{a}} \right) .
So, the sum of roots of the equation x2+px+q=0{x^2} + px + q = 0 is −p- p and the product of roots is qq .
We also know that the roots of the quadratic equation are: sec⁡α\sec \alpha and cos⁡ecα\cos ec\alpha .
Hence, we get, sec⁡α+cos⁡ecα=−p\sec \alpha + \cos ec\alpha = - p and sec⁡αcos⁡ecα=q\sec \alpha \cos ec\alpha = q .
Converting the trigonometric functions secant and cosecant into sine and cosine, we get,
1cos⁡α+1sin⁡α=−p\dfrac{1}{{\cos \alpha }} + \dfrac{1}{{\sin \alpha }} = - p and 1sin⁡αcos⁡α=q\dfrac{1}{{\sin \alpha \cos \alpha }} = q .
Now, we take reciprocals on both sides of the equation 1sin⁡αcos⁡α=q\dfrac{1}{{\sin \alpha \cos \alpha }} = q . So, we get,
⇒sin⁡αcos⁡α=1q\Rightarrow \sin \alpha \cos \alpha = \dfrac{1}{q}
Now, taking the LCM of denominators in equation 1cos⁡α+1sin⁡α=−p\dfrac{1}{{\cos \alpha }} + \dfrac{1}{{\sin \alpha }} = - p .
So, we get, sin⁡α+cos⁡αsin⁡αcos⁡α=−p\dfrac{{\sin \alpha + \cos \alpha }}{{\sin \alpha \cos \alpha }} = - p
Substituting the value of sin⁡αcos⁡α\sin \alpha \cos \alpha in the equation, we get,
⇒sin⁡α+cos⁡α=(−p)×sin⁡αcos⁡α\Rightarrow \sin \alpha + \cos \alpha = \left( { - p} \right) \times \sin \alpha \cos \alpha
⇒sin⁡α+cos⁡α=−pq\Rightarrow \sin \alpha + \cos \alpha = - \dfrac{p}{q}
Squaring both sides of the equation, we get,
⇒(sin⁡α+cos⁡α)2=(−pq)2\Rightarrow {\left( {\sin \alpha + \cos \alpha } \right)^2} = {\left( { - \dfrac{p}{q}} \right)^2}
Now, using the algebraic identity (a+b)2=a2+2ab+b2{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} , we get,
⇒sin⁡2α+2sin⁡αcos⁡α+cos⁡2α=(−pq)2\Rightarrow {\sin ^2}\alpha + 2\sin \alpha \cos \alpha + {\cos ^2}\alpha = {\left( { - \dfrac{p}{q}} \right)^2}
We know that trigonometric identity sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1 . So, we get,
⇒1+2sin⁡αcos⁡α=p2q2\Rightarrow 1 + 2\sin \alpha \cos \alpha = \dfrac{{{p^2}}}{{{q^2}}}
Substituting the value of sin⁡αcos⁡α\sin \alpha \cos \alpha , we get,
⇒1+2q=p2q2\Rightarrow 1 + \dfrac{2}{q} = \dfrac{{{p^2}}}{{{q^2}}}
Multiplying both sides of the equation by q2{q^2} . So, we get,
⇒q2+2q=p2\Rightarrow {q^2} + 2q = {p^2}
Factoring out the common terms, we get,
⇒p2=q(q+2)\Rightarrow {p^2} = q\left( {q + 2} \right)
Hence, p2=q(q+2){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 22 . 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.