Suppose that y varies directly as the square root of x​, and that y equals 27 when x equals 9. Find x when y equals 63.

Question

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Part 1
xequals
  
enter your response here ​(Simplify your​ answer.)

Answer 1

Since \( y \) varies directly as the square root of \( x \), we can express this relationship with the equation:

\[ y = k \sqrt{x} \]

where \( k \) is a constant.

Given that \( y = 27 \) when \( x = 9 \), we can substitute these values into the equation to find \( k \):

\[ 27 = k \sqrt{9} \]

Calculating \( \sqrt{9} \), we find:

\[ \sqrt{9} = 3 \]

So we can rewrite the equation:

\[ 27 = k \cdot 3 \]

Now, solve for \( k \):

\[ k = \frac{27}{3} = 9 \]

Now that we have found \( k \), our equation relating \( y \) and \( x \) is:

\[ y = 9 \sqrt{x} \]

Next, we want to find \( x \) when \( y = 63 \). We substitute 63 for \( y \) in the equation:

\[ 63 = 9 \sqrt{x} \]

To isolate \( \sqrt{x} \), divide both sides by 9:

\[ \sqrt{x} = \frac{63}{9} = 7 \]

Now, square both sides to solve for \( x \):

\[ x = 7^2 = 49 \]

Thus, when \( y = 63 \), \( x \) equals:

\[ \boxed{49} \]