Solving For Z: √13-z² = 6/z - Math Problem

Hey everyone! Today, we're diving into a cool math problem where we need to find the values of 'z' that make two expressions equal. Specifically, we're looking for the values of z that satisfy the equation √13-z² = 6/z. This involves a bit of algebra, some careful consideration of domains, and a dash of problem-solving fun. So, let's get started!

Setting Up the Equation

First, let's clearly state the equation we need to solve. We want to find the values of z that make the following true:

√13-z² = 6/z

To tackle this, we'll need to get rid of the square root and the fraction. The first thing we can do is square both sides of the equation. This eliminates the square root but introduces a potential for extraneous solutions, which we'll need to check later. Squaring both sides gives us:

(√13-z²)² = (6/z)²

This simplifies to:

13 - z² = 36/z²

Now, to get rid of the fraction, we can multiply both sides of the equation by z². This gives us:

z²(13 - z²) = z²(36/z²)

Which simplifies to:

13z² - z⁴ = 36

Transforming into a Quadratic

Now we have a quartic equation (an equation with a term raised to the fourth power). To make this easier to handle, let's rearrange it into a standard form and then make a clever substitution. Moving all terms to one side, we get:

0 = z⁴ - 13z² + 36

This looks intimidating, but we can make it much simpler by using a substitution. Let's let y = z². This transforms our equation into a quadratic equation in terms of y:

y² - 13y + 36 = 0

Now we have a quadratic equation that we can solve using factoring, the quadratic formula, or completing the square. Factoring is often the easiest if it's possible, so let's try that.

Solving the Quadratic Equation

We need to find two numbers that multiply to 36 and add up to -13. These numbers are -4 and -9. So, we can factor the quadratic equation as:

(y - 4)(y - 9) = 0

This gives us two possible solutions for y:

y = 4 or y = 9

Great! But remember, we're not looking for y; we're looking for z. So we need to substitute back z² for y.

Substituting Back and Solving for z

Now we substitute back z² for y:

z² = 4 or z² = 9

Taking the square root of both sides of each equation gives us the potential solutions for z:

z = ±2 or z = ±3

So, we have four potential solutions: z = 2, z = -2, z = 3, and z = -3. But remember what we said earlier about extraneous solutions? We need to check each of these in the original equation to make sure they actually work.

Checking for Extraneous Solutions

Because we squared both sides of the equation, it's possible that some of our solutions might not actually satisfy the original equation. This can happen because squaring can make a negative value positive, thus altering the solutions. Let's check each potential solution.

Case 1: z = 2

Plug z = 2 into the original equation:

√13-(2)² = 6/2

√13-4 = 3

√9 = 3

3 = 3

This is true, so z = 2 is a valid solution.

Case 2: z = -2

Plug z = -2 into the original equation:

√13-(-2)² = 6/-2

√13-4 = -3

√9 = -3

3 = -3

This is not true, so z = -2 is an extraneous solution. We discard it.

Case 3: z = 3

Plug z = 3 into the original equation:

√13-(3)² = 6/3

√13-9 = 2

√4 = 2

2 = 2

This is true, so z = 3 is a valid solution.

Case 4: z = -3

Plug z = -3 into the original equation:

√13-(-3)² = 6/-3

√13-9 = -2

√4 = -2

2 = -2

This is not true, so z = -3 is an extraneous solution. We discard it.

Final Solutions

After checking all potential solutions, we've found that only two values of z satisfy the original equation: z = 2 and z = 3. These are our final, valid solutions!

Domain Considerations

Before we wrap up, it's essential to consider the domain of our original equation. The domain is the set of all possible values of z that we can plug into the equation without causing any mathematical issues, like dividing by zero or taking the square root of a negative number. Let's break this down.

The Square Root Part

We have a square root in our equation: √13-z². For the expression inside the square root to be real (not imaginary), it must be non-negative. That means:

13 - z² ≥ 0

z² ≤ 13

Taking the square root of both sides, we get:

-|√13| ≤ z ≤ √13

So, z must be between -√13 and √13 (approximately -3.61 and 3.61).

The Fraction Part

We also have a fraction in our equation: 6/z. We can't divide by zero, so z cannot be 0. This gives us another restriction on the domain: z ≠ 0.

Combining the Restrictions

Combining these restrictions, the domain of our equation is all real numbers between -√13 and √13, excluding 0. In interval notation, this is:

[-√13, 0) U (0, √13]

This means that any solution we find must fall within this range to be valid. Luckily, our solutions, z = 2 and z = 3, are both within this domain, so we're good to go!

Conclusion

So, guys, we've successfully navigated this equation and found the values of z that make √13-z² equal to 6/z. We set up the equation, squared both sides, transformed it into a quadratic, solved for the potential solutions, checked for extraneous solutions, and considered the domain. The values of z that satisfy the equation are z = 2 and z = 3. Awesome job! Keep practicing these steps, and you'll become a master problem-solver in no time. Math can be fun and rewarding when you break it down step by step. Keep up the great work, and happy solving!