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#1 2005-12-29 06:24:25

katy
Member
Registered: 2005-12-28
Posts: 14

I need help with these problems please..

I am doing some math problems and these are the ones I can not solve..:

1)The sum of 2 numbers is 34.Find the numbers if the sum of their squares is a minimum.
2)Without solving the equation find the discriminant and state the nature of the roots: 2x^2+3x-10
(in question #2 how can you find the discriminant and state the nature of the roots just by looking???)

Many thanks to the people who can help me with this......

Katy

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#2 2005-12-29 06:53:59

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: I need help with these problems please..

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


IPBLE:  Increasing Performance By Lowering Expectations.

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#3 2005-12-29 06:59:08

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: I need help with these problems please..

Plot:

View Image: min.gif

Last edited by krassi_holmz (2005-12-29 07:01:14)


IPBLE:  Increasing Performance By Lowering Expectations.

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#4 2005-12-29 07:04:33

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: I need help with these problems please..

I'm just guessing...
X1/2= (+-sqr(89)-3)/4?

Last edited by krassi_holmz (2005-12-29 07:07:33)


IPBLE:  Increasing Performance By Lowering Expectations.

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#5 2005-12-29 07:09:26

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: I need help with these problems please..

That was joke wink

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


IPBLE:  Increasing Performance By Lowering Expectations.

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#6 2005-12-29 07:12:26

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: I need help with these problems please..

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.


Why did the vector cross the road?
It wanted to be normal.

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#7 2005-12-29 07:30:22

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: I need help with these problems please..

Of course, but I thought that you are not permissed to calculate the discriminant.

And something else :
10>3^2 => discriminant is > 0.


IPBLE:  Increasing Performance By Lowering Expectations.

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#8 2005-12-29 07:44:09

katy
Member
Registered: 2005-12-28
Posts: 14

Re: I need help with these problems please..

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?

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#9 2005-12-29 07:57:03

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: I need help with these problems please..

For the 1b problem.
The parabloa is quadratic surve. So you're right.


IPBLE:  Increasing Performance By Lowering Expectations.

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#10 2005-12-29 08:35:32

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: I need help with these problems please..

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.


A logarithm is just a misspelled algorithm.

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#11 2005-12-29 10:21:51

katy
Member
Registered: 2005-12-28
Posts: 14

Re: I need help with these problems please..

thanks a lot guys...i get them now...you all are smart.
katy.

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#12 2005-12-29 11:04:28

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: I need help with these problems please..

Nah we just pretend to be.


A logarithm is just a misspelled algorithm.

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#13 2005-12-29 11:10:01

katy
Member
Registered: 2005-12-28
Posts: 14

Re: I need help with these problems please..

mikau wrote:

Nah we just pretend to be.

lol...of course you all are....I am the stupid one..hehehe

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#14 2005-12-29 11:20:02

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: I need help with these problems please..

Not stupid. You just need practice. Math isn't hard. It just takes some getting used to.


A logarithm is just a misspelled algorithm.

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#15 2005-12-29 11:24:47

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: I need help with these problems please..

I presentiment katy will become bery good in Maths.


IPBLE:  Increasing Performance By Lowering Expectations.

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#16 2005-12-29 12:35:55

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: I need help with these problems please..

bery good? are you being funny?


A logarithm is just a misspelled algorithm.

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