Comprehensive Review of High School Mathematics Prerequisites

Operations on Fractions

Fractional expressions require careful application of common denominators and algebraic identities to simplify properly.

For a strictly positive real number $x > 0$, consider the algebraic identity verification:

$\frac{\frac{1}{x}}{1 + \frac{1}{x}} + \frac{1 - \frac{1}{x}}{\frac{1}{x}} = \frac{x^2}{x + 1}$

To prove this statement, simplify each fractional term individually before combining. In the first term, rewriting the denominator with a common denominator yields $1 + \frac{1}{x} = \frac{x + 1}{x}$. Substituting this back gives:

$\frac{\frac{1}{x}}{\frac{x + 1}{x}} = \frac{1}{x} \times \frac{x}{x + 1} = \frac{1}{x + 1}$

In the second term, rewriting the numerator with a common denominator yields $1 - \frac{1}{x} = \frac{x - 1}{x}$. Substituting this back gives:

$\frac{\frac{x - 1}{x}}{\frac{1}{x}} = \frac{x - 1}{x} \times \frac{x}{1} = x - 1$

Adding both simplified terms together requires finding a common denominator:

$\frac{1}{x + 1} + (x - 1) = \frac{1 + (x - 1)(x + 1)}{x + 1}$

Applying the standard algebraic identity $(a - b)(a + b) = a^2 - b^2$ to the numerator yields $(x - 1)(x + 1) = x^2 - 1$. Thus, the numerator simplifies to $1 + x^2 - 1 = x^2$, resulting in the final expression:

$\frac{x^2}{x + 1}$

For two non-zero real numbers $x$ and $y$ satisfying $x + y \neq 0$ and $x - y \neq 0$, consider the identity:

$\frac{\frac{1}{x}}{\frac{1}{x} - \frac{1}{y}} - \frac{\frac{1}{y}}{\frac{1}{x} + \frac{1}{y}} = \frac{y^2 + x^2}{y^2 - x^2}$

Simplifying the denominator of the first term gives $\frac{1}{x} - \frac{1}{y} = \frac{y - x}{x y}$. Dividing $\frac{1}{x}$ by this expression yields:

$\frac{\frac{1}{x}}{\frac{y - x}{x y}} = \frac{1}{x} \times \frac{x y}{y - x} = \frac{y}{y - x}$

Simplifying the denominator of the second term gives $\frac{1}{x} + \frac{1}{y} = \frac{y + x}{x y}$. Dividing $\frac{1}{y}$ by this expression yields:

$\frac{\frac{1}{y}}{\frac{y + x}{x y}} = \frac{1}{y} \times \frac{x y}{y + x} = \frac{x}{y + x}$

Subtracting the two resulting fractions with a common denominator of $(y - x)(y + x)$ gives:

$\frac{y(y + x) - x(y - x)}{(y - x)(y + x)}$

Expanding the numerator yields $y^2 + x y - x y + x^2 = y^2 + x^2$. Expanding the denominator using $(a - b)(a + b) = a^2 - b^2$ gives $y^2 - x^2$. The combined fraction is therefore:

$\frac{y^2 + x^2}{y^2 - x^2}$

Operations on Powers

Simplifying expressions involving powers requires systematic application of exponent rules, including power of a product $(a \times b)^n = a^n \times b^n$, power of a power $(a^n)^m = a^{n \times m}$, product of powers $a^n \times a^m = a^{n + m}$, quotient of powers $\frac{a^n}{a^m} = a^{n - m}$, negative exponent rule $a^{-n} = \frac{1}{a^n}$, and power of a quotient $\frac{a^n}{b^n} = \left(\frac{a}{b}\right)^n$

Consider the simplification of the expression $X$:

$X = \frac{b^{16}}{a^{10}} \times \left(\frac{a^3}{b^2}\right)^2 \times \left(\frac{b^2}{a^4}\right)^4$

First, apply the power of a power rule to each factor in parentheses:

$\left(\frac{a^3}{b^2}\right)^2 = \frac{a^{3 \times 2}}{b^{2 \times 2}} = \frac{a^6}{b^4}$

$\left(\frac{b^2}{a^4}\right)^4 = \frac{b^{2 \times 4}}{a^{4 \times 4}} = \frac{b^8}{a^{16}}$

Substituting these back into the expression yields:

$X = \frac{b^{16}}{a^{10}} \times \frac{a^6}{b^4} \times \frac{b^8}{a^{16}}$

Group like base terms together:

$X = \frac{b^{16} \times b^8}{b^4} \times \frac{a^6}{a^{10} \times a^{16}}$

Apply the product rule for exponents to combine terms in the numerator and denominator:

$b^{16} \times b^8 = b^{16 + 8} = b^{24}$

$a^{10} \times a^{16} = a^{10 + 16} = a^{26}$

This gives:

$X = \frac{b^{24}}{b^4} \times \frac{a^6}{a^{26}}$

Apply the quotient rule for exponents:

$\frac{b^{24}}{b^4} = b^{24 - 4} = b^{20}$

$\frac{a^6}{a^{26}} = a^{6 - 26} = a^{-20}$

Using the negative exponent rule $a^{-20} = \frac{1}{a^{20}}$, the expression becomes:

$X = b^{20} \times \frac{1}{a^{20}} = \frac{b^{20}}{a^{20}}$

Finally, applying the quotient power rule yields:

$X = \left(\frac{b}{a}\right)^{20}$

Now consider the expression $Y$:

$Y = \frac{a^3 \times b \times (a^2 \times b^{-2})^4 \times a^5}{a^5 \times b^{-2} \times a^{-1} \times a^2 \times b^{-3}}$

Simplify the exponential term in the numerator first:

$(a^2 \times b^{-2})^4 = (a^2)^4 \times (b^{-2})^4 = a^8 \times b^{-8}$

Rewrite the full numerator and group powers of base $a$ and base $b$:

$\text{Numerator} = a^3 \times b^1 \times a^8 \times b^{-8} \times a^5 = a^{3 + 8 + 5} \times b^{1 - 8} = a^{16} \times b^{-7}$

Simplify and group powers in the denominator:

$\text{Denominator} = a^5 \times b^{-2} \times a^{-1} \times a^2 \times b^{-3} = a^{5 - 1 + 2} \times b^{-2 - 3} = a^6 \times b^{-5}$

Combine the simplified numerator and denominator:

$Y = \frac{a^{16} \times b^{-7}}{a^6 \times b^{-5}} = \left(\frac{a^{16}}{a^6}\right) \times \left(\frac{b^{-7}}{b^{-5}}\right)$

Apply the quotient rule to each base:

$\frac{a^{16}}{a^6} = a^{16 - 6} = a^{10}$

$\frac{b^{-7}}{b^{-5}} = b^{-7 - (-5)} = b^{-7 + 5} = b^{-2}$

Converting the negative exponent $b^{-2} = \frac{1}{b^2}$ gives the simplified result:

$Y = \frac{a^{10}}{b^2}$

Operations on Radicals

Simplifying operations with radicals relies on properties such as $\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$ for $a, b \ge 0$, as well as standard algebraic identity expansions. Note that $\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$. For any real number $a \in \mathbb{R}$, $\sqrt{a^2} = |a|$, and when $a \ge 0$, $\sqrt{a^2} = a$.

To compute and simplify expression $A$:

$A = (2\sqrt{3} + \sqrt{5})(2\sqrt{3} - \sqrt{5})$

Apply the identity $(a + b)(a - b) = a^2 - b^2$ with $a = 2\sqrt{3}$ and $b = \sqrt{5}$:

$A = (2\sqrt{3})^2 - (\sqrt{5})^2$

Calculating each term gives $(2\sqrt{3})^2 = 2^2 \times (\sqrt{3})^2 = 4 \times 3 = 12$ and $(\sqrt{5})^2 = 5$. Subtracting gives:

$A = 12 - 5 = 7$

To compute expression $B$:

$B = (\sqrt{3} + \sqrt{5})^2 - (\sqrt{3} - \sqrt{5})^2$

This expression can be evaluated either by expanding each squared binomial or by applying the difference of two squares identity. Using the expansion of perfect squares $(a \pm b)^2 = a^2 \pm 2ab + b^2$:

$(\sqrt{3} + \sqrt{5})^2 = (\sqrt{3})^2 + (\sqrt{5})^2 + 2\sqrt{3}\sqrt{5} = 3 + 5 + 2\sqrt{15} = 8 + 2\sqrt{15}$

$(\sqrt{3} - \sqrt{5})^2 = (\sqrt{3})^2 + (\sqrt{5})^2 - 2\sqrt{3}\sqrt{5} = 3 + 5 - 2\sqrt{15} = 8 - 2\sqrt{15}$

Subtracting the second expression from the first yields:

$B = (8 + 2\sqrt{15}) - (8 - 2\sqrt{15}) = 8 + 2\sqrt{15} - 8 + 2\sqrt{15} = 4\sqrt{15}$

Alternatively, applying $a^2 - b^2 = (a + b)(a - b)$ with $a = \sqrt{3} + \sqrt{5}$ and $b = \sqrt{3} - \sqrt{5}$ gives $a + b = 2\sqrt{3}$ and $a - b = 2\sqrt{5}$, yielding $B = (2\sqrt{3})(2\sqrt{5}) = 4\sqrt{15}$.

To compute expression $C$:

$C = \sqrt{\sqrt{7} + \sqrt{3}} \times \sqrt{\sqrt{7} - \sqrt{3}}$

Combine the two radical terms using $\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}$:

$C = \sqrt{(\sqrt{7} + \sqrt{3})(\sqrt{7} - \sqrt{3})}$

Inside the root, apply $(a + b)(a - b) = a^2 - b^2$:

$(\sqrt{7} + \sqrt{3})(\sqrt{7} - \sqrt{3}) = (\sqrt{7})^2 - (\sqrt{3})^2 = 7 - 3 = 4$

Taking the square root yields:

$C = \sqrt{4} = 2$

Expansion and Simplification

Expansion converts factored expressions into polynomial forms using distribution and binomial formulas.

To expand and simplify expression $A$:

$A = (2x - 3)^2 + (x + 2)^2$

Expand each squared binomial:

$(2x - 3)^2 = (2x)^2 - 2(2x)(3) + 3^2 = 4x^2 - 12x + 9$

$(x + 2)^2 = x^2 + 2(x)(2) + 2^2 = x^2 + 4x + 4$

Sum the two quadratic expressions:

$A = (4x^2 - 12x + 9) + (x^2 + 4x + 4) = (4x^2 + x^2) + (-12x + 4x) + (9 + 4)$

$A = 5x^2 - 8x + 13$

To expand and simplify expression $B$:

$B = (x - 2)(x^2 + 4x + 4)$

Distribute $(x - 2)$ across the terms of the trinomial:

$B = x(x^2 + 4x + 4) - 2(x^2 + 4x + 4)$

$B = (x^3 + 4x^2 + 4x) - (2x^2 + 8x + 8)$

Combine like terms:

$B = x^3 + (4x^2 - 2x^2) + (4x - 8x) - 8$

$B = x^3 + 2x^2 - 4x - 8$

To expand and simplify expression $C$:

$C = (x + 2)^3 - x(x - 1)^2$

Expand the cubic binomial $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$:

$(x + 2)^3 = x^3 + 3(x^2)(2) + 3(x)(2^2) + 2^3 = x^3 + 6x^2 + 12x + 8$

Expand the second term $x(x - 1)^2$:

$(x - 1)^2 = x^2 - 2x + 1$

$x(x - 1)^2 = x(x^2 - 2x + 1) = x^3 - 2x^2 + x$

Subtract the second polynomial from the first:

$C = (x^3 + 6x^2 + 12x + 8) - (x^3 - 2x^2 + x)$

$C = x^3 + 6x^2 + 12x + 8 - x^3 + 2x^2 - x$

Combine like terms:

$C = (x^3 - x^3) + (6x^2 + 2x^2) + (12x - x) + 8$

$C = 8x^2 + 11x + 8$

Factorization

Factorization rewrites sums and differences of terms into products of simpler factors using common monomial factors and special product identities.

To factorize $A = x^3 + 2x$:

Factor out the greatest common factor $x$:

$A = x(x^2 + 2)$

To factorize $B = x^3 - x$:

First factor out the common factor $x$:

$B = x(x^2 - 1)$

Then apply the difference of squares identity $a^2 - b^2 = (a - b)(a + b)$ to $x^2 - 1$:

$B = x(x - 1)(x + 1)$

To factorize $C = 8x^3 - 1$:

Recognize this as a difference of cubes $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ where $a = 2x$ and $b = 1$:

$8x^3 - 1 = (2x)^3 - 1^3$

$C = (2x - 1)((2x)^2 + (2x)(1) + 1^2)$

$C = (2x - 1)(4x^2 + 2x + 1)$

To factorize $E = 27x^3 + 8$:

Recognize this as a sum of cubes $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ where $a = 3x$ and $b = 2$:

$27x^3 + 8 = (3x)^3 + 2^3$

$E = (3x + 2)((3x)^2 - (3x)(2) + 2^2)$

$E = (3x + 2)(9x^2 - 6x + 4)$

To factorize $D = x^2 - 9 + 2(x - 3)^2$:

First factor $x^2 - 9$ as a difference of squares $(x - 3)(x + 3)$:

$D = (x - 3)(x + 3) + 2(x - 3)^2$

Factor out the common binomial factor $(x - 3)$:

$D = (x - 3) \left[(x + 3) + 2(x - 3)\right]$

Simplify inside the square brackets:

$(x + 3) + 2x - 6 = 3x - 3$

$D = (x - 3)(3x - 3)$

Factor out $3$ from $(3x - 3)$ to obtain the fully factored form:

$D = 3(x - 3)(x - 1)$

Inequalities and Bounding

Bounding (or framing) involves using the rules of inequalities to establish lower and upper bounds for algebraic expressions given interval constraints on variables. Crucial rules include: adding inequalities with the same direction, reversing inequality directions when multiplying or dividing by negative numbers, and reversing inequality directions when taking reciprocals of positive quantities.

Let $x \in [-2, -1]$ (so $-2 \le x \le -1$) and $y \in [2, 5]$ (so $2 \le y \le 5$).

To find bounds for $2x + 3y + 7$:

Multiply $-2 \le x \le -1$ by $2$:

$-4 \le 2x \le -2$

Multiply $2 \le y \le 5$ by $3$:

$6 \le 3y \le 15$

Add the two inequalities side-by-side:

$(-4 + 6) \le 2x + 3y \le (-2 + 15)$

$2 \le 2x + 3y \le 13$

Add $7$ across all terms:

$9 \le 2x + 3y + 7 \le 20$

To find bounds for $2x - 3y$:

From $6 \le 3y \le 15$, multiplying by $-1$ reverses the inequality direction:

$-15 \le -3y \le -6$

Add this to $-4 \le 2x \le -2$ side-by-side:

$(-4 + (-15)) \le 2x - 3y \le (-2 + (-6))$

$-19 \le 2x - 3y \le -8$

To find bounds for the product $x \cdot y$:

Since $x$ is negative ($-2 \le x \le -1$) and $y$ is positive ($2 \le y \le 5$), consider the positive quantity $-x$, which satisfies $1 \le -x \le 2$. Multiplying positive bounds yields:

$1 \times 2 \le (-x) \cdot y \le 2 \times 5$

$2 \le -x y \le 10$

Multiplying by $-1$ reverses the inequality direction:

$-10 \le x y \le -2$

To find bounds for $\frac{x}{y}$:

Express $\frac{x}{y}$ as $x \times \frac{1}{y}$. Taking reciprocals of $2 \le y \le 5$ reverses the inequality direction:

$\frac{1}{5} \le \frac{1}{y} \le \frac{1}{2}$

For the positive variable $-x$ where $1 \le -x \le 2$, multiply side-by-side with $\frac{1}{y}$:

$1 \times \frac{1}{5} \le (-x) \times \frac{1}{y} \le 2 \times \frac{1}{2}$

$\frac{1}{5} \le -\frac{x}{y} \le 1$

Multiplying by $-1$ reverses the inequality direction:

$-1 \le \frac{x}{y} \le -\frac{1}{5}$

To find bounds for $x^2 + y^2$:

Squaring negative numbers in $-2 \le x \le -1$ reverses order:

$(-1)^2 \le x^2 \le (-2)^2 \implies 1 \le x^2 \le 4$

Squaring positive numbers in $2 \le y \le 5$:

$2^2 \le y^2 \le 5^2 \implies 4 \le y^2 \le 25$

Adding these inequalities side-by-side yields:

$(1 + 4) \le x^2 + y^2 \le (4 + 25)$

$5 \le x^2 + y^2 \le 29$

To find bounds for $\sqrt{x + y}$:

First add $-2 \le x \le -1$ and $2 \le y \le 5$ side-by-side:

$(-2 + 2) \le x + y \le (-1 + 5)$

$0 \le x + y \le 4$

Taking the square root across all terms yields:

$\sqrt{0} \le \sqrt{x + y} \le \sqrt{4}$

$0 \le \sqrt{x + y} \le 2$