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AS maths surds question help??

pleaseee can someone help me work this Q out:
simplify (root of x^5 - 2rootx^3)/root x
Original post by zayn.javed
pleaseee can someone help me work this q out:
Simplify (root of x^5 - 2rootx^3)/root x


maths?!?! I do maths!!! Ok, hold on. Mentally calculationg!!!!
root of x^5 = x^(5/2)

therefore (x^(5/2) - 2x^(3/2)) / x^(1/2)

See if you can figure it out
Original post by Alexigender
maths?!?! I do maths!!! Ok, hold on. Mentally calculationg!!!!


What grade is this??
Reply 4
AS maths gets harder
Original post by RhymeAsylumForever
root of x^5 = x^(5/2)

therefore (x^(5/2) - 2x^(3/2)) / x^(1/2)

See if you can figure it out


Then multiply the exponents by 2 (go back to it later) and u would have (as i put it when simplifiying)

x^5 - 2(x^3)
----------------
x
Reply 6
Original post by zayn.javed
pleaseee can someone help me work this Q out:
simplify (root of x^5 - 2rootx^3)/root x


Start with the fact that x5=x  5\sqrt{x^5} = \sqrt{x}^{\;5}, then divide each term in the numerator by the denominator.

For example, x  4x=x  3\frac{\sqrt{x}^{\;4}}{\sqrt{x}}=\sqrt{x}^{\;3} etc
(edited 6 years ago)
Reply 7
Original post by Alexigender
Then multiply the exponents by 2 (go back to it later) and u would have (as i put it when simplifiying)

x^5 - 2(x^3)
----------------
x


6AC14E73-2AAA-4957-A059-CC63199FBEFE.jpg.jpeg would this be the final answer??
Reply 8
Original post by Ed5
Start with the fact that x5=x  5\sqrt{x^5} = \sqrt{x}^{\;5}, then divide each term in the numerator by the denominator.

For example, x  4x=x  3\frac{\sqrt{x}^{\;4}}{\sqrt{x}}=\sqrt{x}^{\;3} etc

(PRESSED SUBMIT BY ACCIDENT HOLD ON)


Would you be able to show me the workings through a picture please?
Original post by zayn.javed
6AC14E73-2AAA-4957-A059-CC63199FBEFE.jpg.jpeg would this be the final answer??


I believe so! This is college math i do in 8th grade
Can you help me with an 8th grade word prob that is bugging the living daylights out of me???
Reply 11
Original post by zayn.javed
would this be the final answer??


Not quite I'm afraid!

Original post by zayn.javed
Would you be able to show me the workings through a picture please?


Okay so start by dividing each term on the numerator:

x52x3x=x5x2x3x\frac{\sqrt{x}^5 - 2\sqrt{x}^3}{\sqrt{x}} = \frac{\sqrt{x}^5}{\sqrt{x}} - \frac{2\sqrt{x}^3}{\sqrt{x}}

Using the example I gave above, can you find what each fraction cancels down to?
Original post by Alexigender
Can you help me with an 8th grade word prob that is bugging the living daylights out of me???


I know its really stupid, but it's not making sense to me.

In an opera theater sections of seating are being laid out. It is planned so each row will have 2 more seats than the one be4 it. 90 people must fit in each section. How many people are in the third row?
(edited 6 years ago)
Reply 13
Original post by Alexigender
I know its really stupid, but it's not making sense to me.

In an opera theater sections of seating are being laid out. It is planned so each row will have 2 more seats than the one be4 it. 90 people must fit in each section. How many people are in the third row?


Alexigender if you need help with a question, please make your own thread.
Original post by Ed5
Okay so start by dividing each term on the numerator:

x52x3x=x5x2x3x\frac{\sqrt{x}^5 - 2\sqrt{x}^3}{\sqrt{x}} = \frac{\sqrt{x}^5}{\sqrt{x}} - \frac{2\sqrt{x}^3}{\sqrt{x}}

Using the example I gave above, can you find what each fraction cancels down to?


54DB9943-B58B-4BF6-83E3-DEB8F54FA01C.jpg.jpeg my brain has just gone dead pleaseee help!! 😭
Reply 15
Original post by zayn.javed
my brain has just gone dead pleaseee help!! 😭


Don't worry! I did this question without converting to fractional powers, which I found much easier.

So the first term simplifies easy enough, knowing that xnx=xn1\frac{x^n}{x}=x^{n-1}:

x5x=x4\frac{\sqrt{x}^5}{\sqrt{x}} = \sqrt{x}^4

The second term is the same but with a coefficient. If this confuses you, just think of it like this:

2x3x=2x3x=2x2\frac{2\sqrt{x}^3}{\sqrt{x}}=2 \cdot\frac{\sqrt{x}^3}{\sqrt{x}}=2\sqrt{x}^2

Combined, we've now simplified the big fraction into:

x42x2\sqrt{x}^4-2\sqrt{x}^2

Now if you just remember that a power is when a term is multiplied by itself that many times, you should be able to simplify this further.
Original post by Ed5
Don't worry! I did this question without converting to fractional powers, which I found much easier.

So the first term simplifies easy enough, knowing that xnx=xn1\frac{x^n}{x}=x^{n-1}:

x5x=x4\frac{\sqrt{x}^5}{\sqrt{x}} = \sqrt{x}^4

The second term is the same but with a coefficient. If this confuses you, just think of it like this:

2x3x=2x3x=2x2\frac{2\sqrt{x}^3}{\sqrt{x}}=2 \cdot\frac{\sqrt{x}^3}{\sqrt{x}}=2\sqrt{x}^2

Combined, we've now simplified the big fraction into:

x42x2\sqrt{x}^4-2\sqrt{x}^2

Now if you just remember that a power is when a term is multiplied by itself that many times, you should be able to simplify this further.


DD2CF569-5CC4-4CF8-96EF-74C67C6D1EB6.jpg.jpeg i can already tell i’ve done this wrong 😭
Reply 17
Original post by zayn.javed
i can already tell i’ve done this wrong 😭


Wyd you just lowered the powers without changing anything 😂

Because two roots multiply to make the real thing, consider x4=xxxx=xx=x2\sqrt{x}^4 = \sqrt{x} \cdot \sqrt{x} \cdot \sqrt{x} \cdot \sqrt{x} = x \cdot x = x^2

So that's the first term simplified. Do the same with the second (it's easier) and you are left with a straight forward polynomial.

It still needs simplifying after that, remember - you can factorise it too.

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