vogelleblanc wrote:
If \(x = u^2 - v^2\), \(y = 2uv\) and \(z = u^2 + v^2\), and if \(x = 11\), what is the value of z?
(1) y = 60
(2) u = 6
my approach towards statement 2:
x=11 and u=6,
thus 11=6^2-v^2
25=v^2
v=5 or v=-5
Since z=u^2+v^2, it does not matter if v is 5 or -5, so z=55
Please correct me if something is wrong!/spoiler]
I do not understand statement 1, please help me
This is a really interesting question.
My approach:
We are given that \(x = u^2 - v^2\), \(x = 11\)
Statement 1 tells us y = 60.
=> \(60 = 2uv\)
Square on both sides\(y^2 = (2uv)^2\)
\((60)^2 = (2uv)^2\)
\(3600 = 4u^2v^2\)
Take \(x = u^2 - v^2\) & square on both sides
we get, \((11)^2 = (u^2 - v^2)^2\)
\(121= u^4 + v^4 - 2u^2v^2\)
Add \(y^2\) on both sides to get a positive value for \(- 2u^2v^2\).
\(121 + 3600 = u^4 + v^4 - 2u^2v^2 + 4u^2v^2\)
\(121 + 3600 = u^4 + v^4 + 2u^2v^2\)
Here, we can write \(u^4 + v^4 + 2u^2v^2\) as \((u^2 + v^2)^2\), since \((a + b)^2 = a^2 + b^2 + 2ab\)
\((u^2 + v^2)^2 = 3721\)
\(u^2 + v^2 = \sqrt{3721}\)
\(z = \sqrt{3721}\)
\(z = 61\)
Statement 1 alone is sufficient!Let's look at Statement 2. It says u = 6.
No need to plug this in the equation we derived from STEM, rather just plug the value in \(x = u^2 - v^2\).
\(11 = 6^2 - v^2\)
\(v^2 = 6^2 - 11\)
\(v^2 = 36 - 11\) = \(v^2 = 25\) Therefore, \(v = +5\) or \(v = -5\)
If \(v = +5\)
then, \(z = 6^2 + 5^2\) (u = 6 from Statement 2)
\(z = 36 + 25\) = \(z = 61\)
If \(v = -5\)
then, \(z = 6^2 + (-5)^2\)
\(z = 36 + 25\) = \(z = 61\)
Statement 2 alone is sufficient!Option D