--- title: "Algebra: Equations and Inequalities" book: "High School Mathematics" subject: math language: en chapter: 2 exercises: 11 source: https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities --- # Chapter 2 — Algebra: Equations and Inequalities Algebra is the art of computing with letters. This chapter reviews and sharpens the two basic moves — [expanding](#def-g10-algebra-expand) and [factoring](#def-g10-algebra-expand) — and uses them to solve [equations](#def-g10-algebra-equation) and inequalities systematically, with the [sign table](#def-g10-algebra-signtable) as the central tool. ## 2.1 Expanding and factoring **Definition 2.1 (Expanding, factoring).** *Expanding* an expression means transforming products into sums; *factoring* means the reverse, transforming a sum into a product. The basic rule is distributivity: $$ k(a + b) = ka + kb, \qquad (a + b)(c + d) = ac + ad + bc + bd . $$ **Example 2.2.** Expand $(2x + 3)(x - 5)$ step by step: $$ \begin{align*} (2x + 3)(x - 5) &= 2x \times x + 2x \times (-5) + 3 \times x + 3 \times (-5) \\ &= 2x^2 - 10x + 3x - 15 \\ &= 2x^2 - 7x - 15 . \end{align*} $$ **Theorem 2.3 (Remarkable identities).** For all [real numbers](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-sets) $a$ and $b$: $$ (a+b)^2 = a^2 + 2ab + b^2, \qquad (a-b)^2 = a^2 - 2ab + b^2, \qquad (a+b)(a-b) = a^2 - b^2 . $$ **Proof.** Each one is a direct expansion. For the first: $$ (a+b)^2 = (a+b)(a+b) = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2 . $$ The second follows by replacing $b$ with $-b$: $(a-b)^2 = a^2 + 2a(-b) + (-b)^2 = a^2 - 2ab + b^2$. For the third: $(a+b)(a-b) = a^2 - ab + ba - b^2 = a^2 - b^2$. ∎ ![Why (a+b)2 = a2 + 2ab + b2: a square of side a + b splits into a square of side a, a square of side b, and two a × b rectangles.](https://one-course.com/images/onecourse/chapters/math-2/g10-algebra/fig-578dad376d43.svg) *Why $(a+b)^2 = a^2 + 2ab + b^2$: a square of side $a + b$ splits into a square of side $a$, a square of side $b$, and two $a \times b$ rectangles.* **Example 2.4.** Used forward ([expanding](#def-g10-algebra-expand)) and backward ([factoring](#def-g10-algebra-expand)): $$ \begin{align*} (3x + 2)^2 &= 9x^2 + 12x + 4, & x^2 - 25 &= (x+5)(x-5), \\ (x - 4)^2 &= x^2 - 8x + 16, & 4x^2 + 4x + 1 &= (2x + 1)^2 . \end{align*} $$ Mental arithmetic benefits too: $101 \times 99 = (100+1)(100-1) = 10000 - 1 = 9999$. **Method 2.5 (Factoring an expression).** Try, in this order: 1. *Common factor:* spot a factor present in every term and pull it out, e.g. $6x^2 + 4x = 2x(3x + 2)$ . The common factor can be a whole bracket: $(x+1)(2x-3) + (x+1)(x+7) = (x+1)(3x+4)$ . 2. *Remarkable identity:* recognize $a^2 - b^2$ , or $a^2 \pm 2ab + b^2$ , e.g. $9x^2 - 16 = (3x)^2 - 4^2 = (3x+4)(3x-4)$ . 3. Combine both, and check the result by [expanding](#def-g10-algebra-expand) it back. **Example 2.6.** Factor $E = (2x + 1)^2 - (x - 3)^2$. This is a difference of two squares $a^2 - b^2$ with $a = 2x+1$ and $b = x-3$: $$ \begin{align*} E &= \bigl[(2x+1) + (x-3)\bigr] \times \bigl[(2x+1) - (x-3)\bigr] \\ &= (3x - 2)(2x + 1 - x + 3) \\ &= (3x - 2)(x + 4). \end{align*} $$ Mind the second bracket: subtracting $x - 3$ means subtracting $x$ *and* adding $3$. ## 2.2 Equations **Definition 2.7 (Equation, solution).** An *equation* is an equality containing an unknown number $x$; a *solution* is a value of $x$ that makes the equality true. Solving the equation means finding *all* its solutions. Two [equations](#def-g10-algebra-equation) are *equivalent* when they have exactly the same solutions. Adding the same number to both sides, or multiplying both sides by the same *nonzero* number, produces an equivalent [equation](#def-g10-algebra-equation). **Example 2.8 (First-degree equation).** Solve $5x - 7 = 2x + 8$, one move at a time: $$ \begin{align*} 5x - 7 &= 2x + 8 \\ 5x - 2x &= 8 + 7 && \text{(add $7 - 2x$ to both sides)} \\ 3x &= 15 \\ x &= 5 && \text{(divide both sides by $3$).} \end{align*} $$ The unique solution is $5$. Check: $5 \times 5 - 7 = 18$ and $2 \times 5 + 8 = 18$. **Theorem 2.9 (Zero-product rule).** A product of [real numbers](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-sets) is zero if and only if at least one of its factors is zero: $$ A \times B = 0 \quad\Longleftrightarrow\quad A = 0 \ \text{ or } \ B = 0 . $$ **Proof.** If $A = 0$ or $B = 0$, the product is clearly $0$. Conversely, suppose $AB = 0$ and $A \neq 0$. Dividing both sides by $A$ gives $B = 0$. ∎ **Method 2.10 (Solving by factoring).** To solve an [equation](#def-g10-algebra-equation) whose right-hand side is $0$ and whose left-hand side can be factored: 1. bring everything to the left so the [equation](#def-g10-algebra-equation) reads $E = 0$ ; 2. factor $E$ into a product of first-degree factors; 3. apply the zero-product rule and solve each small [equation](#def-g10-algebra-equation) ; 4. collect all the solutions found. **Example 2.11.** Solve $(x - 2)(3x + 6) = 0$: either $x - 2 = 0$, giving $x = 2$, or $3x + 6 = 0$, giving $x = -2$. Solutions: $-2$ and $2$. Solve $x^2 = 9x$. Do *not* divide by $x$ (it could be zero!); bring everything to one side and factor: $$ x^2 - 9x = 0 \quad\Longleftrightarrow\quad x(x - 9) = 0 \quad\Longleftrightarrow\quad x = 0 \ \text{ or } \ x = 9 . $$ **Example 2.12 (Quotient equation).** Solve $\dfrac{x - 1}{x + 2} = 0$. A quotient is zero exactly when its numerator is zero *and* its denominator is not. Here $x - 1 = 0$ gives $x = 1$, and $1 + 2 = 3 \neq 0$: the only solution is $1$. The value $x = -2$ is *forbidden* (division by zero) and must always be excluded first. ## 2.3 Inequalities and sign tables **Proposition 2.13 (Rules for inequalities).** Let $a \leq b$. 1. For any $c$ : $a + c \leq b + c$ (adding preserves the order). 2. If $c > 0$ : $ac \leq bc$ (multiplying by a positive number preserves the order). 3. If $c < 0$ : $ac \geq bc$ (multiplying by a negative number *reverses* the order). **Proof.** 1. $(b + c) - (a + c) = b - a \geq 0$. 2. $bc - ac = (b-a)c$ is the product of two nonnegative numbers, so $bc \geq ac$. 3. Now $(b-a)c \leq 0$ as the product of a nonnegative and a negative number, so $bc \leq ac$. ∎ **Example 2.14.** Solve $-2x + 3 > 9$: $$ \begin{align*} -2x + 3 &> 9 \\ -2x &> 6 && \text{(subtract 3)} \\ x &< -3 && \text{(divide by $-2 < 0$: reverse the inequality).} \end{align*} $$ The solution set is $\intoo{-\infty}{-3}$. **Definition 2.15 (Sign table).** A *sign table* records, on one row per factor, the sign ($+$ or $-$) of each factor of an expression as $x$ runs over $\R$, with a $0$ at each value where the factor vanishes; a final row gives the sign of the product, using the rule of signs column by column. **Example 2.16.** Sign of $P(x) = (x - 2)(3 - x)$. The factor $x - 2$ vanishes at $2$ and is positive after; the factor $3 - x$ vanishes at $3$ and is positive before: | $x$ | $-\infty$ | $2$ | | $3$ | $+\infty$ | | --- | --- | --- | --- | --- | --- | | $x - 2$ | $-$ | $0$ | $+$ | | $+$ | | $3 - x$ | $+$ | | $+$ | $0$ | $-$ | | $P(x)$ | $-$ | $0$ | $+$ | $0$ | $-$ | So $P(x) > 0$ exactly on $\intoo{2}{3}$, and $P(x) \leq 0$ on $\intoc{-\infty}{2} \cup \intco{3}{+\infty}$. ![The graph of P(x) = (x-2)(3-x) confirms the sign table: positive between the roots 2 and 3, negative outside.](https://one-course.com/images/onecourse/chapters/math-2/g10-algebra/fig-5304e5f43271.svg) *The graph of $P(x) = (x-2)(3-x)$ confirms the [sign table](#def-g10-algebra-signtable): positive between the roots $2$ and $3$, negative outside.* **Method 2.17 (Solving an inequality with a sign table).** To solve an inequality like $E(x) \geq 0$ or $E(x) < 0$: 1. bring everything to the left-hand side, so the right-hand side is $0$ (never multiply both sides by an expression whose sign is unknown); 2. factor the left-hand side, or put it over a common denominator if it is a quotient; 3. build the [sign table](#def-g10-algebra-signtable) with one row per factor (for a quotient, mark forbidden values with a double bar in the last row); 4. read off the [intervals](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) where the requested sign holds, checking whether each bound is included. **Example 2.18.** Solve $\dfrac{x + 1}{x - 3} \leq 0$. The numerator vanishes at $-1$, the denominator at $3$ (forbidden value). The quotient is negative when the two have opposite signs, which happens for $x$ between $-1$ and $3$. Including the bound where the *numerator* vanishes but excluding the forbidden value: the solution set is $\intco{-1}{3}$. ## 2.4 Exercises **Exercise 2.1 ★.** Expand and simplify: $$ 3(2x - 5) - 2(x - 7), \qquad (x + 4)(2x - 1), \qquad (3x - 2)^2, \qquad (5 - x)(5 + x). $$ **Solution of Exercise 2.1.** $3(2x-5) - 2(x-7) = 6x - 15 - 2x + 14 = 4x - 1$. $(x+4)(2x-1) = 2x^2 - x + 8x - 4 = 2x^2 + 7x - 4$. $(3x-2)^2 = 9x^2 - 12x + 4$. $(5-x)(5+x) = 25 - x^2$. **Exercise 2.2 ★.** Factor: $$ x^2 - 49, \qquad x^2 + 10x + 25, \qquad 7x^2 - 14x, \qquad 16x^2 - 8x + 1 . $$ **Solution of Exercise 2.2.** $x^2 - 49 = (x+7)(x-7)$ (difference of squares). $x^2 + 10x + 25 = (x+5)^2$. $7x^2 - 14x = 7x(x - 2)$ (common factor). $16x^2 - 8x + 1 = (4x - 1)^2$. **Exercise 2.3 ★.** Solve: $$ 4x + 9 = x - 3, \qquad \frac{2x - 1}{3} = 5, \qquad 2(x - 3) = 2x + 1 . $$ (One of them has no solution — explain why.) **Solution of Exercise 2.3.** $4x + 9 = x - 3$: $3x = -12$, so $x = -4$. $\frac{2x-1}{3} = 5$: $2x - 1 = 15$, so $x = 8$. $2(x-3) = 2x + 1$ expands to $2x - 6 = 2x + 1$, i.e. $-6 = 1$: false for every $x$. No solution — the two sides define parallel lines that never meet. **Exercise 2.4 ★.** Solve using the zero-product rule: $$ (x + 5)(2x - 8) = 0, \qquad x^2 - 16 = 0, \qquad x^2 + 6x + 9 = 0 . $$ **Solution of Exercise 2.4.** $(x+5)(2x-8) = 0$: $x = -5$ or $x = 4$. $x^2 - 16 = (x-4)(x+4) = 0$: $x = 4$ or $x = -4$. $x^2 + 6x + 9 = (x+3)^2 = 0$: the single solution $x = -3$. **Exercise 2.5 ★.** Solve the inequalities and write the solution sets as [intervals](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval): $$ 3x - 5 \leq 7, \qquad -4x + 1 > 9, \qquad 2(x - 1) \geq 5x + 4 . $$ **Solution of Exercise 2.5.** $3x - 5 \leq 7$: $3x \leq 12$, so $x \leq 4$: solution set $\intoc{-\infty}{4}$. $-4x + 1 > 9$: $-4x > 8$, dividing by $-4$ reverses the inequality: $x < -2$: solution set $\intoo{-\infty}{-2}$. $2(x-1) \geq 5x + 4$: $2x - 2 \geq 5x + 4$, so $-3x \geq 6$, so $x \leq -2$: solution set $\intoc{-\infty}{-2}$. **Exercise 2.6 ★★.** Factor $E = (x - 1)(x + 3) - (x - 1)(2x - 5)$, then solve $E = 0$. **Solution of Exercise 2.6.** The common factor is $(x-1)$: $$ E = (x-1)\bigl[(x+3) - (2x - 5)\bigr] = (x-1)(x + 3 - 2x + 5) = (x-1)(8 - x). $$ Then $E = 0$ when $x = 1$ or $x = 8$ (zero-product rule). **Exercise 2.7 ★★.** Solve $x^2 = 5x$, then $(2x + 1)^2 = (x - 4)^2$. (For the second one, bring everything to the left and factor the difference of squares.) **Solution of Exercise 2.7.** $x^2 = 5x$: $x^2 - 5x = x(x - 5) = 0$, so $x = 0$ or $x = 5$ (never divide by $x$: the solution $0$ would be lost). $(2x+1)^2 = (x-4)^2$: bring to the left and factor the difference of squares: $$ (2x+1)^2 - (x-4)^2 = \bigl[(2x+1) + (x-4)\bigr]\bigl[(2x+1) - (x-4)\bigr] = (3x - 3)(x + 5). $$ Zero-product rule: $x = 1$ or $x = -5$. **Exercise 2.8 ★★.** Using a [sign table](#def-g10-algebra-signtable), solve $$ (x + 2)(4 - x) > 0, \qquad \frac{2x - 6}{x + 1} \geq 0 . $$ **Solution of Exercise 2.8.** $(x+2)(4-x) > 0$: the factor $x + 2$ vanishes at $-2$ (positive after), $4 - x$ vanishes at $4$ (positive before). The product is positive exactly when both factors are positive, i.e. on $\intoo{-2}{4}$. $\frac{2x-6}{x+1} \geq 0$: numerator zero at $3$, denominator zero at $-1$ (forbidden). [Sign table](#def-g10-algebra-signtable): the quotient is positive when both parts have the same sign, i.e. for $x < -1$ or $x > 3$; it vanishes at $x = 3$. Solution set: $\intoo{-\infty}{-1} \cup \intco{3}{+\infty}$. **Exercise 2.9 ★★.** A streaming service costs $9$ per month, plus a one-time signup fee of $15$; a rival costs $12$ per month with no fee. After how many months does the first service become the cheaper choice overall? Set up and solve an inequality. **Solution of Exercise 2.9.** After $x$ months the first service costs $9x + 15$ and the second $12x$. The first is cheaper when $$ 9x + 15 < 12x \ \Longleftrightarrow\ 15 < 3x \ \Longleftrightarrow\ x > 5 . $$ From the $6$th month on, the first service is the cheaper choice overall. **Exercise 2.10 ★★.** Solve $\dfrac{x+3}{x-1} = 2$. (Multiply both sides by $x - 1$ after excluding the forbidden value, or bring everything to the left over a common denominator.) **Solution of Exercise 2.10.** Forbidden value: $x = 1$. For $x \neq 1$, multiply both sides by $x - 1$: $$ x + 3 = 2(x - 1) = 2x - 2 \ \Longleftrightarrow\ x = 5 . $$ Since $5 \neq 1$, the solution is $x = 5$. Check: $\frac{5+3}{5-1} = \frac84 = 2$. **Exercise 2.11 ★★★.** Let $n$ be an [integer](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-sets). Show that $(n + 1)^2 - n^2$ is always odd, and that the difference between the squares of two consecutive odd numbers is always a multiple of $8$. **Solution of Exercise 2.11.** $(n+1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n + 1$, which is odd for every [integer](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-sets) $n$. Two consecutive odd numbers can be written $2n - 1$ and $2n + 1$. Then $$ (2n+1)^2 - (2n-1)^2 = \bigl[(2n+1)+(2n-1)\bigr]\bigl[(2n+1)-(2n-1)\bigr] = 4n \times 2 = 8n, $$ a multiple of $8$. ## 2.5 Problem: From Babylon to the golden ratio **Problem 2.1.** Weekend problem — sum-and-product problems, completing the square the way al-Khwarizmi drew it, and the equation $x^2 = x + 1$ Four thousand years ago, Babylonian scribes solved this on clay: *find two numbers knowing their sum and their product*. Their trick — look for the numbers symmetrically around the half-sum — still cracks every quadratic [equation](#def-g10-algebra-equation) today, under its modern name, *completing the square*. This problem learns the trick, watches al-Khwarizmi literally complete a square, and puts the method to work on the most celebrated proportion of them all. **Part I — The Babylonian problem.** 1. Warm-up ( [Theorem 2.3](#thm-g10-algebra-identities) ): factor $x^2 - 10x + 25$ , then $9x^2 - 49$ , then — combining an identity with a difference of squares — $x^2 + 6x + 9 - 4$ . 2. Solve $(x + 1)(x + 5) = 0$ and $(3x - 7)(3x + 7) = 0$ ( [Theorem 2.9](#thm-g10-algebra-zeroproduct) ). 3. The scribe’s problem: two numbers have sum $10$ and product $21$ . Following Babylon, write them as $5 - t$ and $5 + t$ (symmetric around the half-sum) and find $t$ , then the two numbers. 4. Same method: sum $14$ , product $33$ . 5. The general recipe: for sum $s$ and product $p$, show that the two numbers are $$\frac s2 - t \quad\text{and}\quad \frac s2 + t, \qquad\text{where}\quad t^2 = \left(\frac s2\right)^2 - p ,$$ and state the condition on $s$ and $p$ for the problem to have a solution. **Part II — Completing the square.** 6. Expand $(x - a)(x - b)$ and conclude: the solutions of $x^2 - sx + p = 0$ are exactly the pairs of numbers with sum $s$ and product $p$ . (The scribes were solving quadratic [equations](#def-g10-algebra-equation) without knowing it.) 7. Verify that $x^2 - 10x + 21 = (x - 5)^2 - 4$ , and solve $x^2 - 10x + 21 = 0$ by [factoring](#def-g10-algebra-expand) the right side as a difference of squares. Compare with question 3. 8. Solve $x^2 + 6x - 7 = 0$ by completing the square. 9. Complete the square in $x^2 + 4x + 5$ , and explain why the [equation](#def-g10-algebra-equation) $x^2 + 4x + 5 = 0$ has *no* real solution. State the general criterion: once written as $(x + h)^2 + k$ , when does a quadratic expression vanish? 10. The name is literal. In ninth-century Baghdad, al-Khwarizmi drew $x^2 + 6x = 7$ as a square of side $x$ with two $3 \times x$ rectangles glued to two of its sides, and *completed the square* with the missing $3 \times 3$ corner. Draw the figure and explain how it turns the [equation](#def-g10-algebra-equation) into $(x + 3)^2 = 16$ — then solve. **Part III — Signs, tables and two old promises.** 11. Build the [sign table](#def-g10-algebra-signtable) of $(x - 3)(x + 2)$ ( [Definition 2.15](#def-g10-algebra-signtable) ) and solve $(x - 3)(x + 2) < 0$ . 12. Solve $x^2 \leq 10x - 21$ (bring everything to one side and reuse question 7). 13. Solve $\dfrac{x - 1}{x + 2} \geq 0$ with a [sign table](#def-g10-algebra-signtable) , minding the forbidden value. 14. An old promise kept: for the farmer of queen Dido’s fence problem, in the Middle School volume, with $20$ m of fence, the area of an $x \times (10 - x)$ pen satisfies $$A(x) = x(10 - x) = 25 - (x - 5)^2 .$$ Verify the identity, and read off it a *proof* of what grade 6 could only conjecture: the maximum area, and the unique $x$ achieving it. 15. Another: prove that $x + \dfrac1x \geq 2$ for every $x > 0$ , with equality only at $x = 1$ . (Multiply by $x$ , gather, recognize a square — the arithmetic–geometric inequality of the Euclid’s-relations weekend problem of the Middle School volume in a new costume.) **Part IV — The golden section.** Cut a segment of length $1$ into a large part $x$ and a small part $1 - x$ so that *whole is to large as large is to small*: $\frac{1}{x} = \frac{x}{1 - x}$. 16. Show that the condition reads $x^2 + x - 1 = 0$ . 17. Solve by completing the square, and keep the root that is a length: $x = \frac{\sqrt5 - 1}{2} \approx 0.618$ . 18. The *golden ratio* is $\varphi = \frac1x = \frac{2}{\sqrt5 - 1}$ . Rationalize the denominator (multiply top and bottom by $\sqrt5 + 1$ , an identity at work) and show $\varphi = \frac{\sqrt5 + 1}{2} \approx 1.618$ . 19. Verify *exactly* the two magic identities of $\varphi$: $$\varphi^2 = \varphi + 1, \qquad \frac{1}{\varphi} = \varphi - 1 .$$ 20. Finale: explain why $\varphi$ is [irrational](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#ex-g10-numbers-classify) (the irrationality weekend problem of the Middle School volume supplies the key fact about $\sqrt5$ ), then compute the ratios $\frac53$ , $\frac85$ , $\frac{13}{8}$ , $\frac{21}{13}$ of consecutive Hemachandra–Fibonacci numbers (the rhythm-counting weekend problem of the Middle School volume) to three decimals. What do they seem to chase? (The proof that they catch it belongs to the chapters on sequences.) **Solution of Problem 2.1.** **1.** $x^2 - 10x + 25 = (x - 5)^2$; $9x^2 - 49 = (3x - 7)(3x + 7)$; $x^2 + 6x + 9 - 4 = (x + 3)^2 - 2^2 = (x + 1)(x + 5)$. **2.** $x = -1$ or $x = -5$; $x = \frac73$ or $x = -\frac73$. **3.** $(5 - t)(5 + t) = 25 - t^2 = 21$ gives $t^2 = 4$, $t = 2$: the numbers are $3$ and $7$ (sum $10$, product $21$: check). **4.** $(7 - t)(7 + t) = 49 - t^2 = 33$: $t = 4$, numbers $3$ and $11$. **5.** $\left(\frac s2 - t\right)\left(\frac s2 + t\right) = \left(\frac s2\right)^2 - t^2 = p$ forces $t^2 = \left(\frac s2\right)^2 - p$. A solution exists exactly when this is $\geq 0$, i.e. when $p \leq \left(\frac s2\right)^2$ — the product of two numbers of given sum can never beat the square of the half-sum (queen Dido’s fence problem, in the Middle School volume knew). **6.** $(x - a)(x - b) = x^2 - (a + b)x + ab$: the [equation](#def-g10-algebra-equation) $x^2 - sx + p = 0$ says precisely “$x$ is one of two numbers with sum $s$ and product $p$”. **7.** $(x-5)^2 - 4 = x^2 - 10x + 25 - 4 = x^2 - 10x + 21$. Difference of squares: $(x - 5 - 2)(x - 5 + 2) = (x - 7)(x - 3) = 0$: solutions $3$ and $7$ — the scribe’s numbers of question 3. **8.** $x^2 + 6x - 7 = (x + 3)^2 - 16 = 0$ gives $(x + 3)^2 = 16$, $x + 3 = \pm 4$: $x = 1$ or $x = -7$. **9.** $x^2 + 4x + 5 = (x + 2)^2 + 1 \geq 1 > 0$ for every $x$: no real solution. In the form $(x + h)^2 + k$: solutions exist exactly when $k \leq 0$ (then $(x+h)^2 = -k$ has the two roots $-h \pm \sqrt{-k}$); when $k > 0$ the expression stays positive. **10.** The square of side $x$ (area $x^2$) with two $3 \times x$ rectangles ($6x$ in all) forms an L of area $x^2 + 6x = 7$; the missing $3 \times 3$ corner (area $9$) completes a big square of side $x + 3$. Adding $9$ to both sides: $(x + 3)^2 = 16$, so $x + 3 = 4$ (lengths are positive) and $x = 1$. The algebra and the picture are the same act. **11.** $(x-3)(x+2)$: negative between the roots. Solution of $< 0$: $x \in \intoo{-2}{3}$. **12.** $x^2 - 10x + 21 \leq 0$, i.e. $(x - 3)(x - 7) \leq 0$: between the roots, ends included: $x \in \intcc{3}{7}$. **13.** Zero at $x = 1$, forbidden at $x = -2$; the quotient is positive outside the roots: $x \in \intoo{-\infty}{-2} \cup \intco{1}{+\infty}$. **14.** $25 - (x - 5)^2 = 25 - x^2 + 10x - 25 = x(10 - x)$: identity verified. Since $(x-5)^2 \geq 0$ always, $A(x) \leq 25$, with equality exactly when $x = 5$: the $5 \times 5$ square pen, maximum $25$ m$^2$ — the grade-6 table, now a two-line proof. **15.** For $x > 0$, multiply the claim by $x$: $x^2 + 1 \geq 2x$, i.e. $x^2 - 2x + 1 = (x - 1)^2 \geq 0$ — true, with equality only at $x = 1$. Dividing back by $x > 0$ preserves everything ([Proposition 2.13](#prop-g10-algebra-ineqrules)). **16.** Cross-multiplying $\frac1x = \frac{x}{1 - x}$ (all lengths positive): $x^2 = 1 - x$, i.e. $x^2 + x - 1 = 0$. **17.** $x^2 + x - 1 = \left(x + \frac12\right)^2 - \frac54 = 0$ gives $x + \frac12 = \pm\frac{\sqrt5}{2}$. The positive root: $x = \frac{\sqrt5 - 1}{2} \approx 0.618$. **18.** $\varphi = \frac{2}{\sqrt5 - 1} = \frac{2(\sqrt5 + 1)}{(\sqrt5 - 1)(\sqrt5 + 1)} = \frac{2(\sqrt5 + 1)}{5 - 1} = \frac{\sqrt5 + 1}{2} \approx 1.618$. **19.** $\varphi^2 = \frac{(\sqrt5 + 1)^2}{4} = \frac{6 + 2\sqrt5}{4} = \frac{3 + \sqrt5}{2}$, and $\varphi + 1 = \frac{\sqrt5 + 3}{2}$: equal. And $\varphi - 1 = \frac{\sqrt5 - 1}{2} = x = \frac1\varphi$ (that is how $\varphi$ was defined). The golden ratio is the number whose square adds one and whose reciprocal subtracts one. **20.** $\sqrt5$ is [irrational](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#ex-g10-numbers-classify) (the irrationality weekend problem of the Middle School volume: $5$ is no perfect square), and adding $1$ then halving preserves irrationality (rational operations, [Problem 1.1](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#pb-g10-numbers-1)). Ratios: $\frac53 \approx 1.667$, $\frac85 = 1.600$, $\frac{13}{8} = 1.625$, $\frac{21}{13} \approx 1.615$: alternately above and below, closing in on $\varphi = 1.618\ldots$ — the rhythm-counting numbers of the rhythm-counting weekend problem of the Middle School volume secretly chase the golden section; the chapters on sequences will catch them in the act.