F4 A.Maths

if sinθ and secθ(0<θ<π/2) are the roots 0f the equation 2x2+kx+1=0,find the value of k.
依題設，方程式
2x2+kx+1=0 的
兩根和
sinθ + secθ = -k/2
兩根積
sinθ secθ = 1/2
tanθ = 1/2
sinθ = 1/√5
cosθ = 2/√5
k = -2(sinθ + secθ)
k = -2(1/√5+ √5/2)
k = -2[(2 + 5)/2√5]
k = -(7/√5)
k = -(7√5)/5

Using product of roots, we have

\sin\theta * \sec\theta = 1/2,
i.e. \tan\theta = 1/2.

Since 0&lt;\theta&lt;\pi/2, we can consider a right-angle triangle with opposite side equal to 1 and adjacent side equal to 2. Then, we have

\sin\theta = 1/\sqrt{5} and \sec\theta = \sqrt{5}/2.

Now, using sum of roots, we have

\sin\theta + \sec\theta = -k/2
1/\sqrt{5} + \sqrt{5}/2 = -k/2
7/(2*\sqrt{5} = -k/2
k = -7/\sqrt{5}
i.e. k = -7\sqrt{5}/5.