replace with θ for more traditional appearance
if sinθ and cosθ are roots of a quadratic, then the quad must be
k(x - sinθ)(x - cosθ) = 0
kx^2 - kxcosθ - kxsinθ + ksinθcosθ = 0
kx^2 - kx(cosθ + sinθ) + ksinθcosθ = 0
compare this with 4x^2 - 5x + a = 0 , then k = 4
and my equation becomes:
4x^2 -4x(cosθ + sinθ) + 4sinθcosθ = 0
further comparison:
-4x(cosθ+sinθ) = -5x
cosθ + sinθ = 5/4
and a = 4sinθcosθ
we know ....
(sinθ + cosθ)^2 = sin^2 θ + cos^2 θ + 2sinθcosθ
25/16 = 1 + (1/2)a
9/16 = a/2
a = 9/8
Yeahhh, you are right.
If sin and cos are the two roots to the equation 4x2-5x+a=0, then the value of a should be
answer= 9/8
2 answers
Thank you so much