What’s the only solution of 2x² + 8x + x² + 16?
You’ve probably seen this expression pop up on a homework sheet or a quick quiz. Practically speaking, the wording feels a bit odd—why not just write it as a single polynomial? But the math is still the same. Let’s break it down, step by step, and see what the “only solution” actually means.
What Is This Equation?
First things first. The expression
2x² + 8x + x² + 16
is just a quadratic polynomial that’s been written in a slightly roundabout way. If you combine like terms, you get
(2x² + x²) + 8x + 16 = 3x² + 8x + 16
So the real question is: what are the values of (x) that make (3x² + 8x + 16 = 0)?
Why It Matters / Why People Care
Quadratics pop up everywhere: from projectile motion to optimizing profit. Knowing how to solve them is a cornerstone of algebra. When a problem asks for “the only solution,” it usually means there’s a single real number that satisfies the equation. That’s a hint that the quadratic might have a repeated root (a double root) or that it simply has no real roots at all.
If you’re stuck on this one, you’re not alone. Many students see a “mis‑written” problem and think they need to rearrange terms or factor in a weird way. The trick is to simplify first, then apply the quadratic formula or check the discriminant.
How It Works (or How to Do It)
1. Combine Like Terms
As we saw, the expression simplifies to:
3x² + 8x + 16 = 0
2. Check the Discriminant
For a quadratic (ax² + bx + c = 0), the discriminant is (Δ = b² - 4ac). It tells you how many real solutions you have:
- (Δ > 0): two distinct real roots
- (Δ = 0): one real root (a repeated or “double” root)
- (Δ < 0): no real roots (two complex conjugates)
Plugging in (a = 3), (b = 8), (c = 16):
Δ = 8² - 4·3·16
= 64 - 192
= -128
Since (-128) is less than zero, there are no real solutions. That’s the short answer.
3. What About Complex Roots?
If you want the full picture, you can solve for the complex roots using the quadratic formula:
x = [-b ± √Δ] / (2a)
With (\Delta = -128):
x = [-8 ± √(-128)] / (2·3)
= [-8 ± i√128] / 6
= [-8 ± i·8√2] / 6
= -4/3 ± (4√2/3)i
So the two complex solutions are:
x = -4/3 + (4√2/3)i
x = -4/3 - (4√2/3)i
But if the problem was asking for “the only solution” in a real‑number context, the answer is: there is none That's the whole idea..
Common Mistakes / What Most People Get Wrong
-
Forgetting to combine like terms.
People sometimes treat the expression as if it were already factored and skip the simplification step. -
Misapplying the quadratic formula.
Plugging the wrong coefficients into the formula leads to nonsense. -
Assuming a solution exists because the problem says “only solution.”
That phrase can be a red flag that the equation has a repeated root or, as in this case, no real roots at all. -
Ignoring the discriminant.
Skipping the discriminant check can make you waste time trying to factor a non‑factorable polynomial Simple, but easy to overlook..
Practical Tips / What Actually Works
-
Always simplify first.
Combine like terms; it turns a confusing expression into a clean quadratic. -
Check the discriminant before rushing to factor.
If (\Delta) is negative, you can skip factoring entirely. -
Use a calculator for the discriminant.
A quick mental math check (e.g., (64 - 192 = -128)) confirms the sign. -
If you’re dealing with a textbook problem, look for a hint.
Many problems that say “only solution” are designed to have a double root. If the discriminant comes out zero, you’ll find a single real root. -
When in doubt, write out the quadratic formula.
Even if you suspect no real roots, having the formula handy ensures you won’t miss the complex ones.
FAQ
Q1: Does “only solution” mean there’s one real number?
Not necessarily. It could mean one solution overall, which could be complex. In this case, the only real solution is none Still holds up..
Q2: Can I factor (3x² + 8x + 16)?
No. Its discriminant is negative, so it has no real linear factors over the reals.
Q3: What if I rewrite the equation as (x²(2 + 1) + 8x + 16 = 0)?
That’s just a fancy way of writing the same thing. It doesn’t change the fact that the discriminant is negative.
Q4: How do I tell if a quadratic has a double root?
If the discriminant equals zero, the quadratic has one real root that repeats.
Q5: Is there a quick test to see if a quadratic has real roots?
Yes: compute (b² - 4ac). Positive means two real roots, zero means one, negative means none.
Closing
So, what’s the only solution of (2x² + 8x + x² + 16)? If you’re looking for a real number, there isn’t one. The equation’s discriminant is negative, so the polynomial never crosses the x‑axis. If you’re comfortable with complex numbers, you can write down the two conjugate solutions, but in the real world, the answer is simply no real solution. That’s the takeaway, and it’s a good reminder to always simplify, check the discriminant, and then decide what kind of roots you’re after That alone is useful..
Easier said than done, but still worth knowing.
Putting It All Together
Let’s walk through the entire process one more time, this time as a quick reference you can drop onto a sticky note or keep in your math notebook.
| Step | What to Do | Why It Matters |
|---|---|---|
| 1. Think about it: combine like terms | (2x^{2}+x^{2}=3x^{2}) → (3x^{2}+8x+16=0) | A clean equation is easier to read and less error‑prone. And |
| 2. Identify (a), (b), (c) | (a=3,; b=8,; c=16) | Needed for the discriminant and quadratic formula. In real terms, |
| 3. Compute the discriminant | (\Delta = 8^{2}-4(3)(16)=64-192=-128) | Tells you the nature of the roots before you dive deeper. Also, |
| 4. Decide on the solution set | (\Delta<0) → no real roots; two complex roots exist. Even so, | Prevents wasted effort trying to factor over (\mathbb{R}). That's why |
| 5. Write the complex solutions (optional) | (x=\frac{-8\pm i\sqrt{128}}{6}) → (-\frac{4}{3}\pm\frac{4i}{3}\sqrt{2}) | Gives the full set of solutions if complex numbers are acceptable. |
Quick “What‑If” Scenarios
-
What if the problem had asked for integer solutions?
Since there are no real solutions, there are obviously no integer solutions either. -
What if the coefficient of (x^{2}) had been (-3) instead of (3)?
The discriminant would be (64-4(-3)(16)=64+192=256), yielding two distinct real roots (x=2) and (x=-\frac{8}{3}) That's the whole idea.. -
What if the problem had a typo and the constant term was (8) instead of (16)?
Then (\Delta=64-4(3)(8)=64-96=-32), still no real roots—so the typo wouldn’t change the conclusion.
Final Takeaway
The equation (2x^{2} + 8x + x^{2} + 16 = 0) simplifies to a single quadratic (3x^{2} + 8x + 16 = 0). Its discriminant is (-128), a negative number that tells us the parabola lies entirely above the x‑axis. Consequently:
- No real solutions exist.
- Two complex conjugate solutions do exist and are given by (-\frac{4}{3}\pm\frac{4i}{3}\sqrt{2}).
If a problem statement insists on “only solution,” double‑check the context: it might refer to a single complex root (as in a repeated root when (\Delta=0)) or simply be a careless phrasing. In this case, the “only solution” in the real number system is none.
So, the next time you encounter a quadratic that seems to be hiding behind a string of terms, remember the three‑step mantra: simplify → check the discriminant → decide on the root type. It saves time, eliminates guesswork, and keeps you from falling into the classic traps of mis‑coefficient, mis‑factoring, or ignoring the discriminant.
