thats funny!! love that!!!! please mark me
(xy)' + (2x)' + (3x^2)' = (4)'
y + xy' + 2 + 6x = 0
xy' = -y -2 -6x
y' = [-y -2 -6x] / x
Now solve y from the original equation and substitue
xy + 2x + 3x^2 = 4 => y = [-2x - 3x^2 + 4] / x
y' = [(-2x - 3x^2 +4) / x - 2 - 6x ] / x
y' = [-2x - 3x^2 + 4 -2x -6x^2 ] x^2 = [ -4x - 9x^2 + 4] / x^2 =
= [-9x^2 - 4x + 4] / x^2
4
is the answer.
:D
jen will have 4 mor epigs than mrty
Answer:
x = 5 and x = -19
Step-by-step explanation:
You're on the right track. It's the "discriminant" that tells you what you want to know here. Before starting, arrange the terms of your quadratic in descending orders of x: 5x^2 + 14x - 19 = 0 (Note that I assumed you meant 14x instead of just 14).
Then the coefficients of this quadratic are a = 5, b = 14 and c = -19.
You are referring to the "quadratic formula." It states this:
-b ± √(b²-4ac)
x = -----------------------
2a
So, we insert the a, b and c values as indicated above:
-14 ± √( 14² - 4[5][-19] ) -14 ± √(196 - 4[5][-19] ) -14 ± √576
x = ----------------------------------- = ---------------------------------- = ----------------------
2(10) 20 20
This comes out to:
x = (-14 + 24) / 2 and x = (-14 - 24) / 2
or:
x = 5 and x = -19
