Consider the following relation.

1−x‾‾‾‾‾√−4y=11−x−4y=1

Step 2 of 3 :  

Evaluate the function found in the previous step at x=−15x=−15.

The relation given is:

\[
\sqrt{1 - x} - 4y = 1
\]

Our task is to evaluate the function found in the previous step at \(x = -15\). Let’s solve this step by step:

### Step 1: Isolate \(\sqrt{1 - x}\)
Starting with the equation:

\[
\sqrt{1 - x} - 4y = 1
\]

Isolate the square root term:

\[
\sqrt{1 - x} = 1 + 4y
\]

### Step 2: Square both sides
Now square both sides to eliminate the square root:

\[
1 - x = (1 + 4y)^2
\]

Expand the right-hand side:

\[
1 - x = 1 + 8y + 16y^2
\]

Now rearrange it to express \(x\) in terms of \(y\):

\[
1 - (1 + 8y + 16y^2) = x
\]

Simplify:

\[
x = -8y - 16y^2
\]

Thus, the function for \(x\) in terms of \(y\) is:

\[
x = -8y - 16y^2
\]

### Step 3: Evaluate at \(x = -15\)
We now evaluate this function when \(x = -15\). Set \(x = -15\) in the equation:

\[
-15 = -8y - 16y^2
\]

Rearranging it:

\[
16y^2 + 8y - 15 = 0
\]

This is a quadratic equation. We can solve it using the quadratic formula:

\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \(a = 16\), \(b = 8\), and \(c = -15\). Now calculate the discriminant:

\[
\Delta = b^2 - 4ac = 8^2 - 4(16)(-15) = 64 + 960 = 1024
\]

Taking the square root of 1024:

\[
\sqrt{1024} = 32
\]

Now plug into the quadratic formula:

\[
y = \frac{-8 \pm 32}{2(16)} = \frac{-8 \pm 32}{32}
\]

Thus, we get two possible values for \(y\):

\[
y_1 = \frac{-8 + 32}{32} = \frac{24}{32} = \frac{3}{4}
\]
\[
y_2 = \frac{-8 - 32}{32} = \frac{-40}{32} = -\frac{5}{4}
\]

### Step 4: Conclusion
When \(x = -15\), the two possible values of \(y\) are:

\[
y = \frac{3}{4} \quad \text{or} \quad y = -\frac{5}{4}
\]

Would you like further clarification on any step?

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Here are some related questions you might find interesting:
1. How do we solve quadratic equations using the quadratic formula?
2. Can quadratic equations have complex solutions?
3. What are some practical applications of solving equations like these?
4. How does squaring both sides of an equation affect the solution?
5. How can we verify the solution for \(y\) by substituting back into the original equation?

**Tip:** Always check your final solutions by substituting them back into the original equation to ensure they work correctly!

Consider the following relation: 1−√(1−x)−4y=1. Evaluate the function found in the previous step at x=−15.

Learn how to solve a quadratic equation with square roots, isolating terms, and applying the quadratic formula to find y when x = -15. This detailed step-by-step solution covers essential algebraic concepts for students in Grades 10-12.

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