Any Algebra Geniuses out there?

[oshea.emma]

Hi guys! How are things....well i'm stuck with a bit of a stickler that i would most appreciate some help on!
The question is this
Find the RANGE! of values of t which is an element of the real numbers for which the quadratic equation (2t -1)x^2 + 5tx+ 2t =0 has real roots.
So I need the roots and I also need to explain why the roots are real when t is an integer.

I tried doing the -b formula... and i got a bit of the way but I couldn't seem to get a value for t other then the a crazy equation that i know can't be right.

MUCHOSSS Appreciate any help! but i can understand if it's too hard
-EMMA!

[zaza]

Hi,

Do the "-b" thing. You should be able to get to the following point:

x = (-5t +- sqrt(9t^2 + 8t))/(4t - 2)

x will only have imaginary roots when the bit in the square root is negative. Hence the tipping point is when the bit in the sqrt = 0.

So, you want to find the point(s) at which 9t^2 + 8t = 0

Thus t(9t+8) = 0 and t=0 or t=-8/9

So if t is between these two, the sqrt will be negative and hence imaginary roots will result. If not, the roots will be real. I'll leave it to you to figure out whether the boundaries of the range should be > or >= ...

zaza

[oshea.emma]

I'll try it out!!
THANKS!

[oshea.emma]

Great answer,
Just 1 question if the quadratic has real roots shouldn't b^2-4ac be greater then 0?

Isn't it true that if b^2-4ac is equal to 0 then it only has 2 equal roots that is a repeated root?

Does this change the answer? (That is should we have to put our b^2-4ac >0 and then square both sides or something?)

Any thoughts on this?
Thanks for your answer!!

[oshea.emma]

Would this be an answer?
Say we have 9t^2 + 8t>0
which gives us t(9t+8)>0
So our range for real roots is t>0 or when t>-8/9

What about the answer to this question
Explain why the roots are real when t is an integer?
I don't know why this would make sense as we seem to have found t>-8/9..
any thoughts?

[zaza]

Actually, it isn't t>-8/9. It's t<=-8/9.

If you fill t=0 in, then your solution for the quadratic becomes -x^2 = 0, so x=0. If t> 0, then as you can see you will always have a real root because the bit in the sqrt >0.
If t= -8/9, then you can work out the possible resultant values of x from the quadratic. They will still be real and hence this is a valid solution. If t < -8/9, then the bit in the sqrt becomes positive again and hence the roots must be real.


So to answer your second question, all values of t>=0 give real roots. That encompasses all positive integers. Also, all values of t<=-8/9 give real roots. That encompasses all negative integers. Hence if t is an integer, it must belong to either of these categories, and hence the roots are real.

zaza

[oshea.emma]

Actually, it isn't t>-8/9. It's t<=-8/9.

Hi Zaza,
thanks again for your explaination... I don't fully get it to be honest. When i use the -b formula i get b^2-4ac equal to 9t^2+8t

I'm looking for the real roots...that is a range of REAL numbers for t.

So according to my maths book: b^2-4ac > 0 for the real roots.
That's cool.

So it makes sense to me to put 9t^2+8t>0... good so far?
Then i take out t and i get t(9t + 8)>0
That means that t is >0 and t>-8/9 for my real roots.

So that would give me my range of t<-8/9 t>0

I'm trying to make some sense of it....Does this way work for you?
Why did you include the equals sign in your answer? that is t>=0?

[zaza]

Let's say t = -1.

Then the bit inside the sqrt is (-1)*(-9+8) = (-1)*(-1) = 1

When t< -8/9, then (9t+8) is negative. So is t, and hence the whole expression becomes positive.
If 0>t>-8/9, then (9t+8) is positive but t is negative, so this part of the expression is negative and hence you are trying to sqrt a negative number...imaginary roots.

Don't forget, t<-8/9 means t is less than -8/9, i.e. is more negative.

As for the real roots, you are looking for real roots to the quadratic itself, not just the "b^2-4ac" part.
If you fill t=0 into your quadratic, then you only get one answer, which is x=0. This is a real solution, and hence t=0 is a valid solution. So t>=0.
If you fill in t=-8/9 into your equation, then your value for x is:

x= -(5*-8/9) +- sqrt(0) / 2*((2*-8/9)-1)

By my calculations, this gives x= -0.8

This is still a real solution, and hence t=-8/9 is valid. So t<=-8/9 also gives real solutions.


Does that make sense?

zaza

[oshea.emma]

Zaza Thanks for your explanation.. it makes a bit more sense now!!
I'll keep tipping away at it!



Hope life's treating you VERY good!

[oshea.emma]

Hi Zaza... I mulled over it for a while and it makes more sense now!
Thanks very much. I understand that t will have real roots outs the range

-8/9<t<0

thanks very much.

[zaza]

Mulling over is usually the way forward. Glad to have helped with the mulling.

Mark this thread resolved using the "Thread Tools" menu at the top of the page.

Cheers,

zaza