Quadratic Help!

Toast · 2006-10-29 23:53:14

Aight, I have 3 moderately challenging worded problems:

1.A right triangle has a hypotenuse of length 41cm and an area of 180cm². What are the lengths of the other two sides?

I got this:

then

substitute:

simplify:

again:

then:

Woat? How in the world do I solve this abomination to the power of 4?

2.Two cars leave Melbourne at the same time and both travel to Albury, a distance of 300km. One car travels 5km/h faster than the other and arrives in Albury 15 minutes before the other. Find the speed of the two cars.

Ok...whaaat? The answers section says it's 75km/h or 80km/h.

3.A person runs 2km and walks 2.25km in 1 hour. If that person ran and walked at the same average speeds for 6km and 2km respectively, the total time taken would be 1 hr 25min. Find the running and walking speeds.

Ok...

make one fraction:

multiply by xy:

multiply by 3 to enable subtraction:

subtract:

Where do u go from here

Thanks

Last edited by Toast (2006-10-30 02:06:09)

mathsyperson · 2006-10-30 01:33:09

In the first question, you're nearly there. You've got a quadratic equation there, it's just that the subject is x² instead of x. But you can still use the quadratic equation to find the values of x², and then you can square root those to get x.

With 2, you made a slight error when changing the information into an equation.

Instead of

, you need . And that should hopefully work out a lot nicer.

For 3, getting the x on its own is quite difficult, but getting the y is easy.
If you turn the 4.75 into a fraction then you can see that it simplifies quite nicely to 3x=xy, and then you can just divide by x (which is allowed because x is non-zero) to see that y=3.
Substituting that into one of the original equations and solving gets you that x=8.

Toast · 2006-10-30 02:15:55

thanks mathsyperson

1. we weren't supposed to learn the quadratic formula this year but meh, I'm familiar with it.
2. yeh i got confused with that
3. can't believe i didn't see that i could divide by 'x' xD.

Jai Ganesh · 2006-10-30 02:26:39

mathsyperson is aboslutely right,
an equation that appears to be of higher degree can be converted into a quadratic for convenience, and be solved. Great work, mathsyperson

Toast · 2006-10-30 03:49:48

Wait, so if the equation is in the form

can I use the quadratic equation to find , for example:

x^1.5 for

x^5 for

x^128 for

man that would be so cool

Last edited by Toast (2006-10-30 03:52:51)

mathsyperson · 2006-10-30 06:32:27

That's exactly right. It's a very useful tool for some areas of mathematics that you'll get into later on. Like 2nd order differential equations. Boy, those are fun.

mathsyperson · 2006-10-30 06:42:57

we weren't supposed to learn the quadratic formula this year but meh, I'm familiar with it.

Well, considering the title of your topic, I'd assume that you'd know how to solve by factorising at least. Now I look at it in more detail, that equation is actually factorisable, but normally when you see equations with coefficients that large, the quadratic equation is almost always the way to go.

Plus, I have no idea how you could be expected to work out that that monstrous quadratic factorises into the equally monstrous (x-81)(x-1600).

Math Is Fun Forum

#1 2006-10-29 23:53:14

Quadratic Help!

#2 2006-10-30 01:33:09

Re: Quadratic Help!

#3 2006-10-30 02:15:55

Re: Quadratic Help!

#4 2006-10-30 02:26:39

Re: Quadratic Help!

#5 2006-10-30 03:49:48

Re: Quadratic Help!

#6 2006-10-30 06:32:27

Re: Quadratic Help!

#7 2006-10-30 06:42:57

Re: Quadratic Help!

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