Algebra Fundamentals
Master the language of mathematics — equations, inequalities, polynomials, and the elegant structures that connect them all.
Key Formulas & Concepts
- Quadratic Formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
- Vieta's Formulas (quadratic): If roots are \(r, s\) then $$r + s = -\frac{b}{a} \quad \text{and} \quad r \cdot s = \frac{c}{a}$$
- Difference of Squares: $$a^2 - b^2 = (a + b)(a - b)$$
- Sum/Difference of Cubes: $$a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$$
- Absolute Value: $$|x| = x \text{ if } x \geq 0;\quad |x| = -x \text{ if } x < 0$$
- Discriminant: $$\Delta = b^2 - 4ac$$ — determines nature of roots
- Factor Theorem: $$p(a) = 0 \iff (x - a) \text{ is a factor of } p(x)$$
- Rational Root Theorem: Possible rational roots of a polynomial are $$\pm \frac{\text{factors of constant}}{\text{factors of leading coefficient}}$$
Find all real values of \(k\) for which the equation \(2x^2 + kx + 8 = 0\) has exactly one real solution.
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Set the discriminant equal to zero: \(k^2 - 4(2)(8) = 0\). Solve for \(k\).
Solve the system: \(x + y + z = 6\), \(xy + yz + xz = 11\), \(xyz = 6\). Find \(x, y, z\).
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These are the elementary symmetric polynomials. Think of \(x, y, z\) as roots of a cubic: \(t^3 - 6t^2 + 11t - 6 = 0\). Factor it.
Find all real \(x\) satisfying \(|2x - 3| + |x + 1| < 7\).
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Consider three intervals: \(x < -1\), \(-1 \leq x < \frac{3}{2}\), and \(x \geq \frac{3}{2}\). Remove absolute values in each case and solve.
Factor completely: \(x^4 - 5x^2 + 4\). Then find all roots.
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Substitute \(u = x^2\) to get \(u^2 - 5u + 4 = (u - 1)(u - 4)\). Then factor each back in terms of \(x\).
The roots of \(x^2 - 7x + k = 0\) differ by 3. Find \(k\) and both roots.
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Let the roots be \(r\) and \(r + 3\). Then \(r + (r+3) = 7\) gives \(r = 2\). Use the product formula to find \(k = r(r+3)\).
Simplify: \(\dfrac{x^3 - 8}{x^2 - 4}\) and state any restrictions on \(x\).
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Factor numerator as difference of cubes: \(x^3 - 8 = (x - 2)(x^2 + 2x + 4)\). Factor denominator as difference of squares.
Simplify: \(\sqrt{6 + 4\sqrt{2}} + \sqrt{6 - 4\sqrt{2}}\).
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Try writing \(6 + 4\sqrt{2}\) as \((a + b)^2\) for some \(a, b\). You'll find \(6 + 4\sqrt{2} = (2 + \sqrt{2})^2\). Do the same for the other term.
Find all functions \(f: \mathbb{R} \rightarrow \mathbb{R}\) such that \(f(x + y) = f(x) + f(y) + 2xy\) for all real \(x, y\), given \(f(1) = 3\).
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Set \(y = 0\) to find \(f(0) = 0\). Set \(y = 1\) to build a recurrence. Guess \(f(x) = x^2 + 2x\) and verify.
For positive reals \(a, b, c\) with \(a + b + c = 1\), prove that \(a^2 + b^2 + c^2 \geq \frac{1}{3}\).
Show Hint
Use the Cauchy-Schwarz inequality or the QM-AM inequality. Alternatively, expand \((a - b)^2 + (b - c)^2 + (c - a)^2 \geq 0\).
Solve for \(x\): \(\log_2(x + 3) + \log_2(x - 1) = 5\).
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Combine logs: \(\log_2((x+3)(x-1)) = 5\), so \((x+3)(x-1) = 32\). Expand and solve the quadratic. Check domain restrictions.
When the polynomial \(p(x) = x^4 + ax^3 + bx^2 + cx + d\) is divided by \((x - 1)\), the remainder is 3. When divided by \((x - 2)\), the remainder is 5. When divided by \((x + 1)\), the remainder is 7. When divided by \((x + 2)\), the remainder is 9. Find \(a + b + c + d\).
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By the Remainder Theorem, \(p(1) = 3\), \(p(2) = 5\), \(p(-1) = 7\), \(p(-2) = 9\). Set up a system of four equations and solve.
Find the number of positive integers \(n \leq 1000\) such that \(\lfloor\sqrt{n}\rfloor\) divides \(n\).
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Let \(\lfloor\sqrt{n}\rfloor = k\), so \(k^2 \leq n < (k+1)^2\). Count the multiples of \(k\) in \([k^2,\, (k+1)^2 - 1]\) for each valid \(k\) from 1 to 31.