A Nice Radical System

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[SyberMath]

Check this one as well:
A Quick and Easy Radical System

[randomjin]

You don't need to deal with 4th powers. Prior to second squaring of the first equation you already have √(x²-y²) so just substitute it to the second equation, isolate the √(x²+y²) and then square it. You will then get that y²=36-12x and from the first equation you get y²=4x-4 ⇒ 36-12x = 4x-4 ⇒ x=5/2. Then you trivially get value for y: y=√6. (negative won't work by the same reason you noted)

[txikitofandango]

2:30 You already have y^2 by itself, no need to isolate x. Just substitute 4x - 4 for y^2 in the second equation 8-)

This means you don't ever get into 4th powers

[echandler]

Nice problem, solutions and posted alternate solutions.

verifying the solution is interesting
√(5/2+√6) - √(5/2-√6) =
[√(5+2√6) - √(5-2√6)]/√2

note:
5=3+2, 6=3*2

so these de-nest easily to
[(√3+√2)-(√3-√2)]/√2 = 2
which satisfy the first equation

[novidsonmychanneljustcomme]

Everybody is commenting about the solution & the way to find it which is the most important part of course. ;) But another interesting aspect I became aware of was the verification if the solution fits. In the second equation it's quite easy bc the square cancels the sqrt and so everything goes right straight forward. But regarding the first equation it didn't seem that trivial at least to me. Though my experience told me that the "key" for expressions like sqrt(blabla...sqrt...blabla) are the 1st and 2nd binomial formulas. 😄 After finding out that sqrt(6)+5/2=(sqrt(3/2)+1)^2 and sqrt(6)-5/2=(sqrt(3/2)-1)^2 I managed to do the verification successfully. But I have to admit that "finding out" was no result of intuition but of solving another system of equations for the general 1st and 2nd binomial formula (to find out a and b for given a^2+2ab+b^2 and a^2-2ab+b^2). 😏😉

[mr_angry_kiddo]

When I bored with solving maths then I do create such things 🙂🙂🙂
If a= tanx +secx
a= sinx/cosx + 1/cosx
a=(sinx+1)/cosx
a= Rationalising nr
a = (1-sin²x)/[cosx(1-sinx)]
a= (cos²x)/[cosx(1-sinx)]
a=cosx/(1-sinx)
a= 1/[(1-sinx)/cosx]
a= 1/(sec-tanx). Rationaling dr
a= tanx +secx
a=a
Edit :😁😁😁😁

[advaithkumar]

a simpler solution:
squaring the first equation we get 4 = 2x - 2sqrt(x^2-y^2) implying sqrt(x^2-y^2) = x-2
and using second equation we get sqrt(x^+y^2) = 6-x
now square and add both of these to get x = 40/16 = 5/2 and hence find y ( ensuring y > 0 for obvious reasons)

[yoav]

Another nice way to solve :from first eq you can get sqrt(x+y)+sqrt(x-y)=y using this new eq and the first eq you see that sqrt(x+y)=(y+2)/2 and sqrt(x-y)=(y-2)/2. Then and also x^2+y^2=((x+y)^2+(x-y)^2)/2 so from eq2 you get eq just with y term you solve for y and then for x

[-basicmaths]

Squar the first'equation We have
√(x^2-y^2)=x-2
x^2-y^2=x^2-4x+4
y^2=4x-4
√(x^2+y^2)+x-2=4
√(x^2+y^2)=6-x
x^2+y^2=x^2-12x+36
y^2=-12x+36
4x-4=-12x+36
16x=40
x=5/2
y^2=6
y=(, √6, -√6)

[math]

very well done Syber, easy to follow and understand

[jjkgeft]

Could you try to solve a question like the

Find ALL the integers between 10000 to such that,
when divided by 7, leave a remainder of 6
when divided by 9, leave a remainder of 5
when divided by 11, leave a remainder of 7
when divided by 13, leave a remainder of 9

Thanks.

[prashantgujar]

Very good explanation. Can we use substitution sqrt(X +Y)= A and sqrt (X-Y) =B ?. (X2 + Y2) = 1/2( ( A2 +B2)2- (A2 -B2)2) and X2 - Y2 = (AB)2

[MrDfrolic]

Negative sqrt(6) is also a solution. If first equation give negative sqrt of both squares

[Blaqjaqshellaq]

One way to look at it is that x=(25/4)^1/2 and y=(24/4)^1/2.

[dehecth]

جميل جدا شرح بارك الله فيك شكرا thank you

[lori]

At 2:39 you could have just remembered √x̅²̅-̅y̅²̅=x-2, therefore √x̅²̅T̶y̅²̅=6-x .
Squaring both sides we get x²+y²=x²-12x+36 or y²=36-12x=12·(3-x). This and the squared thing in 2:24 give x=9-3x+1 so x=⁵/₂ and y=√6̅

[mathsandsciencechannel]

Very nice sir. Keep it up. Love your consistency.

[nuranichandra]

I tried solving these equations by other elegant means but nothing clicked. Yours is the only approach. It is a bit tedious but one has no choice but go through the grind,

[diogenissiganos]

Good old squaring both sides... Always works!

[MathElite]

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