What’s the positive solution of (x^2 - 36 = 5x)?
You’ve probably seen this kind of question on a test or a homework sheet. It looks like a simple algebra problem, but the way you approach it can make a big difference. Let’s dig into the math, walk through the steps, and see why the positive root matters in real life.
What Is the Equation?
At first glance, the expression “(x^2 36 5x)” is a bit confusing. Most people read it as a typo or a missing operator. The intended equation is
[ x^2 - 36 = 5x ]
Here, (x^2) is a square term, (-36) is a constant, and (5x) is a linear term. The goal is to find the value(s) of (x) that make the equation true. In practice, you’ll usually bring everything to one side and solve a standard quadratic equation.
Why This Matters
Quadratic equations pop up everywhere: from calculating projectile motion to optimizing profit in business. Knowing how to isolate the positive root isn’t just a school exercise—it’s a skill that translates to budgeting, engineering, and even game design. When you get stuck on the algebra, you’re missing a tool that can help you solve real problems faster Turns out it matters..
How to Solve It
Let’s walk through the steps. It’s a three‑step process: rearrange, factor or use the quadratic formula, and pick the positive root Small thing, real impact..
1. Rearrange to Standard Form
Move every term to one side so the equation reads
[ x^2 - 5x - 36 = 0 ]
Now we have a classic quadratic: (ax^2 + bx + c = 0) with (a = 1), (b = -5), and (c = -36).
2. Factor or Use the Quadratic Formula
Factoring (if possible)
We look for two numbers that multiply to (-36) and add to (-5). Those numbers are (-9) and (4):
[ (x - 9)(x + 4) = 0 ]
This factorization instantly gives two potential solutions: (x = 9) or (x = -4).
Quadratic Formula (for completeness)
If factoring is tricky, the quadratic formula works every time:
[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]
Plugging in our values:
[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-36)}}{2(1)} = \frac{5 \pm \sqrt{25 + 144}}{2} = \frac{5 \pm \sqrt{169}}{2} = \frac{5 \pm 13}{2} ]
So we again get (x = 9) or (x = -4).
3. Pick the Positive Root
Since the question asks for the positive solution, we choose (x = 9). The negative root (-4) is mathematically correct but irrelevant if only positive values make sense in your context (e.g., a physical distance or a quantity that can’t be negative).
Common Mistakes
- Skipping the sign on the linear term. Forgetting the minus sign before (5x) flips the whole problem.
- Assuming both roots are valid. In many real‑world problems, one root is extraneous because it violates a physical constraint.
- Forgetting to simplify before applying the quadratic formula. If you leave the equation as (x^2 - 36 = 5x) and plug it directly into the formula, you’ll end up with a mess.
Practical Tips
- Check your work: After finding the roots, plug them back into the original equation to confirm they work.
- Look for factorable pairs first: Factoring saves time and reduces the chance of arithmetic errors.
- Use a calculator for the discriminant: Even a simple calculator can quickly confirm whether the square root is an integer (like (13) here) or a decimal.
- Remember the context: If you’re solving for a time, distance, or a quantity that can’t be negative, discard the negative root right away.
FAQ
Q1: What if the quadratic doesn’t factor nicely?
Use the quadratic formula. It always works, though you might get irrational numbers.
Q2: Why does the discriminant need to be positive?
Because the square root of a negative number isn’t a real number. A negative discriminant means no real roots exist Which is the point..
Q3: Can I use a graph to find the positive root?
Absolutely. Plot (y = x^2 - 5x - 36). The x‑intercepts are the roots. The right‑hand intercept is the positive one The details matter here..
Q4: Is there a shortcut if I only need the positive root?
If you’re comfortable with the quadratic formula, just keep the “+” part of (\pm). That gives you the larger root when (a > 0).
Q5: What if the equation is (x^2 + 36 = 5x)?
Move everything: (x^2 - 5x + 36 = 0). Check the discriminant: (25 - 144 = -119). No real roots Worth keeping that in mind. Worth knowing..
Wrapping It Up
Finding the positive solution to (x^2 - 36 = 5x) is a quick win once you know the steps. But rearrange, factor or apply the quadratic formula, then pick the root that fits the problem’s constraints. Keep these shortcuts in your algebra toolkit, and you’ll be ready for any quadratic that comes your way—whether it’s a physics problem, a business model, or a puzzle in a game Which is the point..
Extending the Idea: When the Coefficients Change
What if the numbers aren’t as tidy as 36 and 5? The same workflow still applies, but a few extra tricks can keep the process smooth.
| Situation | What to Do |
|---|---|
| Large constant term (e.Here's the thing — g. , (x^2 - 144 = 7x)) | Divide the whole equation by the greatest common divisor (if any) before factoring. Still, in this example, all terms are divisible by 1, so you move straight to factoring or the formula. |
| Coefficient of (x^2) ≠ 1 (e.g.And , (2x^2 - 36 = 5x)) | First bring everything to one side: (2x^2 - 5x - 36 = 0). Then either factor by grouping or apply the quadratic formula with (a = 2). |
| Non‑integer discriminant (e.Practically speaking, g. Consider this: , (x^2 - 5x - 2 = 0)) | The formula will give an irrational root: (x = \frac{5 \pm \sqrt{33}}{2}). If you only need a decimal approximation, a calculator or a quick long‑division style square‑root estimate will do. Still, |
| Complete the square (useful for vertex form) | Rewrite (x^2 - 5x) as ((x - \frac{5}{2})^2 - (\frac{5}{2})^2). Then the equation becomes ((x - \frac{5}{2})^2 = 36 + (\frac{5}{2})^2). This method is especially handy when you need the axis of symmetry or the vertex of the parabola. |
A Quick Example with a Different Coefficient
Suppose you encounter:
[ 3x^2 - 12 = 9x. ]
- Collect terms: (3x^2 - 9x - 12 = 0).
- Simplify: Divide by 3 → (x^2 - 3x - 4 = 0).
- Factor: ((x - 4)(x + 1) = 0).
- Roots: (x = 4) or (x = -1).
- Select the positive root: (x = 4).
Notice how dividing by the common factor early collapsed the problem to the familiar form we solved earlier.
Visualizing the Positive Root
A graph can reinforce why we discard the negative root in many contexts. But plot the parabola (y = x^2 - 5x - 36). The curve opens upward (since the coefficient of (x^2) is positive) and crosses the x‑axis at (-4) and (9). Think about it: the portion of the graph to the right of the y‑axis represents all positive (x) values. The intersection at (x = 9) is the only point where the parabola meets the axis in that region, confirming algebraically that 9 is the unique positive solution.
If you prefer a quick mental picture, remember that for any quadratic with a positive leading coefficient, the larger root is always the positive one provided the product of the roots (the constant term divided by (a)) is negative. Day to day, in our original equation, the constant term after moving everything is (-36), and (a = 1), so the product of the roots is (-36). A negative product forces the roots to have opposite signs—hence exactly one positive root.
Real‑World Checklists
When you finish a problem, ask yourself:
-
Does the answer make sense in the original scenario?
- A distance can’t be negative.
- A time interval must be non‑negative.
-
Did you verify the solution?
- Substitute back into the original equation.
-
Is there a simpler method you missed?
- Factoring is usually quicker than the formula when possible.
-
Did you simplify the equation before solving?
- Removing common factors often prevents arithmetic slip‑ups.
Final Thoughts
The journey from the equation (x^2 - 36 = 5x) to the clean answer (x = 9) illustrates a core principle of algebra: transform first, solve second. By moving all terms to one side, simplifying, and then choosing the most efficient solving technique—factoring when you can, the quadratic formula when you must—you keep the process organized and error‑free It's one of those things that adds up..
Remember, the positive root isn’t just a number; it’s the solution that respects the constraints of the problem you’re modeling. Whether you’re calculating the height of a projectile, the amount of material needed for a construction project, or the optimal price point in a business model, the same steps apply. Master them, and you’ll be equipped to tackle any quadratic that crosses your path.
Bottom line: Rearrange, simplify, solve, and then validate against the context. Follow this checklist, and the positive root will always emerge clear and confident.
