A Quick and Easy Radical System

This video is about solving a Radical System
If you need to post a picture of your solution or idea:
#ChallengingMathProblems #RadicalSystems

EXPLORE:

[shreyan]

The fact that you upload a video everyday is mind blowing!

[carloshuertas]

To me, solving math problems are fun to do.

[Rundas]

The shortcut using Vieta's formula was genius. Saves a bunch of time if 1 is a root.

[carloshuertas]

It was quick and easy. I got one of the solutions to the math problem. I got x=4 and y=1. Another great explanation!

[veljkozivkovic]

At 4:05 one could rearrange the equation so that 5sqrt(y+3) is on the LHS, so when you square it, you get a quadratic t=sqrt(y)(thera only, sqrt(y) and remainder), so that this quadratic has smaller coefficients, and also seeing that one soluioton is one is much easier(8y-21sqrt(y)+13=0).

[davidseed]

At 2:40 slightly quicker u=sqrty and get
5sqrt(u^2+3)=7+3u
Square and simplify to get
25u^2+75=9u^2+42u+49
16u^2-42u+26=0
u^2-21/8+13/8=0
Use Vieta
(u-1)(u-13/8)=0
So y=1 or (13/8)^2
sqrt(x)= 3-u =2 or 11/8
x=4 or (11/8)^2

[sahilsinghbhandari]

Very nice method by which you solve.
But we can also get integral solution by hit and trial because it is easy to guess the number in this question.

[ashishphatak]

Video get more and more entertaining like reading a mystery novel♥️

[AlephThree]

Very good. I bounded x and y between 0 and 9 from the first equation and quickly saw that (4, 1) was the only integer pair solution. So missed the other one.

[안태영-gw]

My trial: take x = (1/4)*(a - 5/a)^{2} and y = (1/4)*(b - 3/b)^{2}, where a ≥ sqrt(5) and b ≥ sqrt(3).
The purpose is to take off all radicals.

Then sqrt(x) + sqrt(y) = 3 ⇔ (1/2)*(a - 5/a) + (1/2)*(b - 3/b) = 3 …ⓐ
sqrt(x+5) + sqrt(y+3) = 3 ⇔ (1/2)*(a + 5/a) + (1/2)*(b + 3/b) = 5 …ⓑ

ⓐ+ⓑ gives "a + b = 8", and ⓑ-ⓐ becomes "5/a + 3/b = 2" so that "3a + 5b = 2ab".
Substituting 'b = 8 - a', we get "3a + 5(8-a) = 2a(8-a)", so "a^{2} - 9a + 20 = 0".

Thus (a, b) ∈ {(5, 3), (4, 4)}.
Returning them to x and y, we get the final answer (x, y) ∈ {(4, 1), (121/64, 169/64)}.

And very good video btw! Please keep going!

[Dae-Ying-Kim]

* you can try to write
: ??? = x = ??? it might seems fun to be expressed, gonna try it ! *

[SrisailamNavuluri]

35√(y+3)=8y+62=2(4y+31), 35=4y+31 which is odd
√(y+3)=2 which is even
y+3=2^2=4
y=135=4×1+31=35
y=1, x=4 are integer solution

[chaosredefined]

When you had:
-3 sqrt(y) = 7 - 5 sqrt(y+3)

Add 5 sqrt(y+3) to both sides
5 sqrt(y+3) - 3 sqrt(y) = 7 <- Eq (1)

Multiply both sides by 5 sqrt(y+3) + 3 sqrt(y), so we can take advantage of the difference of squares formula
25(y+3) - 9 y = 35 sqrt(y+3) + 21 sqrt(y)
16y + 75 = 35 sqrt(y+3) + 21 sqrt(y) <- Eq (2)

Cancel out the sqrt(y+3) by calculating Eq (2) + 7*Eq (1)
16y + 75 + 35 sqrt(y+3) - 21 sqrt(y) = 35 sqrt(y+3) + 21 sqrt(y) + 49
16y - 42 sqrt(y) + 26 = 0
8y - 21 sqrt(y) + 13 = 0

Since y is (sqrt(y))^2, we can treat this like a quadratic. Conveniently, sqrt(y) = 1 is a solution, so just factorise.
(8 sqrt(y) - 13) (sqrt(y) - 1) = 0
sqrt(y) = 13/8 or sqrt(y) = 1
y = 169/64 or y = 1

Numbers don't need to get as large as they did.

[michaelpurtell]

I did it the way it was done the video. I then spent a day trying to find a “trick“ that would avoid all that squaring. I am disappointed that there actually was none.

[SamsungJ-kknr]

I'd like this exercise because using the two evelated.

[voorth]

The second solution can also be written as: ((11/8)^2, (13/8)^2), which I find a bit more readable...

[jorgepinonesjauch]

Muy interesante, Lo hice un poco más corto, haciendo que a=√(x+5) y b=√(y+3) obteniendo un nuevo sistema de ecuaciones en términos de a y b, para finalmente encontrar los valores de a y b y después los valores de x e y

[pritivarshney]

At 03:20 you could have isolated 5sqrt(y+3) on one side and then squared it would be easier.

[aashsyed]

very awesome
so cool!
you are a pro!

[prabhakarbk]

But I solved this a different way.
Put root x = a². and root y = b²
Then. a+b =3
a² = ( 3-b)² = 9 - 6b +b²

Root (x+5) +root (y+3)=5

Root (x+5) = 5 - root (y+3)
Squaring on both sides we get
x+5 = 25 +y+3 -10 root (y+3)
10 root (y+3) = y+28-5-x
=b²+23-a²
=b²+23 -9-b² +6b
=14+6b
Again squaring on both sides we get
100(y+3) =(14+6b)²
100b² +300 =196+36b² +168b
64b²-168b+104=0
(b-1) (64b -104)=0
b-1 =0
b=1 or root y=1
Hence y=1
a=3-b =3-1=2
So root x=2
X=4