What Is the Positive Solution of (x^2 + 36 = 5x)?
You’ve probably stared at a quadratic equation and thought, “I’m not doing this right.” I’ve been there. When the numbers look messy, the first instinct is to panic. But once you break it down, it’s just a few moves. Let’s walk through the whole process, from the basics to the final answer, and make sure you leave with confidence instead of confusion And that's really what it comes down to..
What Is the Positive Solution of a Quadratic Equation?
In plain talk, a quadratic equation is any equation that can be written in the form
[ ax^2 + bx + c = 0 ]
where (a), (b), and (c) are numbers, and (a) isn’t zero. That's why the solution is the value(s) of (x) that make the equation true. And usually there are two solutions—one might be negative, the other positive, or sometimes both are negative, or both are positive. When the question asks for the positive solution, it means it wants the one that’s greater than zero Easy to understand, harder to ignore. Practical, not theoretical..
Why Do We Care About the Positive Solution?
Because in real life, negative numbers often don’t make sense. Think of distance, time, or money—those are usually non‑negative. If you’re solving for a radius, a length, or a price, you’ll only want the positive root.
The Equation at Hand
We’re looking at
[ x^2 + 36 = 5x ]
The goal is to isolate (x). The first step is to get everything on one side so the right side becomes zero Less friction, more output..
Rearranging
Subtract (5x) from both sides:
[ x^2 - 5x + 36 = 0 ]
Now we have a standard quadratic form with
- (a = 1)
- (b = -5)
- (c = 36)
Checking the Discriminant
The discriminant is (b^2 - 4ac). It tells us how many real solutions we have Worth keeping that in mind. Still holds up..
[ (-5)^2 - 4(1)(36) = 25 - 144 = -119 ]
A negative discriminant means there are no real solutions—only complex ones. That’s a red flag: maybe we mis‑copied the equation? If the problem really is (x^2 + 36 = 5x), then the answer is: *there is no real positive solution.
But let’s assume the intended equation was a typo and the correct one is
[ x^2 + 36 = 5x \quad \Longrightarrow \quad x^2 - 5x + 36 = 0 ]
which, as we saw, has no real roots. If the original problem was actually
[ x^2 - 36 = 5x ]
or
[ x^2 + 36 = 5x ]
the process would be the same; the discriminant would still be negative.
What If the Equation Was Meant to Be (x^2 - 36 = 5x)?
Let’s solve that just in case.
[ x^2 - 36 = 5x \ x^2 - 5x - 36 = 0 ]
Now the discriminant is
[ (-5)^2 - 4(1)(-36) = 25 + 144 = 169 ]
That’s a perfect square ((13^2)). So we can factor or use the quadratic formula.
Using the Quadratic Formula
[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} ]
Plugging in:
[ x = \frac{-(-5) \pm \sqrt{169}}{2(1)} = \frac{5 \pm 13}{2} ]
So the two solutions are
[ x = \frac{5 + 13}{2} = 9 \quad \text{and} \quad x = \frac{5 - 13}{2} = -4 ]
The positive solution is 9 That alone is useful..
Quick Fact Check
If you plug 9 back into the original equation (x^2 - 36 = 5x):
[ 9^2 - 36 = 81 - 36 = 45 \ 5 \times 9 = 45 ]
Works perfectly.
Common Mistakes People Make
- Skipping the sign change when moving terms across the equals sign.
- Using the wrong formula—mixing up the quadratic formula for a linear equation.
- Forgetting to check the discriminant; you might waste time trying to factor a non‑factorable equation.
- Assuming every quadratic has real roots. As we saw, a negative discriminant means the solutions are complex.
- Rounding prematurely. Keep fractions exact until the final step.
Practical Tips That Work
- Always move all terms to one side first. It keeps the equation tidy.
- Compute the discriminant early. If it’s negative, you’re done—no real solutions.
- Factor when possible. It’s faster than the quadratic formula.
- Check your work by plugging back in. One quick substitution can catch a sign error.
- Write down every step. Even if you’re confident, seeing the trail helps avoid slip‑ups.
A Quick Cheat Sheet
| Step | What to Do | Why It Matters |
|---|---|---|
| 1 | Bring everything to one side. | Standard form, easier to spot patterns. |
| 2 | Identify (a), (b), (c). | Needed for the formula and discriminant. |
| 3 | Compute (b^2-4ac). | Tells you whether real solutions exist. |
| 4 | If discriminant ≥ 0, use factorization or formula. | Gives the exact roots. |
| 5 | Pick the positive root if asked. | Matches the problem’s requirement. |
FAQ
Q1: What if the discriminant is zero?
A: That means there’s exactly one real solution (a repeated root). You still use the quadratic formula; the “±” part collapses to a single value That's the part that actually makes a difference. Nothing fancy..
Q2: How do I factor a quadratic that doesn’t look factorable?
A: Use the quadratic formula first. If the result is a nice fraction or integer, you can then factor back out That's the whole idea..
Q3: Why do we care about the sign of the solution?
A: In many real‑world contexts, negative values don’t make sense—like a negative distance or a negative price No workaround needed..
Q4: Can I skip the discriminant and just solve?
A: Sure, but you might waste time trying to factor a non‑factorable equation. The discriminant saves you that hassle It's one of those things that adds up..
Q5: What if my equation has a coefficient other than 1 for (x^2)?
A: The same steps apply; just plug the actual (a) into the formula.
Wrapping It Up
If the equation really was (x^2 + 36 = 5x), the short answer is: there’s no real positive solution. But if the intended problem had a minus sign in front of the 36, the positive root comes out to 9. The key is to set the equation to zero, check the discriminant, and then use the quadratic formula or factoring. Keep these steps in your mental toolbox, and you’ll breeze through any quadratic that comes your way.
Extending the Idea: When the Quadratic Isn’t Isolated
Sometimes the “(x^2) + constant = linear term” pattern appears inside a larger expression, for example:
[ 3\bigl(x^2 + 36\bigr) = 5x + 12. ]
In such cases you still follow the same roadmap, but you have an extra preliminary step: distribute (or expand) before you collect like terms. Here’s the quick workflow:
- Distribute/Expand any parentheses.
[ 3x^2 + 108 = 5x + 12. ] - Bring everything to one side.
[ 3x^2 - 5x + 96 = 0. ] - Identify (a), (b), (c).
[ a = 3,; b = -5,; c = 96. ] - Compute the discriminant.
[ \Delta = (-5)^2 - 4\cdot3\cdot96 = 25 - 1152 = -1127 < 0. ] - Interpret the result. Because (\Delta < 0), the equation has no real solutions—only a pair of complex conjugates.
The extra distribution step is a common source of errors, especially when the coefficient in front of the quadratic term is not 1. Write it out fully, then proceed exactly as you would with a “clean” quadratic Most people skip this — try not to. Nothing fancy..
When the Quadratic Is Embedded in a Fraction
A slightly trickier scenario is when the quadratic sits inside a denominator:
[ \frac{x^2 + 36}{5x} = 1. ]
Multiplying both sides by the denominator (provided (x \neq 0)) clears the fraction:
[ x^2 + 36 = 5x. ]
Now you’re back to the familiar form, and the same analysis applies. On top of that, the caveat is the domain restriction: (x) cannot be zero because you would have divided by zero in the original equation. Always note any restrictions that arise when you multiply or divide by an expression containing the variable.
Real talk — this step gets skipped all the time.
A Real‑World Example: Optimizing a Fence
Suppose a farmer wants to build a rectangular pen using 100 meters of fencing, with one side formed by an existing barn wall (so only three sides need fencing). The area (A) of the pen is (A = x(100 - 2x)), where (x) is the length of the side perpendicular to the barn. To find the dimensions that give a specific area—say, 900 m²—you set up:
[ x(100 - 2x) = 900. ]
Expanding and rearranging yields a quadratic:
[ -2x^2 + 100x - 900 = 0 \quad\Longrightarrow\quad 2x^2 - 100x + 900 = 0. ]
Now apply the steps:
- (a = 2,; b = -100,; c = 900)
- (\Delta = (-100)^2 - 4\cdot2\cdot900 = 10{,}000 - 7{,}200 = 2{,}800) (positive)
- Roots: (\displaystyle x = \frac{100 \pm \sqrt{2{,}800}}{4} = \frac{100 \pm 20\sqrt{7}}{4} = 25 \pm 5\sqrt{7}).
Only the positive root that also respects the physical constraint (0 < x < 50) (you can’t use more than half the fence for the two perpendicular sides) is admissible:
[ x = 25 - 5\sqrt{7} \approx 13.8\text{ m}. ]
The other root, (25 + 5\sqrt{7} \approx 36.2) m, would require a negative length for the side parallel to the barn, which is impossible. This example demonstrates how the same quadratic‑solving checklist guides you through a practical problem, ensuring you respect both algebraic and real‑world constraints.
Most guides skip this. Don't.
Common Pitfalls Revisited
| Pitfall | How to Spot It | Fix |
|---|---|---|
| Dropping a term while moving sides | The resulting constant term looks off (e.In practice, | Write each move on a separate line; double‑check that every term appears on the correct side with the correct sign. g.A quick mental cue—“b squared minus four a c”—helps. If only real numbers are requested, a negative discriminant indeed means “no real solution. |
| Assuming a negative discriminant means “no solution” | Overlooking complex numbers when the problem explicitly allows them. | |
| Rounding early | Approximate numbers appear before you’ve finished simplifying, leading to a slightly off final answer. | Always note domain restrictions whenever you multiply or divide by a variable expression. |
| Dividing by a variable that could be zero | Multiplying both sides by an expression containing (x) without noting (x \neq 0). And ” Otherwise, proceed to write the complex roots. | |
| Mismatched signs in the discriminant | You compute (b^2 + 4ac) instead of (b^2 - 4ac). Day to day, , you end up with (x^2 = 5x) instead of (x^2 - 5x + 36 = 0)). That's why | Clarify the problem’s domain. |
Short version: it depends. Long version — keep reading It's one of those things that adds up..
Checklist for the End‑User
Before you close your notebook, run through this short list:
- Standard Form – Is the equation written as (ax^2 + bx + c = 0)?
- Coefficients Identified – Have you correctly recorded (a), (b), and (c)?
- Discriminant Calculated – Did you compute (\Delta = b^2 - 4ac) correctly?
- Root Type Determined – Does (\Delta) tell you the nature of the roots (real distinct, real repeated, or complex)?
- Roots Found – Have you applied the quadratic formula or factorization accurately?
- Domain Check – Are any values excluded (e.g., division by zero, square‑root of a negative number in a real‑world context)?
- Verification – Does plugging each root back into the original equation satisfy it?
If you answer “yes” to every item, you can be confident that your solution is both mathematically sound and contextually appropriate Worth knowing..
Conclusion
Quadratic equations of the form “(x^2) ± constant = linear term” may look intimidating at first glance, but they are nothing more than a rearranged version of the classic (ax^2 + bx + c = 0). By systematically:
- moving all terms to one side,
- identifying the coefficients,
- evaluating the discriminant,
- solving with the quadratic formula or factoring,
- respecting any sign or domain constraints,
you can demystify any such problem and extract the correct root—real or complex. Remember, the algebraic steps are the same regardless of whether the constant is added or subtracted; the sign merely flips the value of (c) and consequently the discriminant.
Armed with the cheat sheet, the FAQ clarifications, and the practical checklist, you now have a strong mental framework for tackling not only textbook exercises but also real‑world scenarios where quadratics naturally arise. So the next time you encounter a stubborn equation that looks like “(x^2 + 36 = 5x)”, you’ll know exactly how to proceed, why each step matters, and how to verify that your answer truly fits the problem’s requirements. Happy solving!
