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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 wink

Actually I don't know how to solve it without solving it.
smile smile smile

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

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