Q. (y^2 = x^3 – 47).

Problem

Solve the equation:

\(y^2 = x^3 – 47\)

Step 1: Understand what the equation asks

The equation has two variables, \(x\) and \(y\):

\(y^2 = x^3 – 47\)

This means \(y\) is being squared. To solve for \(y\), we need to undo the square.

The opposite operation of squaring is taking the square root.

Step 2: Take the square root of both sides

Start with:

\(y^2 = x^3 – 47\)

Now take the square root of both sides:

\(\sqrt{y^2} = \sqrt{x^3 – 47}\)

The square root of \(y^2\) is not just \(y\). Since both a positive number and a negative number can give the same square, we must include both signs.

For example:

\(5^2 = 25\)

\((-5)^2 = 25\)

So if \(y^2 = 25\), then \(y = 5\) or \(y = -5\).

That is why we write:

\(y = \pm \sqrt{x^3 – 47}\)

Step 3: Write the solution for \(y\)

The equation solved for \(y\) is:

\(y = \pm \sqrt{x^3 – 47}\)

This means there are usually two possible values of \(y\):

\(y = \sqrt{x^3 – 47}\)

and

\(y = -\sqrt{x^3 – 47}\)

Step 4: Find the real-number restriction

If we want real-number solutions, the expression inside the square root must be greater than or equal to zero.

So we need:

\(x^3 – 47 \ge 0\)

Now solve this inequality.

Add \(47\) to both sides:

\(x^3 \ge 47\)

Now take the cube root of both sides:

\(x \ge \sqrt[3]{47}\)

So real values of \(y\) exist only when:

\(x \ge \sqrt[3]{47}\)

Since \(\sqrt[3]{47}\) is about \(3.61\), this also means:

\(x \ge 3.61\) approximately

Step 5: Final answer

The solution for \(y\) is:

\(y = \pm \sqrt{x^3 – 47}\)

For real-number solutions, the domain is:

\(x \ge \sqrt[3]{47}\)

So the full real-number answer is:

\(y = \sqrt{x^3 – 47}\) and \(y = -\sqrt{x^3 – 47}\), where \(x \ge \sqrt[3]{47}\).