AS Maths

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# AS Maths Posts: 76
11 Replies Joined: 3 years ago

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@hrashid_

When you have an expression in brackets, then that bracket has a power, it works like the following: $(2x^3)^3 = 2^3x^{3*3}$

So as you can see, the power times everything in the brackets. So similarly if you have: $x(2x^{\frac{-1}{3}})^4$

Focus on the $(2x^{\frac{-1}{3}})^4$

Simplify that, then times it by x

As for question 5,

imagine if $y = 2x^5$

And they said work out $y^3$

They basically mean just work out $(2x^5)^3$

Which is just $2^{3}x^{15}$

So similarly, try the same thing. Posts: 7
(@lilyy)
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(@further_maths_student)
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@lilyy

When you complete the square, you have the line of symmetry

say for example you had a quadratic and you completed the square and you got:

(x- 5)^2 +  3

Do you know what the minimum point of this quadratic would be? ExamQA+
(@further_maths_student)
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just fixed the expression (@lilyy)
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(@further_maths_student)
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@lilyy

Yeah, literally

so this line of symmetry would just be the minimum point x = 5 (@lilyy)
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@lilyy

Yeah x squared coefficients make no difference - its always the minimum or maximum point (@lilyy)
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(@hrashid_)
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@hrashid_

You need to pretend like they're quadratics

I'll do the first one for you

x^6 + 9x^3 + 8

If you want to make that look like a quadratic (clearly its not right now) you need to pretend x^3 = y

This will make the expression above turn into

y^2 + 9y + 8 which looks much more like a quadratic now

(as y^2 = x^6)

Then you just solve y^2 + 9y + 8 = 0

(y + 8)(y + 1) = 0

y = -8, y = -1

But remember earlier we said x^3 = y

So really

x^3 = -8, x^3 = -1

so x = -2, x = -1

Finished

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