Simplify

[livid forum]

How can I simplify these two? Thanks

[opaque crater]

What's the "simplified" form supposed to look like?

[livid forum]

Im not sure sorry. The instruction only said to simplify

[scenic badger]

=√(7+2√10)
=√[(√5)²+(√2)² + (2)(√5)(√2)]
=√[(√5)+(√2)]²
=√5 + √2

@opaque crater does it seem simplified or does it mean something else?

@livid forum u can simplify the other one like this too

[livid forum]

Thank youu. But I don't rly understand the second line

[next glacier]

@livid forum has given 1 rep to @past kraken

[scenic badger]

Oh so basically what i did was to break it to a²+b²+2ab

So that I can get (a+b)² to get it out of sqrt

[livid forum]

Ohh yeah thank you sm

[opaque crater]

7 = 5 + 2 = sqrt(5)^2 + sqrt(2)^2

[livid forum]

Thank you ❤️❤️❤️

[next glacier]

@livid forum has given 1 rep to @opaque crater

[scenic badger]

Try the second one

[livid forum]

I can't loll. I understand the first one but the second one it's √21 and idk what to do with it

[scenic badger]

Leme try

After lunch so brb

[scenic badger]

Seems like no direct way

Let √[5+√21] = ±(√a+√b)

We need to find it in the form √a + √b that's why

√[5+√21] = ±(√a+√b)
Squaring both sides,
5+√21 = a+b+2√ab

Now comparing both sides

a+b = 5 eq(1)

2√ab = √21
Squaring both sides
4ab = 21
ab = 21/4 eq(2)

Now from a+b = 5
b = 5-a
So using this in equation 2

a(5-a) = 21/4
5a-a² = 21/4
20a-4a² = 21
4a²-20a+21 = 0
Now we can solve this using quadratic formula

D= b²-4ac = 400-4(4)(21)
= 4(100-84)
= 4(16) = 64
Which is greater than 0, so we have real roots

Now (-b±√D)/2a
= (20±√64)/8
= (20±8)/8
So a = 12/8 or 28/8
a = 3/2 or 7/2

[livid forum]

Omg

[scenic badger]

a+b = 5
3/2 + b = 5
b = 7/2

[livid forum]

Do u think the question is wrong

[scenic badger]

So the values of a and b are either 3/2 or 7/2 respectively

[livid forum]

Thank you so much. You are so smart 😭 Which level are you?

I will read that later. I'm out rn

[scenic badger]

So √[5+√21] = √(3/2) + √(7/2)

[scenic badger]

Okay, also if u read it just a simple use of quadratic formula

U seem to be a middle schooler

@opaque crater can u check my calculations, i am not very accurate in my results lol

[livid forum]

Im just starting highschool

Ahaha

[livid forum]

Ok sure

[scenic badger]

Oh so u need to learn quadratic formula

[livid forum]

I learned it

But I didn't know I need it to do that question too

[scenic badger]

So it's a basic use of quadratic formula, we could do part 1 with the formula too but luckily we found a perfect square

[livid forum]

Oh yes

[scenic badger]

Once u read it, it will be easy. We had to write it as √a + √b

[scenic badger]

I am in grade 11

[livid forum]

Ohh

Which country are u from?

[scenic badger]

India

[livid forum]

Makes sense now. You guys are smarter than smart

[scenic badger]

Nah i wish I was smart

The answer is correct

[livid forum]

Where does this part come from btw?

[livid forum]

Bro just solved me 259 lines of a math question, and wishes they are smart

You are

Smart

Wdym

[hollow frost]

Well in a quadratic equation, those are the coefficients of the a, b, and c terms

This is to check whether they have 2 real roots, 1 real root, or 2 imaginary roots

[scenic badger]

U remember we used to find discrimination. So it's b²-4ac

A quadratic equation is in the form ax²+bx+c where a≠0

[livid forum]

Oh ok lemme see again

[livid forum]

Nah

[livid forum]

I'm familiar with that but I don't fully understand I think

[scenic badger]

Also since u have learnt quadratic formula, u might know
b²-4ac = 0 (two equal real roots)
b²-4ac > 0 (two distinct real roots)
b²-4ac < 0 (two imaginary roots)

[livid forum]

I don't know abt that

[hollow frost]

Do you know what it means by roots?

[livid forum]

Thank you sm

[livid forum]

Yeah

[next glacier]

@livid forum has given 1 rep to @past kraken

[hollow frost]

Ok

[livid forum]

But like

[scenic badger]

Tysm

[livid forum]

I never saw that

[scenic badger]

Just entered highschool. Practice will make u familiar with things

[livid forum]

Yep

[scenic badger]

I found an alternative to that question but u might not find the answer this way always.

[livid forum]

Oh really

Can u show me?

[scenic badger]

Yeah

[livid forum]

Thank youu

[livid forum]

Are u sure you aren't busy? Bc I understand the first method that u gave me already, so I don't need the second one, but still interested anyways lol

It's ok tho

[scenic badger]

Basically √[5+√21]
=√[(10/2)+√21]
=√[(7+3)/2+√21]
=√[(√(7/2))²+(√(3/2))²+(2)(√(7/2))(√(3/2))]
=√(√(7/2)+√(3/2))²
=√(7/2)+ √(3/2)

Quadratic equation might be a lengthy method but it's much easier as there is no guessing of what the values would be

[livid forum]

Oh that's a lot easier

Thank you ❤️

[scenic badger]

No it's not

[livid forum]

Why?

[scenic badger]

It took me an infinite amount of time to guess the numbers

[livid forum]

Lollll

[scenic badger]

Literally picked the numbers from the earlier answer

[livid forum]

Really

Hahaah

[scenic badger]

Quadratic is way easier

No guessing

[livid forum]

But alot shorter

[livid forum]

😭 yeah

Just need a bit of luck

[scenic badger]

Shoter in paper bigger in brain

[livid forum]

Fr

[scenic badger]

Stick with quadratic

[livid forum]

Ngl the teacher will suspect me of how did I do it, if I show him the quadratic

Bc it's very long and we never learn that

Have u ever attended in any competitions?

[scenic badger]

I won the district maths competition last year

With 87% score

[livid forum]

Omg

Congrats

[scenic badger]

Thanks

[livid forum]

Do u want to go national?

[scenic badger]

Nah

I lost after district level

[livid forum]

Ohh it's alright

I have never met an Indian who isn't good at math

[scenic badger]

+close