Lesson M1.3: Equations and Inequalities

Solve \(x^2 - 7x + 12 = 0\)

Method: Factoring

Need two numbers with product 12 and sum -7.

Numbers: -3 and -4

\(x^2 - 7x + 12 = (x - 3)(x - 4) = 0\)

Solutions: \(x = 3\) or \(x = 4\)

Verification:

• \(x = 3\): \(9 - 21 + 12 = 0\) ✓
• \(x = 4\): \(16 - 28 + 12 = 0\) ✓

Solve \(\frac{2}{x-1} - \frac{3}{x+2} = 1\)

Step 1: Domain restrictions: \(x \neq 1, -2\)

Step 2: LCD = \((x-1)(x+2)\). Multiply:

\(2(x+2) - 3(x-1) = (x-1)(x+2)\)

Step 3: Expand:

\(2x + 4 - 3x + 3 = x^2 + x - 2\)

\(-x + 7 = x^2 + x - 2\)

\(0 = x^2 + 2x - 9\)

Step 4: Use quadratic formula:

\(x = \frac{-2 \pm \sqrt{4 + 36}}{2} = \frac{-2 \pm \sqrt{40}}{2} = \frac{-2 \pm 2\sqrt{10}}{2} = -1 \pm \sqrt{10}\)

Step 5: Check domain:

\(x = -1 + \sqrt{10} \approx 2.16\) ✓

\(x = -1 - \sqrt{10} \approx -4.16\) ✓

Both are valid (neither equals 1 or -2)

Solutions: \(x = -1 \pm \sqrt{10}\)

Solve \(\sqrt{3x - 2} = 4\)

Step 1: Radical is isolated. Square both sides:

\(3x - 2 = 16\)

Step 2: Solve:

\(3x = 18\)

\(x = 6\)

Step 3: Verify:

\(\sqrt{3(6) - 2} = \sqrt{16} = 4\) ✓

Solution: \(x = 6\)

Solve \(x^2 - 4x - 5 \leq 0\)

Step 1: Factor:

\((x - 5)(x + 1) \leq 0\)

Step 2: Critical points: \(x = -1, 5\)

Step 3: Sign chart:

Step 4: We need product \(\leq 0\) (negative or zero)

Negative in \((-1, 5)\), zero at \(x = -1\) and \(x = 5\)

Solution: \([-1, 5]\)

Solve \(\sqrt{x+3} + \sqrt{x-1} = 4\)

Step 1: Isolate one radical:

\(\sqrt{x+3} = 4 - \sqrt{x-1}\)

Step 2: Square both sides:

\(x + 3 = 16 - 8\sqrt{x-1} + (x-1)\)

\(x + 3 = 15 + x - 8\sqrt{x-1}\)

\(-12 = -8\sqrt{x-1}\)

\(\sqrt{x-1} = \frac{3}{2}\)

Step 3: Square again:

\(x - 1 = \frac{9}{4}\)

\(x = 1 + \frac{9}{4} = \frac{13}{4}\)

Step 4: Verify:

\(\sqrt{\frac{13}{4} + 3} + \sqrt{\frac{13}{4} - 1} = \sqrt{\frac{25}{4}} + \sqrt{\frac{9}{4}} = \frac{5}{2} + \frac{3}{2} = 4\) ✓

Solution: \(x = \frac{13}{4}\)

(This is the same as the example in Section 6.2)

Solve \(\frac{x - 3}{x + 4} \geq 2\)

Following the sign analysis method, we get:

\(\frac{-x - 11}{x + 4} \geq 0\)

Critical points: \(x = -11\) (numerator zero), \(x = -4\) (denominator zero)

After sign analysis, the quotient is non-negative on \([-11, -4)\)

Solution: \([-11, -4)\)

Let \(f(x) = x^3 - 3x + 1\)

(a) Show that \(f\) has exactly three real roots.

Step 1: Analyze the derivative.

\(f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x-1)(x+1)\)

Critical points: \(x = -1\) and \(x = 1\)

Step 2: Classify critical points.

\(f''(x) = 6x\)

• At \(x = -1\): \(f''(-1) = -6 < 0\) → local maximum
• At \(x = 1\): \(f''(1) = 6 > 0\) → local minimum

Step 3: Evaluate \(f\) at critical points.

\(f(-1) = (-1)^3 - 3(-1) + 1 = -1 + 3 + 1 = 3 > 0\)

\(f(1) = (1)^3 - 3(1) + 1 = 1 - 3 + 1 = -1 < 0\)

Step 4: Analyze end behavior.

\(\lim_{x \to -\infty} f(x) = -\infty\) and \(\lim_{x \to +\infty} f(x) = +\infty\)

Step 5: Apply the Intermediate Value Theorem.

Since \(f\) is continuous:

• \(f(-\infty) = -\infty < 0\) and \(f(-1) = 3 > 0\) → root in \((-\infty, -1)\)
• \(f(-1) = 3 > 0\) and \(f(1) = -1 < 0\) → root in \((-1, 1)\)
• \(f(1) = -1 < 0\) and \(f(+\infty) = +\infty\) → root in \((1, +\infty)\)

Since a cubic has at most 3 real roots, \(f\) has exactly three real roots. ∎

(b) Prove that exactly one root lies in \((0, 1)\).

\(f(0) = 0 - 0 + 1 = 1 > 0\)

\(f(1) = 1 - 3 + 1 = -1 < 0\)

By IVT, there exists at least one \(c \in (0, 1)\) with \(f(c) = 0\).

For uniqueness: On \((0, 1)\), we have \(f'(x) = 3(x^2 - 1) < 0\) (since \(x^2 < 1\)).

Thus \(f\) is strictly decreasing on \((0, 1)\), so there can be at most one root.

Combined: Exactly one root in \((0, 1)\). ∎

(c) Newton's method convergence from \(x_0 = 0.5\).

Newton's method: \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\)

At \(x_0 = 0.5\):

• \(f(0.5) = 0.125 - 1.5 + 1 = -0.375\)
• \(f'(0.5) = 3(0.25) - 3 = -2.25\)

Since:

1. \(f\) is smooth (polynomial)
2. \(f'(x) \neq 0\) on \((0, 1)\) (we showed \(f' < 0\) there)
3. The root is simple (not repeated)
4. \(x_0 = 0.5\) is close to the root

Newton's method converges quadratically to the unique root in \((0, 1)\).

Theoretical justification: If \(x^*\) is a simple root and \(f''\) is continuous near \(x^*\), then for \(x_0\) sufficiently close to \(x^*\), Newton's method converges with order 2. ∎

Find all \(\lambda\) for which \(A - \lambda I = \begin{pmatrix} 3 - \lambda & 1 \\ 4 & 3 - \lambda \end{pmatrix}\) is singular.

A matrix is singular if and only if its determinant is zero.

Step 1: Compute the determinant.

\(\det(A - \lambda I) = (3 - \lambda)(3 - \lambda) - (1)(4)\)

\(= (3 - \lambda)^2 - 4\)

\(= 9 - 6\lambda + \lambda^2 - 4\)

\(= \lambda^2 - 6\lambda + 5\)

Step 2: Set equal to zero and solve (characteristic equation).

\(\lambda^2 - 6\lambda + 5 = 0\)

Factor: \((\lambda - 5)(\lambda - 1) = 0\)

Step 3: Find eigenvalues.

\(\lambda = 5\) or \(\lambda = 1\)

Verification of properties:

• Sum of eigenvalues: \(5 + 1 = 6 = \text{tr}(A) = 3 + 3\) ✓
• Product of eigenvalues: \(5 \cdot 1 = 5 = \det(A) = 9 - 4 = 5\) ✓

Answer: The eigenvalues are \(\lambda = 1\) and \(\lambda = 5\).

The equation \(x = \cos(x)\) has a unique solution in \([0, \pi/2]\).

(a) Prove existence using IVT.

Define \(g(x) = x - \cos(x)\).

\(g(0) = 0 - \cos(0) = 0 - 1 = -1 < 0\)

\(g(\pi/2) = \pi/2 - \cos(\pi/2) = \pi/2 - 0 = \pi/2 > 0\)

Since \(g\) is continuous on \([0, \pi/2]\) and \(g(0) < 0 < g(\pi/2)\), by IVT, there exists \(c \in (0, \pi/2)\) such that \(g(c) = 0\), i.e., \(c = \cos(c)\). ∎

(b) Show the iteration is a contraction on \([0, 1]\).

For \(\phi(x) = \cos(x)\) to be a contraction on \([0, 1]\), we need \(|\phi'(x)| < 1\) for all \(x \in [0, 1]\).

\(\phi'(x) = -\sin(x)\)

\(|\phi'(x)| = |\sin(x)|\)

For \(x \in [0, 1]\):

• At \(x = 0\): \(|\sin(0)| = 0\)
• At \(x = 1\): \(|\sin(1)| \approx 0.841\)

Since \(\sin\) is increasing on \([0, 1]\) (as \(1 < \pi/2\)):

\(|\phi'(x)| = \sin(x) \leq \sin(1) \approx 0.841 < 1\) for all \(x \in [0, 1]\)

Therefore, \(\phi\) is a contraction with Lipschitz constant \(L = \sin(1) < 1\). ∎

(c) Linear convergence and \(\cos'(x^*)\).

For fixed-point iteration \(x_{n+1} = \phi(x_n)\), the convergence is linear with rate \(|\phi'(x^*)|\) when \(0 < |\phi'(x^*)| < 1\).

The error satisfies: \(|x_{n+1} - x^*| \approx |\phi'(x^*)| \cdot |x_n - x^*|\)

At the fixed point \(x^* \approx 0.739\):

\(|\cos'(x^*)| = |\sin(x^*)| = \sin(0.739) \approx 0.673\)

This means:

• Each iteration reduces the error by a factor of approximately 0.673
• The convergence is linear (not quadratic like Newton's method)
• After \(n\) iterations, error \(\approx (0.673)^n \cdot |\text{initial error}|\)

Interpretation: Since \(\cos'(x^*) \neq 0\), we get only linear convergence. If we wanted faster convergence, we could use Newton's method on \(g(x) = x - \cos(x)\), which would give quadratic convergence. ∎

Solve the inequality that determines when \(f(x) = ax^2 + bx + c\) (with \(a > 0\)) satisfies \(f(x) > 0\) for all real \(x\).

Step 1: Understand the geometry.

Since \(a > 0\), the parabola opens upward. For \(f(x) > 0\) for all \(x\), the parabola must lie entirely above the x-axis.

This happens if and only if the parabola has no real roots.

Step 2: Apply the discriminant condition.

The quadratic \(ax^2 + bx + c = 0\) has no real roots when:

\(\Delta = b^2 - 4ac < 0\)

Step 3: State the condition.

$$b^2 - 4ac < 0$$

Or equivalently:

$$b^2 < 4ac$$

Answer: \(f(x) = ax^2 + bx + c > 0\) for all \(x \in \mathbb{R}\) if and only if \(a > 0\) and \(b^2 < 4ac\).