Solving For Q A Step By Step Guide To Making Q The Subject

Hey guys! Ever found yourself staring at a mathematical equation, feeling like you're trying to decipher an ancient scroll? Well, you're not alone! Math can be tricky, but breaking it down step by step can make it way less intimidating. Today, we're going to tackle a classic problem: making 'q' the subject of the relation in the equation $\bf{t=\sqrt{\frac{ pq }{ r }}- r ^2 q}$. This basically means we want to rearrange the equation so that it reads 'q = [some expression involving other variables]'. Sounds like a mission? Let’s get started!

Understanding the Challenge

Before we dive into the steps, let's understand what we're up against. The equation $\bf{t=\sqrt{\frac{ pq }{ r }}- r ^2 q}$ looks complex because 'q' appears in two different places: inside a square root and as a standalone term. This means we'll need to use a combination of algebraic manipulations to isolate 'q'. We'll be using techniques like isolating terms, squaring both sides, factoring, and potentially the quadratic formula. So, buckle up, and let's get our hands dirty with some algebra!

Step 1: Isolating the Square Root

The first step in making 'q' the subject is to isolate the square root term. This will allow us to get rid of the square root in the next step. To do this, we need to add $\bf{r^2q}$ to both sides of the equation. Remember, whatever we do to one side of the equation, we must do to the other to maintain the balance. So, we have:

$\bf{t + r^2q = \sqrt{\frac{pq}{r}}}$

Now, the square root term is nicely isolated on the right-hand side. This sets us up perfectly for the next step, where we'll eliminate the square root altogether.

Step 2: Eliminating the Square Root

Okay, now for the fun part: getting rid of that pesky square root! To do this, we'll square both sides of the equation. Squaring a square root cancels it out, which is exactly what we want. Squaring the left-hand side, $\bf{(t + r^2q)}$, requires a bit more care since it's a binomial. We need to remember the formula $\bf{(a + b)^2 = a^2 + 2ab + b^2}$. Applying this, we get:

$\bf{(t + r^2q)^2 = t^2 + 2tr^2q + r^4q^2}$

Squaring the right-hand side, $\bf{\sqrt{\frac{pq}{r}}}$, is much simpler. The square root just disappears, leaving us with:

$\bf{\frac{pq}{r}}$

So, our equation now looks like this:

$\bf{t^2 + 2tr^2q + r^4q^2 = \frac{pq}{r}}$

We've successfully eliminated the square root, but we're left with a quadratic equation in terms of 'q'. Don't worry, we'll tackle this next!

Step 3: Rearranging into a Quadratic Equation

Now that we've squared both sides, we have a quadratic equation in 'q'. A quadratic equation is an equation of the form $\bf{ax^2 + bx + c = 0}$, where 'x' is the variable. To solve for 'q', we need to rearrange our equation into this standard form. First, let’s multiply both sides of the equation by $\bf{r}$ to get rid of the fraction:

$\bf{r(t^2 + 2tr^2q + r^4q^2) = pq}$

This simplifies to:

$\bf{rt^2 + 2tr^3q + r^5q^2 = pq}$

Next, we want to move all the terms to one side of the equation to set it equal to zero. Let's subtract $\bf{pq}$ from both sides:

$\bf{r^5q^2 + 2tr^3q - pq + rt^2 = 0}$

Now, we can rearrange the terms to group the 'q' terms together:

$\bf{r^5q^2 + (2tr^3 - p)q + rt^2 = 0}$

We now have a quadratic equation in the standard form $\bf{aq^2 + bq + c = 0}$, where:

• $\bf{a = r^5}$
• $\bf{b = 2tr^3 - p}$
• $\bf{c = rt^2}$

This sets us up for the final step: using the quadratic formula to solve for 'q'.

Step 4: Applying the Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations. It states that for an equation of the form $\bf{ax^2 + bx + c = 0}$, the solutions for 'x' are given by:

$\bf{x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$

In our case, 'x' is 'q', and we have already identified 'a', 'b', and 'c'. Let's plug those values into the quadratic formula:

$\bf{q = \frac{-(2tr^3 - p) \pm \sqrt{(2tr^3 - p)^2 - 4(r^5)(rt^2)}}{2r^5}}$

This looks like a mouthful, but we're almost there! Now, let's simplify this expression step by step.

Step 5: Simplifying the Quadratic Formula Solution

First, let's expand the term inside the square root:

$\bf{(2tr^3 - p)^2 = 4t^2r^6 - 4tr^3p + p^2}$

And the second term inside the square root:

$\bf{4(r^5)(rt^2) = 4r^6t^2}$

Now, substitute these back into the square root:

$\bf{\sqrt{4t^2r^6 - 4tr^3p + p^2 - 4r^6t^2}}$

Notice that the $\bf{4t^2r^6}$ and $\bf{-4r^6t^2}$ terms cancel each other out, leaving us with:

$\bf{\sqrt{p^2 - 4tr^3p}}$

So, our equation for 'q' now looks like this:

$\bf{q = \frac{-(2tr^3 - p) \pm \sqrt{p^2 - 4tr^3p}}{2r^5}}$

We can further simplify this by factoring out a 'p' from the square root:

$\bf{\sqrt{p^2 - 4tr^3p} = \sqrt{p(p - 4tr^3)} = \sqrt{p} \sqrt{p - 4tr^3}}$

Substituting this back into our equation, we get:

$\bf{q = \frac{p - 2tr^3 \pm \sqrt{p(p - 4tr^3)}}{2r^5}}$

And there you have it! We have successfully made 'q' the subject of the relation. This final expression gives us two possible solutions for 'q', one with the plus sign and one with the minus sign.

Conclusion

Making 'q' the subject of the relation $\bf{t=\sqrt{\frac{ pq }{ r }}- r ^2 q}$ was a challenging but rewarding journey. We started by isolating the square root, then squared both sides to eliminate it. This led us to a quadratic equation, which we solved using the quadratic formula. Finally, we simplified the result to obtain our solution for 'q'. Remember, the key to tackling complex math problems is to break them down into smaller, manageable steps. Keep practicing, and you'll become a math whiz in no time! Guys, math isn't scary; it's just a puzzle waiting to be solved. Keep at it!