<span>1/(4p)(x-h)^2+k=0
</span><span>1/(4p)(x-h)^2 = -k
</span>
<span>k(4p)(x-h)^2+1=0
4kp (x^2 - 2xh + h^2) + 1 = 0
4kp x^2 - 8kph x + 4kph^2+1 = 0
D = (-8kph)^2 - 4(4kp)(4kph^2+1) = 64(kph)^2 - 64(kph)^2 - 16kp
D = -16kp < 0
SO discriminant is always less than 0
</span>
Let x and y be the 2 parts of 15 ==> x + y=15 (given)
Reciprocal of x and y ==> 1/x +1/y ==> 1/x + 1/y = 3/10 (given)
Let's solve 1/x + 1/y = 3/10 . Common denominator = 10.x.y (reduce to same denominator)
==> (10y+10x)/10xy = 3xy/10xy ==> 10x+10y =3xy
But x+y = 15 , then 10x+10y =150 ==> 150=3xy and xy = 50
Now we have the sum S of the 2 parts that is S = 15 and
their Product = xy =50
Let's use the quadratic equation for S and P==> X² -SX +P =0
Or X² - 15X + 50=0, Solve for X & you will find:
The 1st part of 15 is 10 & the 2nd part is 5
Answer:
look at the workey??
Step-by-step explanation:
Answer:
D,0,2,-2
Step-by-step explanation:
2x^5-3x^3-20x=0
x(2x^4-3x^2-20)=0
x=0
or 2x^4-3x^2-20=0
put x²=t
2t²-3t-20=0
-20×2=-40
8-5=3
8×-5=-40
2t²-(8-5)t-20=0
2t²-8t+5t-20=0
2t(t-4)+5(t-4)=0
(t-4)(2t+5)=0
t=4
x²=4
x=2,-2
t=-5/2
x²=-5/2
it gives imaginary root. so real rational roots are 0,2,-2
(12x3 - 14x2 - 40x) / (2x-5)
Rewriting we have:
(2x (2x-5) (3x + 4)) / (2x-5)
We simplify similar terms:
2x * (3x + 4)
We rewrite:
6x2 + 8x
Answer:
The simplified expression is given in this case by:
6x2 + 8x
