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Topic review (newest first)
- mikau
- 2005-12-30 11:35:55
bery good? are you being funny?
- krassi_holmz
- 2005-12-30 10:24:47
I presentiment katy will become bery good in Maths.
- mikau
- 2005-12-30 10:20:02
Not stupid. You just need practice. Math isn't hard. It just takes some getting used to.
- katy
- 2005-12-30 10:10:01
mikau wrote:
Nah we just pretend to be.
lol...of course you all are....I am the stupid one..hehehe
- mikau
- 2005-12-30 10:04:28
Nah we just pretend to be.
- katy
- 2005-12-30 09:21:51
thanks a lot guys...i get them now...you all are smart.
katy.
- mikau
- 2005-12-30 07:35:32
2 real roots when the discriminant is positve. 2 imaginary roots when the discriminant is negative. Two equal roots when the discriminant is zero.
If you forget what the descriminant is:
You can find the roots of a quadratic equation of the form:
ax^2 + bx + c
using the quadratic formula:
(-b +- √ b^2 - 4ac)/2a
the part of the formula inside the radical side is the descriminant.
√ b^2 - 4ac
when the descriminant (b^2 - 4ac) is positive, it has two real roots. When it is negative, it will require the square root of a negative number and thus only has imaginary roots. When the discriminant is 0 the roots will be -b/2a + 0, and -b/2a - 0 which are obviously equal.
- krassi_holmz
- 2005-12-30 06:57:03
For the 1b problem.
The parabloa is quadratic surve. So you're right.
- katy
- 2005-12-30 06:44:09
thanks a lot guys....
I have one more question:
1)Explain when a parabola will have 2 real roots, 2 imaginary roots, and 2 equal real roots.Explain what characteristics the discriminant will have for the above cases
My answer is:
discriminant>0 (2 real roots)
discriminant<0(2 imaginary roots)
discriminant=0(2 equal real roots)
is it right?
- krassi_holmz
- 2005-12-30 06:30:22
Of course, but I thought that you are not permissed to calculate the discriminant.
And something else :
10>3^2 => discriminant is > 0.
- mathsyperson
- 2005-12-30 06:12:26
For 2) it doesn't ask you to find the values of x, just the discriminant.
The discriminant is the bit inside the square root of the quadratic equation:
In this case, the discriminant would be 3² - 4*2*(-10) = 89.
Because it is positive, this shows that both roots will be real and distinct.
- krassi_holmz
- 2005-12-30 06:09:26
That was joke 
Actually I don't know how to solve it without solving it. 
- krassi_holmz
- 2005-12-30 06:04:33
I'm just guessing...
X1/2= (+-sqr(89)-3)/4?
- krassi_holmz
- 2005-12-30 05:59:08
Plot:
- krassi_holmz
- 2005-12-30 05:53:59
x+y=34 =>
y=34-x
Substitute in 1:
Let {is} means "must be"
x^2+(34-x)^2 {is} min =>
x^2+34^2-68x+x^2 {is} min =>
2x^2-68x {is} min =>
2(x^2 - 34x) {is} min =>
x^2 - 34x {is} min =>
x^2 - 2.17.x {is} min =>
(here is the thin moment...)
x^2-2.17x+17^2 {is} min =>
(x-17)^2 {is} min =>
(x-17)^2=0
x=17
y=34-17=17