Bunuel wrote:
10. Given that \(5x=125-3y+z\) and \(\sqrt{5x}-5-\sqrt{z-3y}=0\), then what is the value of \(\sqrt{\frac{45(z-3y)}{x}}\)?
A. 5
B. 10
C. 15
D. 20
E. Can not be determined
Rearranging both expressions we'll get: \(5x-(z-3y)=125\) and \(\sqrt{5x}-\sqrt{z-3y}=5\). Denote \(\sqrt{5x}\) as \(a\) and \(\sqrt{z-3y}\) as \(b\).
So we have that \(a^2-b^2=125\) and \(a-b=5\). Now, \(a^2-b^2=(a-b)(a+b)=125\) and as \(a-b=5\) then \((a-b)(a+b)=5*(a+b)=125\) --> \(a+b=25\). Thus we get two equations with two unknowns: \(a+b=25\) and \(a-b=5\) --> solving for \(a\) --> \(a=15=\sqrt{5x}\) --> \(x=45\). Solving for \(b\) -->\(b=10=\sqrt{z-3y}\)
Finally, \(\sqrt{\frac{45(z-3y)}{x}}=\sqrt{\frac{45*10^2}{45}}=10\).
Answer: B.
This is how I went about solving this problem:
Given \(\sqrt{5x} - 5 - \sqrt{z-3y}= 0\):
Rearrange and square
\((\sqrt{5x})^2 = (\sqrt{z-3y}+5)^2\)
\(|5x| = z-3y + 25 + 10\sqrt{z-3y}\)
Assuming that x is positive and given \(5x=125-3y+z\), set the two equations equal to each other
\(125 - 3y + z = z-3y + 25 + 10\sqrt{z-3y}\)
\(100 = 10\sqrt{z-3y}\)
\(100 = z-3y\)\(5x = 125-3y+z\)
\(5x = 125 + 100\)
\(x = 45\)Thus by substitution:
\(= \sqrt{\frac{45(z-3y)}{x}}\)
\(= \sqrt{\frac{45(100)}{45}}\)
\(= \sqrt{100}\)
\(= 10\)
However, notice that I had to assume that |5x| = 5x
Why is this permissible in this problem?