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secondary 2 | Maths
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Nancy
Nancy

secondary 2 chevron_right Maths chevron_right Singapore

surds

Date Posted: 3 years ago
Views: 274
J
J
3 years ago
Let (25 + 10√5)^⅓ + (25 - 10√5)^⅓ = x

x³ = [ (25 + 10√5)^⅓ + (25 - 10√5)^⅓ ]³

x³ = ((25 + 10√5)^⅓)³
+ 3((25 + 10√5)^⅓)² (25 - 10√5)^⅓
+ 3(25 + 10√5)^⅓ ((25 - 10√5)^⅓)²
+ ((25 - 10√5)^⅓)³

(Property used is (a+b)³ = a³ + 3a²b + 3ab² + b³)

= 25 + 10√5
+ 3(25 + 10√5)^⅓ (25 - 10√5)^⅓ ( (25 + 10√5)^⅓ + (25 - 10√5)^⅓ )
+ 25 - 10√5

= 50 + 3[(25+10√5)(25-10√5)]^⅓ x

= 50 + 3[25² - (10√5)²]^⅓ x

= 50 + 3(625 - 500)^⅓ x

= 50 + 3(125)^⅓ x

= 50 + 3(5)x

= 50 + 15x


So, x³ = 50 + 15x
x³ - 15x - 50 = 0

When x = 5, x³ - 15x - 50
= 5³ - 15(5) - 50
= 125 - 75 - 50
= 0

By the Factor Theorem, (x - 5) is a factor.

Factorise (either mentally or long division)

(x - 5)(x² + 5x + 10) = 0

(x - 5) = 0
x = 5

or

x² + 5x + 10 = 0

For x² + 5x + 10,
discriminant = 5² - 4(1)(10)
= 25 - 40
= -15 < 0

Since discriminant < 0, there are no real roots/solutions for this equation.


∴ (25 + 10√5)^⅓ + (25 - 10√5)^⅓ = 5
J
J
3 years ago
Alternatively,

(25 + 10√5)^⅓ + (25 - 10√5)^⅓

= [⅛(200 + 80√5)]^⅓ + [⅛(200 - 80√5)]^⅓

= [⅛(125 + 75√5 + 75 + 5√5)]^⅓ + [⅛(125 - 75√5 + 75 - 5√5)]^⅓

= [⅛(5³ + 3x5x5x√5 + 3x5x5 + √5 x √5 x √5)]^⅓ + [⅛(5³ - 3x5x5x√5 + 3x5x5 - √5 x √5 x √5)]^⅓

= [⅛(5³ + 3(5²)√5 + 3(5)(√5)² + (√5)³)]^⅓ + [⅛(5³ - 3(5²)√5 + 3(5)(√5)² - (√5)³)]^⅓

= [⅛(5 + √5)³)]^⅓ + [⅛(5 - √5)³]^⅓


(Property used is (a+b)³ = a³ + 3a²b + 3ab² + b³)

= (⅛)^⅓ ((5 + √5)³)^⅓ + (⅛)^⅓ ((5 - √5)³)^⅓

= ½(5+√5) + ½(5 - √5)

= ½(5 + √5 + 5 - √5)

= ½(10)

= 5

See 3 Answers

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Kahwai
Kahwai's answer
232 answers (A Helpful Person)
1st
Using

The formula

Know that

The square in between can be cancelled

Redo the sum

It is not that difficult
Kahwai
Kahwai
3 years ago
See the new one
J
J
3 years ago
-ab should be simplified to the following

i.e

-ab = - (25 + 10√5)^⅓ (25 - 10√5)^⅓

= - (25² - 10² x 5)^⅓

= - (125)^⅓

= -5


But a² and b² still have the exponentials.

i.e a² = (25 + 10√5)^⅔
b² = (25 - 10√5)^⅔


a² + b² will not simplify to (25² + (10√5)²) x 2,
Kahwai
Kahwai
3 years ago
You know how to become perfect square easyvproblrm please go and learn

A + B Perfect square is a square plus b square mijus two ab
J
J
3 years ago
We all know that.

The point is, you should be getting

50 = x((a+b)² - 2ab - ab)

50 = x(x² - 2(125)^⅓ - (125)^⅓)

50 = x (x² - 2(5) - 5)

50 = x(x² - 15)

50 = x³ - 15x

x³ - 15x - 50 = 0


Which is not what you wrote above
.
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Kahwai
Kahwai's answer
232 answers (A Helpful Person)
Solve x if you are okay
J
J
3 years ago
ab = 5, not 125 .

But this method is correct. You should end up with x³ - 15x - 50 = 0 however.
J
J
3 years ago
Your idea is is similar to mine (see main comments section)

I used (a + b)³ = a³ + 3a²b + 3ab² + b³
Kahwai
Kahwai
3 years ago
Your method is lengthy and too naggy
J
J
3 years ago
You've skipped a lot of steps. That's why
Kahwai
Kahwai
3 years ago
You don’t even know the tricks
The intent of the teacher is to know the property of surfs

I didn’t skip steps

This is how should be done

The thinking processes matter

I suggest if you want to help put in answer not just comments so that we won’t help
J
J
3 years ago
Using (a + b)³ = a³ + 3a²b + 3ab² + b³ is equally valid.

The idea is to realise that the middle terms
can be changed into 3(ab)(a+b), which can be easily simplified.

You realised that a³ + b³ = (a + b)(a² - ab + b²) is derived from this, don't you?

Your method in fact, incurred an extra step by having to use a² + b² = (a+b)² - 2ab , which is extra work.

So you used 2 identities :

a³ + b³ = (a + b)(a² - ab + b)²
a² + b² = (a + b)² - 2ab

Compared to my single identity :
(a + b)³ = a³ + 3a²b + 3ab² + b³
= a³ + b³ + 3ab(a+b)

Yet, you say that my method is lengthy and naggy when yours is more complicated but basically uses the same principle of conjugates to reach the same conclusion of x³ - 15x - 50 = 0

My point is, if you want to help, make sure your workings are correct and they lead logically to the next step.
J
J
3 years ago
It is my choice whether to post as answer or to write in the comments.

It is far easier to write in comments than to post an answer which would appear as a wall of text.

Handwriting issues are very common here, as we can see from yours (pretty illegible I must say)
Kahwai
Kahwai
3 years ago
Student has to learn
This is Just guide

I am not spoon feeding them because during exam no one can help them
J
J
3 years ago
So it remains as a fact that you did skip steps, mainly the expansion of terms and also the evaluation of the conjugates.
(i.e (a + b)^⅓(a - b)⅓ = ((a+b)(a-b))^⅓
= (a² - b²)^⅓

You are free to help or post an alternative answer as a tutor. Perhaps the student likes yours better. If you're so worried about having duplicate answers, then you are free to check the main comments section first.


This is Olympiad Math so knowledge of (a+ b)³ is to be expected, along with a³ + b³ = (a + b) (a² -ab + b²) The latter is only taught in sec 3 for A Math)


a² + b² = (a + b)² - 2ab is also expected as it's already covered in sec 2.
Kahwai
Kahwai
3 years ago
Pls stop writing
Kahwai
Kahwai
3 years ago
If you keep on doing this noone could help the students and not all students are learning at the same methodologies

If you want to write so much I rather you do more math if you are really helpful
J
J
3 years ago
I've been in this app long enoug, and have been helping students all the way.

I also try to point out working mistakes that people have, so the students do not risk writing the wrong thing or being misinformed.

Your first picture contained working errors so I was being nice by pointing it out to you.

But now you want to come and call my method too lengthy and naggy, when all I did was to show full working, and it was equally valid.

Whereas yours is brief writings (hardly legible and with mistakes). In fact, this photo here, your second attempt, contained mistakes again (how is the last step 50 = x(x² - 45) ???).

So you have no right to say anything about lengthy and when you have provided inaccurate and incomplete answers in the first place.


You should learn to accept your errors instead of being so defensive. And also, double check your working and proofread before posting.

Saves you embarrassment and keeps your credibility up.
J
J
3 years ago
Also, telling someone that he doesn't even know the tricks is just premature judgment and jumping to conclusions.


No one said your method is wrong. I agree with it.

You should also realise that for mathematics, there are multiple approaches to each problem. No one method is superior or inferior to others.

What more, when our answers use the same principle of binomial expansion.

The whole point is to be exposed to different techniques and methods (especially for Olympiad Math)
J
J
3 years ago
If you've noticed, many students do not know even where to start or have conceptual gaps. They often ask for elaborate and detailed explanation.

The goal should be to provide working as clear and logical/intuitive as possible.

Guidelines/brief methods only confuse many of them, which would eventually prompt more questions

So, full workings aren't spoon-feeding in this context.
50 = x . ( x.x -15)

Take x is 5

Then the Rhs is equal to lhs
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Kahwai
Kahwai's answer
232 answers (A Helpful Person)