Functions

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/showthread.php?t=6162768

Here the roots satisfy

x^8+6x^4-7=0

and in order to solve this equation I'm telling you to make a substitution

u=x^4

. What does the equation become in terms of

u

I understood the last thread substitution. But, I am just lost on this one

(edited 4 years ago)

Reply 7

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?

Reply 8

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

(x^4)^2 + 6(x^4) - 7 = 0

... so the substitution

u=x^4

brings it to

u^2+6u-7=0

Reply 9

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.

(edited 4 years ago)

Reply 10

Original post by RDKGames

Same process so I'm not sure why you're stuck on this one.

That equation is

(x^4)^2 + 6(x^4) - 7 = 0

... so the substitution

u=x^4

brings it to

u^2+6u-7=0

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.

Reply 11

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

u=-7

u=1

then

x^4 = -7

x^4 = 1

must hold. Solve these two for

x

and you're done.

Reply 12

Original post by MM2002

Also why was x^4 chosen. Would x^2 work aswell?

x^2

is not going to simplify things as much for us. Try substituting in

u=x^2

, you get

u^4 + 6u^2 - 7 =0

which not as nice as a quadratic we obtain via

x^4

.

Generally, any equation of the form

a[f(x)]^{2n} + b [f(x)]^n + c = 0

where

n \in \mathbb{Z}

,

can be brought down to a simple quadratic

au^2+bu+c=0

where

u=[f(x)]^n

. If this simplified quadratic has roots

u_1, u_2

then our original solutions for

x

satisfy:

[f(x)]^n = u_1

[f(x)]^n = u_2

Rearranging these two equations for

x

leads us to our solutions.

(edited 4 years ago)

Reply 13

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

Reply 14

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?

Reply 15

I got 1 and -7 square root to the ^4 which shows up as error as its negative. How did they get the -1.

Reply 16

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

x^4 = 1

Reply 17