y varies inversely as the square root of x.‬ ‭ When x = 9, y = 2.‬ ‭ (a)‬‭ Find y in terms of x (b) Find y when x = 36 (c)Write x in terms of y When y is multiplied by 0.5, x is multiplied by‬‭ k.‬ ‭ (d) Find the value of k

Given that \( y \) varies inversely as the square root of \( x \), we can write the relationship as:

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

where \( k \) is a constant.

### (a) Find \( y \) in terms of \( x \)

To determine \( k \), we use the given condition:

When \( x = 9 \), \( y = 2 \).

Substituting these values into the equation:

\[
2 = \frac{k}{\sqrt{9}}
\]

Since \( \sqrt{9} = 3 \), the equation becomes:

\[
2 = \frac{k}{3}
\]

Solving for \( k \):

\[
k = 2 \times 3 = 6
\]

Thus, the equation relating \( y \) and \( x \) is:

\[
y = \frac{6}{\sqrt{x}}
\]

### (b) Find \( y \) when \( x = 36 \)

Substitute \( x = 36 \) into the equation \( y = \frac{6}{\sqrt{x}} \):

\[
y = \frac{6}{\sqrt{36}} = \frac{6}{6} = 1
\]

So, \( y = 1 \) when \( x = 36 \).

### (c) Write \( x \) in terms of \( y \)

Starting from the equation \( y = \frac{6}{\sqrt{x}} \), solve for \( x \):

\[
y = \frac{6}{\sqrt{x}}
\]

Multiply both sides by \( \sqrt{x} \):

\[
y \sqrt{x} = 6
\]

Square both sides to eliminate the square root:

\[
y^2 x = 36
\]

Solve for \( x \):

\[
x = \frac{36}{y^2}
\]

### (d) Find the value of \( k \) when \( y \) is multiplied by 0.5

When \( y \) is multiplied by 0.5, the new \( y \) becomes:

\[
y' = 0.5y
\]

Using the equation \( y = \frac{6}{\sqrt{x}} \), the new equation becomes:

\[
0.5y = \frac{6}{\sqrt{x'}}
\]

Substituting \( y = \frac{6}{\sqrt{x}} \):

\[
0.5 \times \frac{6}{\sqrt{x}} = \frac{6}{\sqrt{x'}}
\]

Simplify to find the relationship between \( x \) and \( x' \):

\[
\frac{3}{\sqrt{x}} = \frac{6}{\sqrt{x'}}
\]

Cross-multiply:

\[
3 \sqrt{x'} = 6 \sqrt{x}
\]

Divide both sides by 3:

\[
\sqrt{x'} = 2 \sqrt{x}
\]

Square both sides:

\[
x' = 4x
\]

Thus, when \( y \) is multiplied by 0.5, \( x \) is multiplied by 4. Therefore, \( k = 4 \).

---

Would you like any further details or have any questions?

Here are 5 related questions you might find interesting:

1. If \( y \) varies directly as the square root of \( x \), how would the equation change?
2. How would the relationship change if \( y \) varies inversely as \( x \) instead of the square root of \( x \)?
3. What happens to \( y \) if \( x \) is doubled in the original relationship?
4. Can you graph the relationship between \( y \) and \( x \) for the given equation?
5. How would the inverse variation be represented if \( y \) varies inversely as the cube root of \( x \)?

**Tip:** Inverse variation problems often require careful attention to how variables change relative to each other. Practice by comparing different forms of variation like direct, inverse, and joint to build a solid understanding.

Explore how to solve inverse variation problems using square root functions. Learn the relationship between y and x in inverse variation, with step-by-step examples and solutions. Suitable for high school students studying algebra and functions.

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