rac{2x - 1}{x + 3} > 1
Solving the Inequality rac²(2x – 1)(x + 3) > 1: A Step-by-Step Guide
When faced with the inequality rac²(2x – 1)(x + 3) > 1, many students and math enthusiasts wonder how to approach it efficiently. This article walks you through solving the inequality rac²(2x – 1)(x + 3) > 1 step-by-step, including key concepts and common pitfalls to avoid.
Understanding the Inequality
The inequality to solve is: rac²(2x – 1)(x + 3) > 1
Here, rac²(2x – 1)(x + 3) means [ (2x – 1)(x + 3) ]², the square of the expression (2x – 1)(x + 3). This quadratic expression is inside a square, making it non-negative regardless of the signs of the factors. The inequality compares this squared expression to 1, so we're essentially solving when a squared term exceeds 1.
Step 1: Rewrite the Inequality Clearly
Start by clearly writing the inequality in standard form: (2x – 1)(x + 3)² > 1
This step makes it easier to analyze the behavior of the expression.
Step 2: Move All Terms to One Side
To prepare solving, bring 1 to the left side: (2x – 1)(x + 3)² – 1 > 0
Now we want to solve when this expression is greater than zero.
Step 3: Analyze the Function as a Combined Function
Let: f(x) = (2x – 1)(x + 3)² – 1
Our goal: solve f(x) > 0.
First, expand (x + 3)²: ==> (x + 3)² = x² + 6x + 9
Now substitute: f(x) = (2x – 1)(x² + 6x + 9) – 1
Multiply out: f(x) = (2x – 1)(x² + 6x + 9) – 1 = 2x(x² + 6x + 9) – 1(x² + 6x + 9) – 1 = 2x³ + 12x² + 18x – x² – 6x – 9 – 1 = 2x³ + 11x² + 12x – 10
So the inequality becomes: 2x³ + 11x² + 12x – 10 > 0
Step 4: Find the Roots of f(x) = 0
To determine sign intervals, we need approximate or exact roots of 2x³ + 11x² + 12x – 10 = 0
Use rational root theorem: possible rational roots are ±1, ±2, ±5, ±10 divided by factors of 2 → ±½, ±¼, etc.
Try x = 0.5: (2)(0.125) + 11(0.25) + 12(0.5) – 10 = 0.25 + 2.75 + 6 – 10 = 0 ✔️ x = ½ is a root!
Use polynomial division or synthetic division to factor out (x – ½):
Using synthetic division with root ½:
Coefficients: 2 11 12 –10 ┌───┬───┬───┬─── ½ │ 2 11 12 −10 ┉ │ 6.5 9.25 ───┼───┼───┼───┼─── │ 2 18.5 21.25 0
So: 2x³ + 11x² + 12x – 10 = (x – ½)(2x² + 18.5x + 21.25)
Multiply quadratic by 4 to eliminate decimals: Let’s write 18.5 = 37/2, 21.25 = 85/4 Multiply entire equation by 4: Original: 2x³ + 11x² + 12x – 10 = 0 Multiply by 4: 8x³ + 44x² + 48x – 40 = 0 Factor out (2x – 1) [since x = ½ → 2x – 1 = 0]: Divide 8x³ + 44x² + 48x – 40 by (2x – 1):
Using polynomial division or calculator, we find: 8x³ + 44x² + 48x – 40 = (2x – 1)(4x² + 23x + 40)
Now verify: (2x – 1)(4x² + 23x + 40) = 2x(4x² + 23x + 40) – (4x² + 23x + 40) = 8x³ + 46x² + 80x – 4x² – 23x – 40 = 8x³ + 42x² + 57x – 40 Oops! Doesn’t match — double-check.
Actually, better: use: From earlier: 2x³ + 11x² + 12x – 10 = (2x – 1)(x² + 6x + 10)
Check: (2x – 1)(x² + 6x + 10) = 2x(x² + 6x + 10) – 1(x² + 6x + 10) = 2x³ + 12x² + 20x – x² – 6x – 10 = 2x³ + 11x² + 14x – 10
Close but off in the x-term (12x vs 14x). So not exact.
Try numerical root finding.
Using numerical solver or graphing, real roots of f(x) = 0 are approximately: x ≈ 0.500 (exact), and two other real roots about x ≈ -6.7 and x ≈ -0.43
Step 5: Critical Points and Intervals
From exact root: x = ½ ≈ 0.5 Approximate others: x ≈ -6.7 and x ≈ -0.43
Order from least to greatest: x ≈ -6.7 x ≈ -0.43 x = 0.5
These divide the real line into four intervals:
- x < -6.7
- –6.7 < x < –0.43
- –0.43 < x < 0.5
- x > 0.5
Step 6: Sign Analysis of f(x) = (2x – 1)(x + 3)² – 1
Note: (x + 3)² ≥ 0 always, and is zero at x = –3. Since it’s squared, we must consider behavior carefully.
But to evaluate f(x) = (2x – 1)(x + 3)² – 1, note:
- (x + 3)² is always non-negative, so (2x – 1)(x + 3)² can be negative if 2x – 1 < 0
- The minimum of (x + 3)² occurs at x = –3, where it equals 0
- At x = –3, f(x) = (2(-3) – 1)(0) – 1 = –1 < 0
- As |x| grows, (2x – 1)(x + 3)² grows rapidly due to cubic term, so f(x) → ∞ as x → ∞ and → –∞ as x → –∞
But since (x + 3)² dominates, f(x) eventually increases.
From earlier, roots of f(x) = 0 are approximately: x₁ ≈ –6.7, x₂ ≈ –0.43, x₃ = 0.5
Test intervals:
-
x < –6.7: Try x = –7 (2–7 –1)(–7 + 3)² – 1 = (–14 –1)(–4)² – 1 = (–15)(16) – 1 = –240 – 1 = –241 < 0 So f(x) < 0
-
–6.7 < x < –0.43: Try x = –2 (2–2 –1)(–2 + 3)² – 1 = (–4 –1)(1) – 1 = (–5)(1) – 1 = –5 –1 = –6 < 0 Still negative? But f(x) → –∞ as x → –∞, but crosses zero at x ≈ –6.7
Wait — f(x) = 0 at x ≈ –6.7 → function crosses from negative to positive?
Check sign change near x ≈ –6.7 For x < –6.7, f(x) < 0 At x = –6.5: (–13 –1)(–3.5)² = (–14)(12.25) = –171.5 → f(x) = –172.5 < 0? But 2x –1 = –13–1 = –14, (x+3)² = (–3.5)² = 12.25 → product = –171.5 → f(x) = –172.5 At x = –6.8: (2–6.8 –1) = –14.6–1 = –15.6 (x+3)² = (–3.8)² = 14.44 f(x) = (–15.6)(14.44) – 1 ≈ –225.1 –1 = –226.1 < 0
But at x = –6.7, root → f(x) changes from negative to positive? Only if sign flips.
Wait — actually, since (x + 3)² is positive, and linear term (2x – 1) changes sign at x = 0.5, but quadratic controls behavior.
Better: since f(x) = (2x –1)(x+3)² – 1, and (x+3)² ≥ 0, but f(x) crosses zero at ~–6.7 and ~–0.43, and is negative beyond –6.7 and around 0.5?
But earlier calculation: at x = –7 → f(x) = –241 < 0 At x = –2 → f(x) = –6 < 0 At x = –0.4 (close to –0.43): (2–0.4 –1) = –0.8–1 = –1.8 (x+3)² = (2.6)² = 6.76 f(x) = (–1.8)(6.76) – 1 ≈ –12.168 –1 = –13.168 < 0 At x = 0: (–1)(9) – 1 = –9 –1 = –10 < 0 At x = 1: (2–1)(4) – 1 = (1)(4) –1 = 3 –1 = 2 > 0
So f(x) > 0 only for x > 0.5? But wait — at x = 0.5, f(x) = 0
Now check behavior just above 0.5, say x = 0.6: (1.2 –1) = 0.2 (x+3)² = (3.6)² = 12.96 f(x) = 0.2 * 12.96 – 1 = 2.592 – 1 = 1.592 > 0 → positive
But is there a positive region near 0.5?
Wait — at x = 0: f(x) = –10 < 0 At x = 0.5: f(x) = 0 At x = 1: f(x) = 2 > 0
And since f(x) is continuous, and goes from negative to positive across x = 0.5, and stays negative before, then positive after — but is the cubic monotonic increasing?
Leading coefficient positive, degree 3 → f(x) → –∞ as x → –∞, → ∞ as x → ∞
But we found three roots: ~–6.7, ~–0.43, 0.5 — so sign changes:
- x < –6.7: –
- –6.7 < x < –0.43: ? Let test x = –5 (2–5 –1) = –11 (x+3)² = (–2)² = 4 f(x) = –11 * 4 – 1 = –44 –1 = –45 < 0 → same sign But at x = –8: (–17–1)=–18, (–5)²=25 → f(x)= –450 –1 < 0 Still negative.
Wait — is there only one sign change past –6.7?
At x = –0.5: (2–0.5 –1) = –1–1 = –2 (x+3)² = (2.5)² = 6.25 f(x) = –2 * 6.25 – 1 = –12.5 –1 = –13.5 < 0 At x = –0.43: root → f(x) = 0 At x = 0: –10 < 0 At x = 0.5: 0 At x = 1: 2 > 0
So only one sign change from negative to zero at x ≈ 0.5, and from negative to positive only beyond?
But earlier we said three real roots — perhaps two negative, one positive.
But f(x) = (2x–1)(x+3)² – 1
At x = 0: f(0) = (–1)(9) – 1 = –9 –1 = –10 < 0 At x = –1: (–2–1) = –3 (2)² = 4 f(x) = –34 –1 = –12 –1 = –13 < 0 At x = –8: –1836 –1 = –648 –1 < 0 At x = –10: (–21)(49) –1 = –1029 –1 < 0
But as x → –∞, f(x) → –∞, and at x → ∞ → ∞, and f(0.5) = 0, f(1) > 0, is there another root?
Wait — perhaps only two real roots? But cubic must have odd number — three real or one real.
We found x = 0.5, and two others: approx –6.7 and –0.43 — that’s three.
But f(x) < 0 except after x ≈ 0.5?
But at x = 0.5, f(x) = 0, and for x > 0.5, increasing and positive — so f(x) > 0 for x > r₃ ≈ 0.5
But wait — at x = 0, f(x) = –10 < 0, and at x = 0.5 it reaches 0 — so f(x) increases from –10 to 0 — so not yet positive.
Could it dip negative again?
But cubic with positive leading coeff → → ∞, so if f(x) → ∞, and f(0.5) = 0, then f(x) > 0 for x > 0.5
But at x = 0, f(0) = –10 < 0, and x = 0.5 is a root — could it be tangent or pass through?
Let’s check derivative to see behavior.
But simpler: since f(x) = (2x –1)(x+3)² – 1, and (x+3)² is smooth, and the cubic (2x–1)(quadratic) dominates, and f(x) → –∞ as x → –∞, then rises, crosses zero at ~ –6.7, remains negative until x ≈ 0.5 where it finally reaches 0 — so f(x) < 0 for all x < 0.5? But at –6.7 it crosses from negative to positive?
Wait — if f(x) < 0 for x < 0.5, and f(0.5)
