Question

if sin and cos are the two roots to the equation 32x^2-ax+9=0, then value of a can be

Answer 1

x^2 - a/32 x + 9/32 = 0
(x - sin t )(x-cos t) = 0
x^2 - (sin t + cos t) x + sin t cos t = 0
so
a/32 = sin t + cos t
9/32 = sin t cos t ..... so sin t = 9 / 32cos t
a/32 = 8 / 32cos t + cos t = 40/32 cos t
a = 40 cos t
unfortunately I do not know what t is.

Answer 2

thank u!!

Answer 3

I did a similar question for somebody yesterday, but with Jishka' non-functional Search feature, I can't find it

So:
Using the sum and product of roots of a quadratic property:
Since sint and cost are roots of the equation,
the sum of the roots = sint + cost = a/32
and the product of the roots = sintcost = 9/32

(sint + cost)^2 = sin^2 t + 2sintcost + cos^2 t) , but sin^2 t + cos^2 t = 1
a^2 / 1024 = 1 + 2(9/32)
a^2 / 1024 = 25/16
a^2 = 1600
a = ±√1600 = ± 40

Answer 4

thank u!!!!!