What’s the positive solution of x² – 36 = 5x?
You’ve probably seen that little algebraic puzzle pop up on homework sheets, in test prep apps, or even as a meme where someone pretends to “solve for x” with a magic wand. The short answer is a single number, but getting there reveals a handful of tricks that are worth keeping in your back‑of‑the‑napkin toolbox.
What Is This Equation, Anyway?
At its core we’re looking at a quadratic equation – an expression where the highest power of x is 2. In plain English the problem reads:
“Find the value(s) of x that make x² minus 36 equal to 5x.”
If you rearrange the terms so everything sits on one side, it becomes the classic “standard form” most textbooks love:
x² – 5x – 36 = 0
That’s the version we’ll work with, because once the equation is set to zero you can apply a handful of reliable methods: factoring, completing the square, or the quadratic formula. The question specifically asks for the positive solution, so after we find both roots we’ll just pick the one that’s greater than zero Not complicated — just consistent. That alone is useful..
Why It Matters / Why People Care
You might wonder why anyone spends time on a single‑digit algebra problem. The truth is that quadratic equations show up everywhere—from physics (projectile motion) to finance (compound interest) to everyday puzzles (when you try to figure out the optimal dimensions of a garden).
If you can spot the pattern “ax² + bx + c = 0” and know how to solve it, you instantly access a shortcut for dozens of real‑world scenarios. Miss the step, and you’ll end up guessing or, worse, trusting a calculator that may have mis‑entered a sign Took long enough..
In practice, mastering this particular equation also sharpens two skills that pay off later:
- Rearranging terms without losing a sign – a common source of “why does my answer look wrong?” moments.
- Choosing the right method – sometimes factoring is fastest, other times the quadratic formula is the only viable path.
How It Works (or How to Do It)
Below are three ways to crack the problem. Pick the one that feels most natural; the answer will be the same.
1. Factoring the Polynomial
First, write the equation in standard form:
x² – 5x – 36 = 0
Now look for two numbers that multiply to –36 and add to –5 It's one of those things that adds up..
- 9 × (–4) = –36
- 9 + (–4) = 5, but we need –5, so flip the signs: –9 + 4 = –5.
That gives us the factor pair (x – 9)(x + 4) = 0 Simple, but easy to overlook..
Set each factor equal to zero:
x – 9 = 0 → x = 9
x + 4 = 0 → x = –4
Two solutions appear, but the question only wants the positive one. The positive solution is 9.
2. Completing the Square
If you’re in a mood for a more “show‑your‑work” approach, completing the square is a neat trick Took long enough..
Start with the standard form again:
x² – 5x – 36 = 0
Move the constant term to the other side:
x² – 5x = 36
Take half of the coefficient of x (which is –5), square it, and add to both sides:
(–5/2)² = 25/4
x² – 5x + 25/4 = 36 + 25/4
The left side now factors into a perfect square:
(x – 5/2)² = 144/4 + 25/4 = 169/4
Take the square root of both sides (remember the ±):
x – 5/2 = ± √(169/4) = ±13/2
Solve for x:
x = 5/2 ± 13/2
That yields two values:
- x = (5 + 13)/2 = 18/2 = 9
- x = (5 – 13)/2 = –8/2 = –4
Again, the positive root is 9 Still holds up..
3. The Quadratic Formula
When factoring feels like hunting for a needle in a haystack, the quadratic formula never fails:
[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} ]
Here a = 1, b = –5, c = –36.
Plug in:
[ 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} ]
[ \sqrt{169} = 13 ]
So:
[ x = \frac{5 \pm 13}{2} ]
Two outcomes:
- (x = \frac{5 + 13}{2} = 9)
- (x = \frac{5 - 13}{2} = -4)
Positive solution again lands on 9 The details matter here..
Common Mistakes / What Most People Get Wrong
- Dropping the negative sign – When you move –36 to the other side, it becomes +36. Forgetting that flips the whole problem.
- Mismatched signs in factoring – It’s easy to pair the wrong numbers (e.g., 12 and –3 give –36 but add to 9, not –5).
- Skipping the “±” when taking square roots – That’s how the negative root disappears, leaving you with an incomplete answer set.
- Dividing by the wrong coefficient – In the quadratic formula, the denominator is 2a. If a ≠ 1 and you forget to multiply by a, you’ll get a completely off answer.
- Assuming the positive root is always the larger number – In some equations the positive root can be smaller (think x² – 10x + 9 = 0, where the positive root is 1). Always check the sign, not the magnitude.
Practical Tips / What Actually Works
- Write the equation in standard form first. Seeing ax² + bx + c = 0 on the page triggers the right mental checklist.
- Test the factor pair quickly: list divisors of |c|, then see which pair gives you b when you consider signs.
- Keep a “quick‑check” box: after you find a root, plug it back into the original equation. One substitution catches most sign slip‑ups.
- Use the discriminant (b² – 4ac) as a sanity gauge. If it’s negative, you know there’s no real solution; if it’s a perfect square, factoring will likely work.
- When in doubt, fall back on the quadratic formula. It’s a reliable safety net, and you’ll never forget it if you practice a few times a week.
FAQ
Q1: Could there be more than one positive solution?
A: For a quadratic, there are at most two real solutions. If both are positive, the constant term c must be positive (so the parabola opens upward and crosses the x‑axis twice on the right side). In our case, one root is negative, so only 9 counts.
Q2: What if the equation were x² + 5x – 36 = 0?
A: Flip the sign of b. Factoring gives (x + 9)(x – 4) = 0, so the positive root would be 4 Less friction, more output..
Q3: Do I need a calculator for the quadratic formula?
A: Not for this problem— √169 is a clean 13. In general, a basic scientific calculator or even a phone’s calculator app handles the square root and division easily.
Q4: How can I remember the quadratic formula?
A: Think of the “rock‑bottom” rhyme: “Minus b, plus or minus the square root, over two a.” Or picture the formula as a tiny “bowl” holding the discriminant It's one of those things that adds up. Practical, not theoretical..
Q5: Is there a graphical way to see the positive solution?
A: Plot y = x² – 5x – 36. The curve crosses the x‑axis at –4 and 9. The right‑hand crossing is the positive solution you’re after.
That’s it. The positive solution to x² – 36 = 5x is 9. Here's the thing — knowing how to get there—whether you factor, complete the square, or fire up the quadratic formula—gives you a flexible toolkit for any similar problem that pops up down the road. Now go ahead and try it on a few variations; the pattern will start to feel second nature. Happy solving!
A Few “What‑If” Variations to Cement the Idea
| Variation | How to Tackle It | Positive Solution |
|---|---|---|
| x² – 36 = –5x | Move the term: x² + 5x – 36 = 0 → (x + 9)(x – 4)=0 | 4 |
| 2x² – 10x – 72 = 0 | Divide by 2 → x² – 5x – 36 = 0 (same as original) | 9 |
| x² + 5x + 6 = 0 | Factor → (x + 2)(x + 3)=0 → both roots negative → no positive solution | |
| x² – 5x – 36 = k (where k is a constant) | Bring k to the left, then factor or use the formula. But the sign of k shifts the parabola up or down, which can create, destroy, or move the positive root. Now, | |
| x² – 5x = 36 | Same as the original but with the constant on the right. Rearrange → x² – 5x – 36 = 0 → 9 again. |
Seeing the same structure reappear in different guises helps you recognize the underlying pattern: the positive root is the one that lies to the right of the y‑axis on the number line. Once you’ve internalized that, the algebraic steps become almost automatic.
When to Choose One Method Over Another
| Situation | Best Method | Why |
|---|---|---|
| Coefficients are small integers | Factoring | Quick, no arithmetic errors, reinforces number‑sense. |
| Leading coefficient a ≠ 1 | Quadratic formula or completing the square | Factoring can become messy; the formula works regardless of a. Think about it: |
| Discriminant is a perfect square | Factoring (if you spot the pair) or formula | Guarantees rational roots, so both approaches are safe. In real terms, |
| Discriminant is not a perfect square | Quadratic formula | Gives exact irrational or complex roots; factoring won’t help. |
| You need a visual check | Graphing (calculator or software) | Instantly shows where the curve crosses the x‑axis; great for confirming sign. |
A Mini‑Exercise Set (Try It Before You Finish)
- Solve x² – 7x – 30 = 0 and state the positive root.
- Determine the positive solution of 3x² – 12x – 9 = 0.
- For x² + 4x + 4 = 0, does a positive root exist? Explain.
Answers:
- (x – 10)(x + 3)=0 → 10.
- Divide by 3 → x² – 4x – 3=0 → x = [4 ± √(16 + 12)]/2 = [4 ± √28]/2 → positive root = (4 + √28)/2 ≈ 4.65.
- (x + 2)²=0 → the only root is –2, so no positive solution.
Final Thoughts
The journey from the original statement “find the positive solution of x² – 36 = 5x” to the crisp answer 9 illustrates a broader lesson: understanding the structure of a quadratic beats rote memorization. By:
- Re‑arranging the equation into standard form,
- Checking the discriminant for feasibility,
- Choosing the most efficient solving technique (factor, complete the square, or formula), and
- Verifying the result by substitution or a quick graph,
you build a reliable, repeatable workflow that applies to any quadratic you encounter.
So the next time a test, homework, or real‑world problem throws a quadratic your way, you’ll know exactly which steps to take, why they work, and how to spot the positive root without second‑guessing yourself.
Happy solving, and may your quadratics always resolve cleanly!
5. Completing the Square – A Visual Shortcut
While the quadratic formula is the universal hammer, completing the square gives you a geometric intuition that often speeds up mental checks. Let’s revisit the original problem with this lens:
[ x^{2}-5x-36=0 ]
-
Move the constant term to the other side
[ x^{2}-5x = 36 ] -
Halve the linear coefficient and square it
[ \left(\frac{-5}{2}\right)^{2}= \frac{25}{4} ] -
Add this square to both sides (the left‑hand side becomes a perfect square)
[ x^{2}-5x+\frac{25}{4}=36+\frac{25}{4} ] [ \left(x-\frac{5}{2}\right)^{2}= \frac{169}{4} ] -
Take the square root
[ x-\frac{5}{2}= \pm \frac{13}{2} ] -
Solve for (x)
[ x = \frac{5}{2} \pm \frac{13}{2} ] giving the two candidates (x = 9) and (x = -4). The positive solution is, once again, 9 Less friction, more output..
The advantage of this method is that it makes the symmetry of the parabola explicit: the vertex sits at (x = \frac{5}{2}), and the distance to each root is (\frac{13}{2}). When you can spot that the number under the radical is a perfect square (here (169 = 13^{2})), the arithmetic collapses to a handful of mental steps Easy to understand, harder to ignore. Turns out it matters..
6. When “Positive” Isn’t Enough – Checking the Domain
In many applied contexts—physics, economics, biology—the variable you’re solving for must satisfy additional constraints beyond “being greater than zero.” For instance:
| Context | Typical Constraint | How It Affects the Solution |
|---|---|---|
| Time (e., t ≥ 0) | Non‑negative | Discard any root that is negative; if both are non‑negative, keep the one that makes sense for the scenario (often the larger one). |
| Length or distance | Positive real numbers | Same as time, but a zero length may be physically impossible, so you may need (x>0). g. |
| Population | Integer ≥ 1 | After finding the real root, round to the nearest whole number and verify that it satisfies the original model. |
| Probability | 0 ≤ p ≤ 1 | If a quadratic yields a root outside this interval, it’s not a valid probability and must be rejected. |
Quick note before moving on.
Practical tip: after you have the algebraic root, plug it back into the original model (or a simplified version) to ensure it respects the domain. This “sanity check” catches sign errors, extraneous solutions from squaring both sides, and other subtle pitfalls.
7. A Quick Reference Cheat Sheet
| Goal | Step‑by‑Step |
|---|---|
| Identify the positive root | 1. 4. |
| Check your work | Substitute the candidate back into the original equation; the left‑hand side should equal the right‑hand side (within rounding error if you used decimals). |
| Factor when possible | Look for two numbers whose product is (ac) and whose sum is (b). 3. Which means rewrite, factor by grouping, then set each factor to zero. |
| Validate against the problem’s context | Ensure the root satisfies any extra conditions (e.g.Still, 2. Compute the discriminant (D=b^{2}-4ac). Write the equation in standard form (ax^{2}+bx+c=0). Choose the sign that yields (x>0). |
| Complete the square | Move (c) to the RHS, add ((b/2a)^{2}) to both sides, rewrite left side as ((x+\frac{b}{2a})^{2}), solve. This leads to use (x=\dfrac{-b\pm\sqrt{D}}{2a}). , (x>0), integer, within a specific interval). |
Keep this sheet printed or bookmarked—you’ll find yourself reaching for it whenever a quadratic pops up in a test, a homework set, or a real‑world calculation.
Conclusion
The “positive solution of (x^{2}-36=5x)” is 9, but the true value of the exercise lies in the toolbox you now possess. By:
- Re‑arranging equations into the canonical quadratic form,
- Choosing the most efficient method—factoring, the quadratic formula, or completing the square—based on the coefficients,
- Interpreting the discriminant to anticipate the nature of the roots, and
- Verifying the solution against both the original algebraic statement and any real‑world constraints,
you have transformed a single problem into a repeatable strategy. Whether you’re tackling a high‑school test, a college‑level physics lab, or a data‑analysis script that models growth, these steps will guide you to the correct positive root quickly and confidently Practical, not theoretical..
So the next time a quadratic appears, remember: the algebra tells you where the roots are, the number line tells you which one is positive, and a quick plug‑in tells you you’re right. With that three‑step sanity check, you’ll never be caught off guard again The details matter here. Worth knowing..
Happy solving—and may every quadratic you meet resolve to a clean, positive answer!
