# FunctionsWatch

Announcements
#1
F(x)= x^8 +6x^4 -7

How would I find all the real roots of this function?
0
4 months ago
#2
(Original post by MM2002)
F(x)= x^8 +6x^4 -7

How would I find all the real roots of this function?
Let and the equation becomes... what in terms of u?
0
#3
Quartic graph. But how would that effect the other terms?
0
4 months ago
#4
(Original post by MM2002)
Quartic graph. But how would that effect the other terms?
gets replaced by .

is the same as so it gets replaced by what..?
0
#5
What do you mean by u? So, are we left with (x^4)^2 + (6x^2)^2 -7
0
4 months ago
#6
(Original post by MM2002)
What do you mean by u? So, are we left with (x^4)^2 + (6x^2)^2 -7
I'm telling you to make a substitution, just like you made a substitution in your last thread: https://www.thestudentroom.co.uk/sho....php?t=6162768

Here the roots satisfy and in order to solve this equation I'm telling you to make a substitution . What does the equation become in terms of ?
0
#7
(Original post by RDKGames)
I'm telling you to make a substitution, just like you made a substitution in your last thread: https://www.thestudentroom.co.uk/sho....php?t=6162768

Here the roots satisfy and in order to solve this equation I'm telling you to make a substitution . What does the equation become in terms of ?
I understood the last thread substitution. But, I am just lost on this one
Last edited by MM2002; 4 months ago
0
#8
(Original post by MM2002)
I understood the last thread substitution. But, I am just on this one
Also why was x^4 chosen. Would x^2 work aswell?
0
4 months ago
#9
(Original post by MM2002)
I understood the last thread substitution. But, I am just on this one
Same process so I'm not sure why you're stuck on this one.

That equation is ... so the substitution brings it to .
0
#10
The two values of u are -7 or 1. Whats the next step? On the last thread I would of just inserted this value into the power.
Last edited by MM2002; 4 months ago
0
#11
(Original post by RDKGames)
Same process so I'm not sure why you're stuck on this one.

That equation is ... so the substitution brings it to .
The two values of u are -7 or 1. Whats the next step? On the last thread I would of just inserted this value into the power.
0
4 months ago
#12
(Original post by MM2002)
The two values of u are -7 or 1. Whats the next step? On the last thread I would of just inserted this value into the power.
Just backtrack. Honestly, this process is the same all the time and shouldn't cause you issues...

Since you know that
or
then
or must hold. Solve these two for and you're done.
0
4 months ago
#13
(Original post by MM2002)
Also why was x^4 chosen. Would x^2 work aswell?
is not going to simplify things as much for us. Try substituting in , you get which not as nice as a quadratic we obtain via .

Generally, any equation of the form

where ,

can be brought down to a simple quadratic where . If this simplified quadratic has roots then our original solutions for satisfy:

or

Rearranging these two equations for leads us to our solutions.
Last edited by RDKGames; 4 months ago
0
#14
Is there a name for this method? And Im not getting one of the answers correct according to my textbook.
It says 1 which I got but Im not getting -1 which is another answer
0
4 months ago
#15
(Original post by MM2002)
Is there a name for this method? And Im not getting one of the answers correct according to my textbook.
Not really, this is quite a general technique across mathematics, but what you're dealing with here are disguised quadratics and it's important you can identity them.

Well, what answers do you get?
0
#16
I got 1 and -7 square root to the ^4 which shows up as error as its negative. How did they get the -1.
0
4 months ago
#17
(Original post by MM2002)
I got 1 and -7 square root to the ^4 which shows up as error as its negative. How did they get the -1.
-1 is one of the solutions to .
0
#18
I got 1 and -7 square root to the ^4 which shows up as error as its negative. How did they get the -1

(Original post by RDKGames)
Not really, this is quite a general technique across mathematics, but what you're dealing with here are disguised quadratics and it's important you can identity them.

Well, what answers do you get?
0
#19
Then how is 1 a solution? I know I said I had it earlier is because I thought 1 square root to the power of 4 equaled 1.
0
4 months ago
#20
(Original post by MM2002)
Then how is 1 a solution? I know I said I had it earlier is because I thought 1 square root to the power of 4 equaled 1.
by square rooting both sides. We cannot have because this has no real solutions. So we reduce to

. Again square root both sides and you get .
0
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• University of Warwick
Thu, 20 Feb '20
• St George's, University of London
Thu, 20 Feb '20
• University of Hertfordshire
Sat, 22 Feb '20

### Poll

Join the discussion

Yes (224)
67.47%
No (108)
32.53%