A puzzle?

[carnyx]

If x = sqrt(y^2 + z^2) , and y = z

then solve for x =1

[speakertoanimals]

What's so puzzling?

[petergriffin]

Ah I see your puzzle

Fogive my rust math, but this is how I did it

x=sqrt(Y^2+Z^2)
if z=y the it becomes
x=sqrt(2y^2)
thus x=1
1=sqrt2y^2
1=2y
1/2=y

Thus y and z are 1 but if you then do this at the equation at top you get

1=sqrt(0.5^2+0.5^2)
1=sqrt(0.25+0.25)
1=sqrt(0.5)
1=0.707107

which is clearly not true.

What I don't understand is why this is so, I fear the math is not strictly correct but I cannot see where I have gone wrong.

[speakertoanimals]

1=sqrt2y^2
1=2y

Because square root (2 y squared) is NOT the same as square root ( (2y) all squared).

So, what is the puzzle?

[Progenitor A]

Jan 11, 2011 14:13:10 GMT 1 petergriffin said:

Ah I see your puzzle

Fogive my rust math, but this is how I did it

x=sqrt(Y^2+Z^2)
if z=y the it becomes
x=sqrt(2y^2)
thus x=1
1=sqrt2y^2
1=2y
1/2=y

Thus y and z are 1 but if you then do this at the equation at top you get

1=sqrt(0.5^2+0.5^2)
1=sqrt(0.25+0.25)
1=sqrt(0.5)
1=0.707107

which is clearly not true.

What I don't understand is why this is so, I fear the math is not strictly correct but I cannot see where I have gone wrong.

I cannot see wher your math sis at fault eitther, but here is another way of approaching the problem

1 =(x² + z²)^0.5
squaring bothe sides:
1²=x² + z²
But 1² = 1
therefore
1=x² + z²
If x =z, then
1=x² + x²
1=2x²
x=0.5^0.5
Therefore
1=2x² becomes
1=2(0.5^0.5)²
1=2(0.5)
1=1

[speakertoanimals]

I cannot see wher your math sis at fault eitther

Which doesn't bode well, I have already shiown where his maths went wrong.

My question still stands -- what is the supposed problem?

x = sqrt(y^2 + z^2) with y=z and x=1.

Basic algebra:(step-by-step for those who are a bit rusty!)

1) Substitute in for z:

x = sqrt(2y^2)

2) Substitute in for x:

sqrt(2y^2) = 1

3) Square both sides (remembering we will be adding in the extra solution x=-1 by doing this!)

2y^2 = 1

4) divide both sides by 2:

y^2 = (1/2)

5) Take square root of both sides:

y = plus or minus (1/sqrt(2)).

We get two solutions, but this is fine, the original equation contains only y^2, so both are valid. TO check, subsititute back into original, and check they both work.

So, what's the puzzle?

[petergriffin]

I did say my math was rusty, and doing the math while eating does lead to errors, the correct result is that where x=1 y = root 2. Speakertoanimals has shown the correct method (as always).

The easy way to visualise this is:- this is basically the formula for righ angled triangle but where z=y if you thing of a square of side 1 then the diagonal would be root 2. To renforce this the sine and cosine of angle 45 dgrees is root 2.

So no puzzle, it just the way you do the algerbra, it very easy to make mistakes which give the wrong answer, as I did.

Using the same process you can show 1=2, but again this has an error in the algerbra.

Hope this sheds more light on the answers.