Q. \( z^2 + y^2 z + x^3 – 3 = 0. \)

Problem

Solve the equation:

\(z^2 + y^2z + x^3 – 3 = 0\)

Step 1: Identify the variable we want to solve for

The equation is:

\(z^2 + y^2z + x^3 – 3 = 0\)

This equation contains the variables \(x\), \(y\), and \(z\).

The expression has a \(z^2\) term and a \(z\) term, so it is natural to solve it as a quadratic equation in \(z\).

That means we treat \(x\) and \(y\) as constants while solving for \(z\).

Step 2: Compare the equation with the standard quadratic form

The standard quadratic form is:

\(az^2 + bz + c = 0\)

Our equation is:

\(z^2 + y^2z + x^3 – 3 = 0\)

Now match each part:

The coefficient of \(z^2\) is:

\(a = 1\)

The coefficient of \(z\) is:

\(b = y^2\)

The constant term is:

\(c = x^3 – 3\)

So we have:

\(a = 1\)

\(b = y^2\)

\(c = x^3 – 3\)

Step 3: Use the quadratic formula

For a quadratic equation of the form:

\(az^2 + bz + c = 0\)

the quadratic formula is:

\(z = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\)

Now substitute:

\(a = 1\)

\(b = y^2\)

\(c = x^3 – 3\)

This gives:

\(z = \frac{-y^2 \pm \sqrt{(y^2)^2 – 4(1)(x^3 – 3)}}{2(1)}\)

Step 4: Simplify the expression inside the square root

The expression inside the square root is called the discriminant.

It is:

\((y^2)^2 – 4(1)(x^3 – 3)\)

First simplify \((y^2)^2\):

\((y^2)^2 = y^4\)

Now simplify the product:

\(-4(1)(x^3 – 3) = -4(x^3 – 3)\)

Distribute \(-4\):

\(-4(x^3 – 3) = -4x^3 + 12\)

So the discriminant becomes:

\(y^4 – 4x^3 + 12\)

Step 5: Substitute the simplified discriminant back into the formula

We had:

\(z = \frac{-y^2 \pm \sqrt{(y^2)^2 – 4(1)(x^3 – 3)}}{2(1)}\)

After simplifying the square root part, we get:

\(z = \frac{-y^2 \pm \sqrt{y^4 – 4x^3 + 12}}{2}\)

This is the equation solved for \(z\).

Step 6: Write the two solution branches

The symbol \(\pm\) means there are two possible solutions.

The first solution uses the plus sign:

\(z = \frac{-y^2 + \sqrt{y^4 – 4x^3 + 12}}{2}\)

The second solution uses the minus sign:

\(z = \frac{-y^2 – \sqrt{y^4 – 4x^3 + 12}}{2}\)

Step 7: Check the real-number condition

If we want real-number values of \(z\), the expression inside the square root must be greater than or equal to zero.

The expression inside the square root is:

\(y^4 – 4x^3 + 12\)

So the real-number condition is:

\(y^4 – 4x^3 + 12 \ge 0\)

This condition tells us which values of \(x\) and \(y\) give real values of \(z\).

We can also rewrite the condition:

\(y^4 + 12 \ge 4x^3\)

Divide both sides by \(4\):

\(\frac{y^4 + 12}{4} \ge x^3\)

Equivalently:

\(x^3 \le \frac{y^4 + 12}{4}\)

Taking the cube root gives:

\(x \le \sqrt[3]{\frac{y^4 + 12}{4}}\)

So real values of \(z\) exist when:

\(x \le \sqrt[3]{\frac{y^4 + 12}{4}}\)

Step 8: Verify by substitution conceptually

The original equation is quadratic in \(z\):

\(z^2 + y^2z + x^3 – 3 = 0\)

The quadratic formula gives all values of \(z\) that satisfy a quadratic equation.

Since we correctly identified:

\(a = 1\)

\(b = y^2\)

\(c = x^3 – 3\)

and substituted these into the quadratic formula, the two expressions found for \(z\) solve the original equation.

Final answer

The equation solved for \(z\) is:

\(z = \frac{-y^2 \pm \sqrt{y^4 – 4x^3 + 12}}{2}\)

So the two solution branches are:

\(z = \frac{-y^2 + \sqrt{y^4 – 4x^3 + 12}}{2}\)

and

\(z = \frac{-y^2 – \sqrt{y^4 – 4x^3 + 12}}{2}\)

For real-number solutions, the condition is:

\(y^4 – 4x^3 + 12 \ge 0\)

Equivalently:

\(x \le \sqrt[3]{\frac{y^4 + 12}{4}}\)