How to Solve x³ + x² = 2: A Step‑by‑Step Guide for Students
If you're looking to learn how to solve x³ + x² = 2, this guide walks you through the entire process—from understanding the problem to finding all real and complex solutions. Whether you’re a high‑school student tackling algebra for the first time or a learner brushing up on polynomial equations, the methods described here will give you a clear roadmap and build confidence in handling cubic equations.
Introduction
The equation x³ + x² = 2 is a classic example of a cubic polynomial set equal to a constant. Solving it involves moving all terms to one side, factoring where possible, and then applying the appropriate root‑finding techniques. Still, the goal is to find every value of x that satisfies the equality, including both real and imaginary roots. In this article we’ll break down the problem into manageable steps, explain the underlying theory, and provide practical tips to avoid common mistakes And that's really what it comes down to..
This changes depending on context. Keep that in mind.
Step 1: Rearrange the Equation
The first move is to bring every term onto a single side so the equation is in standard polynomial form.
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Subtract 2 from both sides
[ x^{3} + x^{2} - 2 = 0 ] -
Write the polynomial in descending order
[ f(x) = x^{3} + x^{2} - 2 ]
Now the problem is to find the roots of the cubic polynomial f(x) The details matter here..
Step 2: Apply the Rational Root Theorem
The Rational Root Theorem helps you list possible rational roots quickly. For a polynomial (a_nx^n + a_{n-1}x^{n-1} + \dots + a_0), any rational root (\frac{p}{q}) must satisfy:
- p divides the constant term (‑2) → possible values: ±1, ±2
- q divides the leading coefficient (1) → possible values: ±1
Thus, the only rational candidates are ±1 and ±2.
Step 3: Test the Candidates
Plug each candidate into f(x) to see if it yields zero Simple, but easy to overlook..
| Candidate | f(x) = x³ + x² – 2 | Result |
|---|---|---|
| x = 1 | 1³ + 1² – 2 = 0 | Root |
| x = –1 | (–1)³ + (–1)² – 2 = –1 + 1 – 2 = –2 | Not a root |
| x = 2 | 8 + 4 – 2 = 10 | Not a |
Root | | x = –2 | (–2)³ + (–2)² – 2 = –8 + 4 – 2 = –6 | Not a root |
Only x = 1 satisfies the equation, so we have found one real rational root.
Step 4: Factor the Polynomial Using Synthetic Division
Since (x = 1) is a root, the Factor Theorem guarantees that ((x - 1)) is a factor of (f(x)). We divide (x^{3} + x^{2} - 2) by ((x - 1)) to reduce the cubic to a quadratic.
Synthetic division setup (root = 1):
| 1 | 1 | 0 | –2 | |
|---|---|---|---|---|
| 1 | 1 | 2 | 2 | |
| 1 | 2 | 2 | 0 |
Bring down the leading coefficient (1). Multiply by the root (1) and add to the next coefficient. Repeat.
The bottom row gives the coefficients of the quotient polynomial: (x^{2} + 2x + 2) with a remainder of 0.
Thus, the factorization is: [ x^{3} + x^{2} - 2 = (x - 1)(x^{2} + 2x + 2) ]
Step 5: Solve the Remaining Quadratic
Now solve (x^{2} + 2x + 2 = 0) using the quadratic formula: [ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} ] with (a = 1), (b = 2), (c = 2).
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Calculate the discriminant: [ \Delta = b^{2} - 4ac = 2^{2} - 4(1)(2) = 4 - 8 = -4 ]
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Because (\Delta < 0), the two remaining roots are complex conjugates: [ x = \frac{-2 \pm \sqrt{-4}}{2} = \frac{-2 \pm 2i}{2} = -1 \pm i ]
Step 6: State the Complete Solution Set
Combining the real root from Step 3 with the complex roots from Step 5, the equation (x^{3} + x^{2} = 2) has three solutions (counting multiplicity), as guaranteed by the Fundamental Theorem of Algebra:
[ \boxed{x = 1,\quad x = -1 + i,\quad x = -1 - i} ]
Verification (Optional but Recommended)
It is always good practice to substitute the roots back into the original equation (x^{3} + x^{2} = 2) It's one of those things that adds up..
- For (x = 1): (1^{3} + 1^{2} = 2) ✓
- For (x = -1 + i): [ (-1+i)^{2} = 1 - 2i - 1 = -2i ] [ (-1+i)^{3} = (-2i)(-1+i) = 2i - 2i^{2} = 2i + 2 = 2 + 2i ] [ (2 + 2i) + (-2i) = 2 \quad \text{✓} ]
- For (x = -1 - i): The calculation is the complex conjugate of the above, yielding 2 ✓.
Common Pitfalls to Avoid
- Forgetting to set the equation to zero before applying the Rational Root Theorem or synthetic division.
- Misidentifying the constant term when listing (p/q) candidates (here the constant is (-2), not (+2)).
- Stopping after finding the first real root. A cubic always has three roots (real or complex); the quadratic factor must be solved to complete the answer.
- Arithmetic errors in synthetic division. Double-check the "multiply and add" steps, especially with negative signs.
Conclusion
Solving (x^{3} + x^{2} = 2) demonstrates the standard workflow for tackling cubic equations with rational coefficients: **rearrange, list rational candidates, test for a root, factor by division, and solve the resulting quadratic.Mastering this process not only prepares you for higher-degree polynomials but also reinforces the beautiful interplay between algebraic manipulation and the Fundamental Theorem of Algebra—every polynomial equation of degree (n) has exactly (n) solutions in the complex number system. ** In this case, the single rational root (x = 1) unlocked the factorization, revealing a pair of complex conjugate roots, (-1 \pm i). With practice, these steps become second nature, turning intimidating cubics into routine exercises.
