# Alevel Maths - Finding the Roots of functions

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#1
Hi everyone! please could someone explain to me the the part of the solution working that I drew an arrow to? I really dont understand that bit. thanks!

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Attachment 958264
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2 months ago
#2
(Original post by tomas530)
Hi everyone! please could someone explain to me the the part of the solution working that I drew an arrow to? I really dont understand that bit. thanks!
It is explicitly rewriting the function into a form that shows f(x) is actually a hidden quadratic. Remember that quadratics are of the form

a(variable)^2 + b(variable) + c

So in your case the (variable) is just x^3
2
2 months ago
#3
(Original post by tomas530)
Hi everyone! please could someone explain to me the the part of the solution working that I drew an arrow to? I really dont understand that bit. thanks!

Attachment 958262
Attachment 958264
Hi! What they did was take out a factor of x^3 from all terms with an x^3 in to make the factorising easier!

To think about it a different way, for the x^6 + 7x^3 - 8 =0, they in effect replaced any x^3 with “k” (just a constant like “x” is), to get k^2 + 7k - 8 = 0, which you’ll notice is a lot easier to factorise than anything with a power of 6
Then they factorise the new equation for (k-1)(k+8)=0
Now replace the k with x^3 again and solve the brackets
Last edited by laurawatt; 2 months ago
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#4
(Original post by laurawatt)
Hi! What they did there was pull a factor of x^3 out, to make the factorising easier!

To think about it a different way, for the x^6 + 7x^3 - 8 =0, they in effect replaced any x^3 with “k” (just a constant like “x” is), to get k^2 + 7k - 8 = 0, which you’ll notice is a lot easier to factorise than anything with a power of 6
Then they factorise the new equation for (k-1)(k+8)=0
Now replace the k with x^3 again and solve the brackets
Ah thats perfect. thank you to both of you for your help : )
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#5
(Original post by laurawatt)
Hi! What they did there was pull a factor of x^3 out, to make the factorising easier!

To think about it a different way, for the x^6 + 7x^3 - 8 =0, they in effect replaced any x^3 with “k” (just a constant like “x” is), to get k^2 + 7k - 8 = 0, which you’ll notice is a lot easier to factorise than anything with a power of 6
Then they factorise the new equation for (k-1)(k+8)=0
Now replace the k with x^3 again and solve the brackets
Hi, Im practising some more of these questions and came across this one:

Find all the roots of the following function:

m(x) = x^3 + 5x^2 -24x

But Here you cant just pull out a factor of x to the power of something. So how should I start? Thanks
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2 months ago
#6
(Original post by tomas530)
Hi, Im practising some more of these questions and came across this one:

Find all the roots of the following function:

m(x) = x^3 + 5x^2 -24x

But Here you cant just pull out a factor of x to the power of something. So how should I start? Thanks
Hi take out a factor of x to get x(... + ..... - ......) =0
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#7
(Original post by tomas530)
Hi, Im practising some more of these questions and came across this one:

Find all the roots of the following function:

m(x) = x^3 + 5x^2 -24x

But Here you cant just pull out a factor of x to the power of something. So how should I start? Thanks
Oh I think ive got it. I just pull out x as a factor, right?
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2 months ago
#8
(Original post by tomas530)
Oh I think ive got it. I just pull out x as a factor, right?
Yep
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#9
(Original post by laurawatt)
Yep
brill. thanks Laura!
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2 months ago
#10
(Original post by laurawatt)
Hi! What they did there was pull a factor of x^3 out, to make the factorising easier!

To think about it a different way, for the x^6 + 7x^3 - 8 =0, they in effect replaced any x^3 with “k” (just a constant like “x” is), to get k^2 + 7k - 8 = 0, which you’ll notice is a lot easier to factorise than anything with a power of 6
Then they factorise the new equation for (k-1)(k+8)=0
Now replace the k with x^3 again and solve the brackets
Slight care is needed with the wording. Here x^3 is not 'pulled out as a factor' at all.

However, in the most recent example, indeed a factor of x needs to be pulled out first.
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2 months ago
#11
(Original post by RDKGames)
Slight care is needed with the wording. Here x^3 is not 'pulled out as a factor' at all.

However, in the most recent example, indeed a factor of x needs to be pulled out first.
I agree, slight error in my wording as x^3 was not a factor for all of the terms - thanks!
Last edited by laurawatt; 2 months ago
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#12
Alright, I have done a few more correctly, but... here comes a hard one. Could one of you please help me with this new question before I make a frustrating mistake?

The question is:

Find all real roots of the following function:

m(x) = 2x^2/3 + 2x^1/3 -12

Would i need to convert them into surds? Or should I 'factor out' ^1/3 ?

Thanks

PS: sorry for so many questions
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2 months ago
#13
(Original post by tomas530)
Alright, I have done a few more correctly, but... here comes a hard one. Could one of you please help me with this new question before I make a frustrating mistake?

The question is:

Find all real roots of the following function:

m(x) = 2x^2/3 + 2x^1/3 -12

Would i need to convert them into surds? Or should I 'factor out' ^1/3 ?

Thanks

PS: sorry for so many questions
let y = x1/3
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#14
(Original post by the bear)
let y = x1/3
how though?
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2 months ago
#15
(Original post by tomas530)
how though?
Rewrite the question in terms of y (given that y = x^(1/3)). This will give you a quadratic in y.
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#16

this doesn't look right 😩
0
2 months ago
#17
.
1
2 months ago
#18
(Original post by tomas530)

this doesn't look right 😩
what must you do to x1/3 to make it become x2/3 ?
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2 months ago
#19
(Original post by the bear)
what must you do to x1/3 to make it become x2/3 ?
Rub out the 1 and write 2 in its place?
1
2 months ago
#20
(Original post by RDKGames)
Rub out the 1 and write 2 in its place?
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