MM2002
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F(x)= x^8 +6x^4 -7

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

How would I find all the real roots of this function?
Let u=x^4 and the equation x^8+6x^4-7=0 becomes... what in terms of u?
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MM2002
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Quartic graph. But how would that effect the other terms?
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RDKGames
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(Original post by MM2002)
Quartic graph. But how would that effect the other terms?
x^4 gets replaced by u.

x^8 is the same as (x^4)^2 so it gets replaced by what..?
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MM2002
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What do you mean by u? So, are we left with (x^4)^2 + (6x^2)^2 -7
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RDKGames
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(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 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 ?
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MM2002
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(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 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
Last edited by MM2002; 4 months ago
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MM2002
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(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?
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RDKGames
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(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.
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MM2002
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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
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MM2002
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(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.
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RDKGames
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(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 or u=1
then
x^4 = -7 or x^4 = 1 must hold. Solve these two for x and you're done.
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RDKGames
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(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 or [f(x)]^n = u_2

Rearranging these two equations for x leads us to our solutions.
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MM2002
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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
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RDKGames
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(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?
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MM2002
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#16
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I got 1 and -7 square root to the ^4 which shows up as error as its negative. How did they get the -1.
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RDKGames
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(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.
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MM2002
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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?
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MM2002
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#19
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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.
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RDKGames
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(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.
x^4 = 1 \implies x^2 = \pm \sqrt{1} by square rooting both sides. We cannot have x^2 = -\sqrt{1} because this has no real solutions. So we reduce to

x^2 = +\sqrt{1} = 1. Again square root both sides and you get x = \pm \sqrt{1} = \pm 1.
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