A quadratic function is a polynomial function of degree two, typically written in standard form f(x) = ax² + bx + c, where a, b and c are real numbers and a ≠ 0. (If a were zero, the x² term would vanish and you would be back to a line.)
3.1.1 — Analyzing a Quadratic Function by Sketching. The graph of a quadratic function is a parabola. It opens upward or downward depending on the sign of the leading coefficient a. The tip of the parabola, labelled V in the book’s diagrams, is the vertex, with coordinates (h, k). The vertical line through the vertex is the axis of symmetry, and the vertex is the turning point where the graph changes direction. If a > 0, the vertex is a minimum point; if a < 0, it is a maximum point. To sketch a quadratic function, find the x-intercept, the y-intercept and the vertex; to analyze the sketch, say whether the vertex is a minimum or a maximum and name the intervals where the function is increasing or decreasing.
Example 1 (p. 35). Sketch and analyze y = −x² − 2x + 3. The y-intercept is y = 3 (put x = 0). The x-intercepts come from −x² − 2x + 3 = 0, i.e. x² + 2x − 3 = 0, i.e. (x + 3)(x − 1) = 0, so x = −3 and x = 1. The vertex: h = −b/(2a) = −(−2)/(2(−1)) = −1, and k = −(−1)² − 2(−1) + 3 = 4, so the vertex is (−1, 4). Since a = −1 < 0, it is a maximum point, and the function is increasing on (−∞, −1) and decreasing on (−1, ∞).
دو درجی تفاعل درجہ دو کا کثیر رقمی تفاعل ہے، عام صورت f(x) = ax² + bx + c ہے، جہاں a ≠ 0 ہو۔ اس کا گراف پیرابولا کہلاتا ہے۔
پیرابولا کی نوک کو ورٹیکس (h, k) کہتے ہیں۔ ورٹیکس سے گزرنے والی عمودی خط محور تماثل ہے۔ اگر a > 0 ہو تو ورٹیکس کم سے کم نقطہ ہے، اور اگر a < 0 ہو تو زیادہ سے زیادہ نقطہ ہے۔ خاکہ بنانے کے لیے x-تقاطع، y-تقاطع اور ورٹیکس معلوم کریں، پھر بتائیں کہ تفاعل کہاں بڑھ رہا ہے اور کہاں گھٹ رہا ہے۔
مثال ۱: y = −x² − 2x + 3 کا y-تقاطع 3، x-تقاطع −3 اور 1، ورٹیکس (−1, 4) ہے جو زیادہ سے زیادہ نقطہ ہے۔ تفاعل (−∞, −1) پر بڑھتا ہے اور (−1, ∞) پر گھٹتا ہے۔
Completing the square rewrites a quadratic function f(x) = ax² + bx + c in vertex form f(x) = a(x − h)² + k, where the vertex is (h, k), with h = −b/(2a) and k = c − b²/(4a). Once it is in this form, the maximum or minimum is staring at you: the square (x − h)² is never negative, so if a > 0, the minimum value of f(x) at x = h is k; if a < 0, the maximum value of f(x) at x = h is k.
Example 2 (p. 35). Find the maximum or minimum of f(x) = −2x² + 4x + 3 by completing the square. Take −2 out of the x-terms: f(x) = −2(x² − 2x) + 3. Complete the square inside: x² − 2x + 1 − 1 = (x − 1)² − 1, so f(x) = −2[(x − 1)² − 1] + 3 = −2(x − 1)² + 5. Here a = −2 < 0, so the maximum value is 5, at x = 1.
Example 3 (p. 36). Find the maximum or minimum of f(x) = x² − 2x − 3. Here a = 1, b = −2, c = −3, so h = −(−2)/(2·1) = 1 and k = −3 − (−2)²/(4·1) = −4. Since a = 1 > 0, the minimum value at x = 1 is −4. (Check by completing the square: (x − 1)² − 4, same answer.)
مربع مکمل کرنے سے دو درجی تفاعل f(x) = ax² + bx + c کو ورٹیکس صورت f(x) = a(x − h)² + k میں لکھا جاتا ہے، جہاں h = −b/(2a) اور k = c − b²/(4a) ہے۔ مربع (x − h)² کبھی منفی نہیں ہوتا، اس لیے a > 0 ہو تو x = h پر کم سے کم قدر k ہے، اور a < 0 ہو تو زیادہ سے زیادہ قدر k ہے۔
مثال ۲: f(x) = −2x² + 4x + 3 = −2(x − 1)² + 5، اس لیے زیادہ سے زیادہ قدر 5 ہے، جو x = 1 پر حاصل ہوتی ہے۔ مثال ۳: f(x) = x² − 2x − 3 کی کم سے کم قدر −4 ہے، جو x = 1 پر حاصل ہوتی ہے۔
Quadratic functions are typically not one-to-one over their entire domain — a parabola takes most y-values twice, once on each side of the vertex. To find an inverse, we must restrict the domain to a portion where the function is one-to-one. Commonly we restrict to x ≥ h (the right side of the vertex) or x ≤ h (the left side), where h is the x-coordinate of the vertex. The inverse is then found the usual way: swap x and y, solve for y, and pick the branch whose outputs lie in the restricted domain.
Example 4 (pp. 36–37). Find the inverse of f(x) = x² + 4x + 3, x ≥ −2, and its domain and range. Swap: x = y² + 4y + 3, so y² + 4y + 3 − x = 0, and by the quadratic formula y = −2 ± √(1 + x). Now choose: the inverse’s outputs must lie in the domain of f, which is [−2, ∞). Test x = 0: the two candidates are −1 and −3; only −1 lies in [−2, ∞), so the negative branch is discarded and f⁻¹(x) = −2 + √(1 + x). For the domain and range: f(x) = (x + 2)² − 1, so its minimum is −1, giving Domain f = [−2, ∞), Range f = [−1, ∞), and therefore Domain f⁻¹ = [−1, ∞), Range f⁻¹ = [−2, ∞) — domain and range swap, exactly as they should for an inverse.
📌 Book wording note (p. 37): the book justifies discarding −3 by saying it does not lie “in the range of f”. The deciding set is really the domain of f, [−2, ∞) — the set the inverse’s outputs must belong to. Both descriptions happen to keep −1 and drop −3 here, but the domain version is the rule to use in Exercise 3.1 Q3.
دو درجی تفاعل پورے ڈومین پر یک رخی نہیں ہوتا، کیونکہ پیرابولا ایک y-قدر دو جگہ دیتا ہے۔ معکوس نکالنے کے لیے ڈومین کو ورٹیکس کے ایک طرف محدود کرتے ہیں، عام طور پر x ≥ h یا x ≤ h۔ پھر x اور y کو آپس میں بدل کر y کے لیے حل کرتے ہیں، اور وہ شاخ چنتے ہیں جس کے آؤٹ پٹ محدود ڈومین میں آئیں۔
مثال ۴: f(x) = x² + 4x + 3، x ≥ −2 کا معکوس f⁻¹(x) = −2 + √(1 + x) ہے۔ f کا ڈومین [−2, ∞) اور رینج [−1, ∞) ہے، اس لیے معکوس کا ڈومین [−1, ∞) اور رینج [−2, ∞) ہے — معکوس میں ڈومین اور رینج آپس میں بدل جاتے ہیں۔
📌 کتابی نوٹ: کتاب −3 کو رد کرنے کی وجہ رینج بتاتی ہے، لیکن اصل اصول یہ ہے کہ معکوس کا آؤٹ پٹ f کے ڈومین میں ہونا چاہیے۔
3.3 — Absolute Value. The absolute value of x is defined as |x| = x if x ≥ 0, and −x if x < 0. In words: it is the distance from zero, so it is never negative — |5| = 5 and |−5| = 5. Its graph is the familiar V-shape with its point at the origin.
3.3.1 — Absolute Value Quadratic Equations. To solve them, all answers must be substituted back into the original equation to verify whether they are valid or not. Sometimes “extraneous” solutions appear which are not valid and must be eliminated from the final answer. The usual move is to split |expression| = d into expression = d or expression = −d, then check every root you get.
Example 5 (p. 37). Solve |x² − 4| = 5. Split: x² − 4 = 5 or x² − 4 = −5, so x² = 9 or x² = −1. The second case gives x = ±√(−1), imaginary — not a real solution, so it is dropped for real-valued work. Check the survivors: for x = 3, |9 − 4| = 5 ✓; for x = −3, |9 − 4| = 5 ✓. Hence the solution set is {±3}.
مطلق قدر |x| کی تعریف یہ ہے: اگر x ≥ 0 ہو تو |x| = x، اور اگر x < 0 ہو تو |x| = −x۔ یعنی یہ صفر سے فاصلہ ہے، اس لیے کبھی منفی نہیں ہوتی۔ اس کا گراف V کی شکل کا ہوتا ہے۔
مطلق قدر والی مساواتیں حل کرتے وقت ہر جواب کو اصل مساوات میں رکھ کر جانچنا ضروری ہے، کیونکہ کبھی کبھی اضافی (extraneous) حل نکل آتے ہیں جنہیں نکالنا ہوتا ہے۔ عام طریقہ: |اظہار| = d کو اظہار = d یا اظہار = −d میں توڑیں، پھر ہر جذر کی جانچ کریں۔
مثال ۵: |x² − 4| = 5 سے x² = 9 یا x² = −1 ملتا ہے۔ دوسری صورت حقیقی نہیں، اس لیے حل {±3} ہے، اور دونوں کی جانچ درست ہے۔
Absolute value quadratic inequalities are inequalities that involve a quadratic expression within absolute value bars, generally of the forms |ax² + bx + c| < d, > d, ≤ d or ≥ d. The method the book uses is to unwrap the bars into a double inequality, split it into two ordinary quadratic inequalities, solve each by critical values and test points, and then intersect the two solution sets — both conditions must hold at once.
Example 6 (pp. 38–39). Solve |x² − 6x − 4| < 3. Unwrap: −3 < x² − 6x − 4 < 3. Split: (i) x² − 6x − 1 > 0 and (ii) x² − 6x − 7 < 0. For (i), x² − 6x − 1 = 0 gives x = 3 ± √10 (about −0.16 and 6.16); testing x = −1 (true), x = 0 (false) and x = 7 (true) gives (−∞, 3 − √10) ∪ (3 + √10, ∞). For (ii), x² − 6x − 7 = (x + 1)(x − 7), and testing gives (−1, 7). Intersecting: (−1, 3 − √10) ∪ (3 + √10, 7). Notice the middle piece between the two roots of (i) is cut out, and the ends are cut by (ii) — that is what an intersection does.
مطلق قدر والی نامساوات میں دو درجی اظہار مطلق قدر کی علامتوں میں ہوتا ہے، جیسے |ax² + bx + c| < d۔ حل کا طریقہ: علامتیں کھول کر دوہری نامساوات بنائیں، اسے دو عام دو درجی نامساواتوں میں توڑیں، ہر ایک کو نازک اقدار (critical values) اور آزمائشی نقطوں سے حل کریں، پھر دونوں حلوں کا تقاطع لیں، کیونکہ دونوں شرطیں ایک ساتھ پوری ہونی چاہییں۔
مثال ۶: |x² − 6x − 4| < 3 سے −3 < x² − 6x − 4 < 3 بنتا ہے۔ پہلی نامساوات کا حل (−∞, 3 − √10) ∪ (3 + √10, ∞) اور دوسری کا (−1, 7) ہے۔ تقاطع (−1, 3 − √10) ∪ (3 + √10, 7) ہے۔
Some equations do not look like degree 2, but they can be reduced to a quadratic equation. The book discusses rational and radical equations. 3.4.1 — Rational equations contain one or more rational expressions, typically with a variable in the denominator. The move is to multiply through by the common denominator (remembering the values that make it zero are forbidden), then solve the resulting quadratic.
Example 7 (p. 40). Solve 1/x + 2/(x + 1) = 1, x ≠ 0, x ≠ −1. Multiply by x(x + 1): (x + 1) + 2x = x(x + 1), so 3x + 1 = x² + x, i.e. x² − 2x − 1 = 0. By the quadratic formula, x = 1 ± √2. Neither value is forbidden, so the solution set is {1 ± √2}.
3.4.2 — Radical equations involve radical expressions of the variable. To solve one, first obtain an equation free from radicals (usually by squaring, sometimes twice). Warning: every solution of a radical equation is a solution of the radical-free equation, but the new equation has solutions that are not solutions of the original — such extra roots are called extraneous roots, and only checking in the original equation separates them.
Example 8 (p. 41). Solve √(x + 8) + √(x + 3) = √(12x + 13). Square once: x + 8 + x + 3 + 2√((x + 8)(x + 3)) = 12x + 13, so √((x + 8)(x + 3)) = 5x + 1. Square again: x² + 11x + 24 = 25x² + 10x + 1, so 24x² − x − 23 = 0, i.e. (24x + 23)(x − 1) = 0, giving x = −23/24 or x = 1. On checking, −23/24 is extraneous (it would make 5x + 1 negative while a square root cannot be negative), so the solution set is {1}.
کچھ مساواتیں بظاہر درجہ دو کی نہیں لگتیں، لیکن انہیں دو درجی مساوات میں بدلا جا سکتا ہے۔ ناطق مساواتوں میں متغیر مخرج میں ہوتا ہے؛ مشترکہ مخرج سے ضرب دے کر دو درجی مساوات بنائیں، اور ممنوع اقدار (جن سے مخرج صفر ہو) کو یاد رکھیں۔
جذری مساواتوں میں جزر کی علامتیں ہوتی ہیں؛ مربع لے کر جزر ختم کریں، کبھی دو بار مربع لینا پڑتا ہے۔ خبردار: مربع لینے سے اضافی جڑیں پیدا ہو سکتی ہیں، اس لیے ہر جواب کو اصل مساوات میں رکھ کر جانچنا لازم ہے۔
مثال ۷: 1/x + 2/(x + 1) = 1 کا حل {1 ± √2} ہے۔ مثال ۸: √(x + 8) + √(x + 3) = √(12x + 13) سے x = 1 ملتا ہے؛ x = −23/24 اضافی جڑ ہے۔
We now solve problems which, when expressed symbolically, lead to quadratic equations in one variable. The book’s two-step method: (i) suppose the unknown quantities to be x or y, etc.; (ii) translate the problem into symbols and form the equation or inequality satisfying the given conditions. Then solve — and remember that a mathematically correct root can still be inadmissible in the story (a negative breadth, a negative time), so the story gets the last word.
Example 9 (p. 42). The length of a room is 3 metres greater than its breadth, and its area is 180 square metres. Let breadth = x, so length = x + 3, and x(x + 3) = 180, i.e. x² + 3x − 180 = 0, i.e. (x + 15)(x − 12) = 0. Breadth cannot be negative, so x = −15 is rejected; breadth = 12 m and length = 15 m.
Example 10. A company’s weekly profit function (in thousands of dollars) is P(x) = −x² + 2x + 3, where x is the number of laptops produced (in hundreds). “At least $4,000 profit” means P(x) ≥ 4, so −x² + 2x + 3 ≥ 4, i.e. (x − 1)² ≤ 0. A square is never negative, so this holds only when (x − 1)² = 0, i.e. x = 1. The company makes exactly $4,000 profit when 100 laptops are produced, and there is no production level where profit is more than $4,000 — the maximum of P is 4, at x = 1, which is the same fact in vertex language.
عملی مسائل کو علامتوں میں بدل کر دو درجی مساواتیں بنائی جاتی ہیں۔ طریقہ: (۱) نامعلوم مقدار کو x فرض کریں، (۲) مسئلے کو علامتوں میں لکھ کر مساوات یا نامساوات بنائیں، پھر حل کریں۔ یاد رکھیں: ریاضی کا درست جواب کہانی میں غلط ہو سکتا ہے، جیسے منفی چوڑائی، اس لیے آخر میں کہانی کی شرط بھی دیکھیں۔
مثال ۹: کمرے کی چوڑائی 12 میٹر اور لمبائی 15 میٹر ہے (منفی جڑ −15 رد کر دی گئی)۔ مثال ۱۰: منافع P(x) = −x² + 2x + 3 ہے؛ 4000 ڈالر یا زیادہ منافع صرف 100 لیپ ٹاپ پر ملتا ہے، اس سے زیادہ کبھی نہیں، کیونکہ زیادہ سے زیادہ منافع ہی 4000 ہے۔
Below are all questions of Exercises 3.1 and 3.2 from Unit 3 of the official book (printed pages 39–40 and 42–43), each solved step by step. The book has no multiple-choice section, no review exercise and no inquisitive/numerical section in this unit, so none is invented here. Counting rule: one numbered question = one answer block; its parts are solved inside that block. That gives 4 + 7 = 11 blocks, covering 42 parts in total.
⚠️ Two answers worth a second look: Ex. 3.1 Q4(vi) collapses to “all real numbers” because one of its two cases is always true — that is correct, not a mistake. Ex. 3.2 Q1(iv) has two roots, 1/a + 1/b and 2/(a + b); if a = b the second root equals the forbidden value 1/a and must be dropped (flagged in the answer). Ex. 3.2 Q1(v) has no solution at all.
(i) x² + 6x + 13 = (x + 3)² − 9 + 13 = (x + 3)² + 4. a > 0, so minimum 4 at x = −3.
(ii) x² + 4x = (x + 2)² − 4. Minimum −4 at x = −2.
(iii) −x² + 8x + 13 = −(x² − 8x) + 13 = −(x − 4)² + 16 + 13 = −(x − 4)² + 29. a < 0, so maximum 29 at x = 4.
(iv) −x² − 3x − 5 = −(x² + 3x) − 5 = −(x + 3/2)² + 9/4 − 5 = −(x + 3/2)² − 11/4. Maximum −11/4 at x = −3/2.
(v) 3x² + 6x − 13 = 3(x² + 2x) − 13 = 3(x + 1)² − 3 − 13 = 3(x + 1)² − 16. Minimum −16 at x = −1.
(vi) −2x² − x + 21 = −2(x² + x/2) + 21 = −2(x + 1/4)² + 2(1/16) + 21 = −2(x + 1/4)² + 169/8. Maximum 169/8 (21.125) at x = −1/4.
Domain is R for every part (a polynomial has no restriction); the sketch needs vertex, intercepts and opening direction.
(i) x² − 4x: vertex (2, −4), minimum. x-intercepts 0 and 4, y-intercept 0. Increasing on (2, ∞), decreasing on (−∞, 2). Range [−4, ∞).
(ii) x² − 5x + 6: vertex (5/2, −1/4), minimum. x-intercepts 2 and 3 (factors (x − 2)(x − 3)), y-intercept 6. Range [−1/4, ∞).
(iii) −x² + 2x − 8: vertex (1, −7), maximum. No real x-intercepts (discriminant 4 − 32 < 0, and the parabola opens downward below the axis), y-intercept −8. Range (−∞, −7].
(iv) x² − 4x + 4 = (x − 2)²: vertex (2, 0), minimum, touching the axis. x-intercept 2 (double root), y-intercept 4. Range [0, ∞).
(v) x² + 2x − 8.3: vertex (−1, −9.3), minimum. x-intercepts from x = −1 ± √9.3 ≈ 2.05 and −4.05, y-intercept −8.3. Range [−9.3, ∞).
(vi) 6 − x − x²: vertex (−1/2, 25/4), maximum. x-intercepts −3 and 2 (factors −(x + 3)(x − 2)), y-intercept 6. Range (−∞, 25/4].
Method for every part: swap x and y, solve for y, keep the branch whose outputs lie in the given (restricted) domain; then domain and range swap for the inverse.
(i) x² − 3, x ≤ 0: Domain (−∞, 0], Range [−3, ∞). Inverse: y = −√(x + 3) (negative branch, since outputs must be ≤ 0). Domain f⁻¹ [−3, ∞), Range (−∞, 0].
(ii) x² + 6x + 4 = (x + 3)² − 5, x < −3: Domain (−∞, −3), Range (−5, ∞). Inverse: y = −3 − √(x + 5) (outputs < −3). Domain (−5, ∞), Range (−∞, −3).
(iii) 2x² − 8x + 11 = 2(x − 2)² + 3, x ≥ 2: Domain [2, ∞), Range [3, ∞). Inverse: y = 2 + √((x − 3)/2). Domain [3, ∞), Range [2, ∞).
(iv) 3x² − 2x + 6, x ≥ 5: f(5) = 71, and f increases on [5, ∞), so Domain [5, ∞), Range [71, ∞). Inverse: from 3y² − 2y + 6 = x, y = [1 + √(3x − 17)]/3 (positive branch; check x = 71 gives (1 + 14)/3 = 5 ✓). Domain [71, ∞), Range [5, ∞).
(v) 2(x − 3)² + 1, x ≥ 3: Domain [3, ∞), Range [1, ∞). Inverse: y = 3 + √((x − 1)/2). Domain [1, ∞), Range [3, ∞).
(vi) −3(x + 4)² − 5, x < −4: Domain (−∞, −4), Range (−∞, −5). Inverse: y = −4 − √(−(x + 5)/3) (outputs < −4). Domain (−∞, −5), Range (−∞, −4).
(i) x² + 1 is always positive, so x² + 1 = 5 gives x² = 4: {±2}. (The case = −5 is impossible.)
(ii) Absolute value zero means the inside is zero: x² + 5x + 4 = (x + 1)(x + 4) = 0: {−1, −4}.
(iii) x² − 6x + 8 = 4 gives x² − 6x + 4 = 0, x = 3 ± √5; x² − 6x + 8 = −4 gives x² − 6x + 12 = 0, discriminant < 0, no real roots. Answer {3 − √5, 3 + √5}.
(iv) Note x² − x + 1 > 0 always (discriminant < 0), so both cases are allowed. Plus case: 3x² − 7x + 2 = x² − x + 1 gives 2x² − 6x + 1 = 0, x = (3 ± √7)/2. Minus case: 3x² − 7x + 2 = −x² + x − 1 gives 4x² − 8x + 3 = 0, i.e. (2x − 1)(2x − 3) = 0, x = 1/2, 3/2. All four check in the original: {(3 − √7)/2, (3 + √7)/2, 1/2, 3/2}.
(v) −5 < x² − 4 < 5 gives x² < 9 (the other side, x² > −1, is always true): (−3, 3).
(vi) Case > 4: x² − 3x − 2 > 0 gives x < (3 − √17)/2 or x > (3 + √17)/2. Case < −4: x² − 3x + 6 > 0, discriminant 9 − 24 < 0, so this holds for every real x. Union with the first case is therefore all real numbers, R — surprising, but correct; the second case alone already covers everything.
(vii) First, x + 2 ≥ 0, so x ≥ −2 (an absolute value cannot equal a negative). Then −(x + 2) ≤ x² − 5x + 6 ≤ x + 2: the right side gives x² − 6x + 4 ≤ 0, i.e. 3 − √5 ≤ x ≤ 3 + √5; the left side gives (x − 2)² ≥ 0, always true. Intersect with x ≥ −2: [3 − √5, 3 + √5].
(viii) −4 < 2x² − 3x − 5 < 4: right side gives 2x² − 3x − 9 < 0, i.e. (2x + 3)(x − 3) < 0, so −3/2 < x < 3; left side gives 2x² − 3x − 1 > 0, i.e. x < (3 − √17)/4 or x > (3 + √17)/4. Intersect: (−3/2, (3 − √17)/4) ∪ ((3 + √17)/4, 3).
(i) 1/(3x) + 2x/3 = 1. Multiply by 3x: 1 + 2x² = 3x, so 2x² − 3x + 1 = (2x − 1)(x − 1) = 0: {1/2, 1}, both allowed.
(ii) Multiply by x(x + 1): x² + (x + 1)² = (5/2)x(x + 1). Times 2: 4x² + 4x + 2 = 5x² + 5x, so x² + x − 2 = (x + 2)(x − 1) = 0: {−2, 1}, both allowed.
(iii) Left side combines to (3x + 4)/((x + 1)(x + 2)). Cross-multiplying with 7/(x + 5): (3x + 4)(x + 5) = 7(x + 1)(x + 2), so 3x² + 19x + 20 = 7x² + 21x + 14, i.e. 2x² + x − 3 = (2x + 3)(x − 1) = 0: {−3/2, 1}, both allowed.
(iv) Rewrite as [a/(ax − 1) − b] + [b/(bx − 1) − a] = 0. Each bracket simplifies to (a + b − abx) over its denominator, so (a + b − abx)[1/(ax − 1) + 1/(bx − 1)] = 0. First factor: x = (a + b)/ab = 1/a + 1/b. Second factor: 1/(ax − 1) + 1/(bx − 1) = 0 gives (a + b)x = 2, so x = 2/(a + b). ⚠️ Edge case: if a = b, the second root equals 1/a, which is forbidden, so it must be dropped and only 1/a + 1/b (= 2/a) survives.
(v) No solution: the left side is (2x² + 2)/(x² − 1) = 2 + 4/(x² − 1), which can never equal 2, since 4/(x² − 1) is never zero. The equation 2x² + 2 = 2x² − 2 gives 2 = −2, impossible. Solution set: ∅.
(vi) Let t = √(x² + 5x + 1) ≥ 0. Then x² + 5x = t² − 1, and the equation becomes 3t² − 3 − 2t = 2, i.e. 3t² − 2t − 5 = (3t − 5)(t + 1) = 0, so t = 5/3 (t = −1 rejected, a square root is not negative). Then x² + 5x + 1 = 25/9, i.e. 9x² + 45x − 16 = 0, giving x = (−45 ± 51)/18: {1/3, −16/3}. Both check (each gives t = 5/3).
(vii) Domain x ≥ −4. Isolate √(2x + 8) = 7 − √(x + 5) and square: 2x + 8 = 54 + x − 14√(x + 5), so 14√(x + 5) = 46 − x, which needs 46 − x ≥ 0. Square again: 196x + 980 = x² − 92x + 2116, i.e. x² − 288x + 1136 = 0, with discriminant 280², giving x = 284 or x = 4. x = 284 fails the 46 − x ≥ 0 condition (extraneous); x = 4 checks: 4 + 3 = 7 ✓. Answer {4}.
(viii) Domain x ≥ 2. Square: 3x + 4 = 4 + 2x − 4 + 4√(2x − 4), so x + 4 = 4√(2x − 4). Square again: x² + 8x + 16 = 32x − 64, i.e. (x − 4)(x − 20) = 0. Both check: x = 4 gives 4 = 2 + 2 ✓; x = 20 gives 8 = 2 + 6 ✓. Answer {4, 20}.
(ix) Domain x ≥ −2. Square: 2x + 9 + 2√((x + 7)(x + 2)) = 6x + 13, so √((x + 7)(x + 2)) = 2x + 2, needing x ≥ −1. Square again: x² + 9x + 14 = 4x² + 8x + 4, i.e. 3x² − x − 10 = (3x + 5)(x − 2) = 0. x = −5/3 fails x ≥ −1 (extraneous); x = 2 checks: 3 + 2 = 5 = √25 ✓. Answer {2}.
(x) Domain x ≥ 3. Write √(x + 5) = 2 + √(x − 3) and square: x + 5 = 4 + x − 3 + 4√(x − 3), so √(x − 3) = 1, giving x = 4. Check: √9 − √1 = 3 − 1 = 2 ✓. Answer {4}.
Let n be the number of sheep; price each = 9000/n. At Rs. 100 less each, (9000/n − 100)(n + 3) = 9000. Expand: 9000 + 27000/n − 100n − 300 = 9000, so 27000/n − 100n − 300 = 0. Divide by 100 and multiply by n: 270 − n² − 3n = 0, i.e. n² + 3n − 270 = (n + 18)(n − 15) = 0. Reject n = −18; the farmer bought 15 sheep (at Rs. 600 each; check: at Rs. 500 each, Rs. 9000 buys 18 sheep, 3 more ✓).
Let d dozen; price per dozen = 2400/d. Then (d + 2)(2400/d − 0.50) = 2400. Expand: 2400 + 4800/d − 0.5d − 1 = 2400, so 4800/d − 0.5d − 1 = 0. Multiply by 2d: 9600 − d² − 2d = 0, i.e. d² + 2d − 9600 = 0. Discriminant 4 + 38400 = 38404 = 196², so d = (−2 + 196)/2 = 97 dozen (the negative root −99 is rejected).
Let speed = v km/h; time = 48/v hours. Slower: 48/(v − 2) = 48/v + 2. Divide by 2 and multiply by v(v − 2): 24v = 24(v − 2) + v(v − 2), so 24v = 24v − 48 + v² − 2v, i.e. v² − 2v − 48 = (v − 8)(v + 6) = 0. Reject v = −6; v = 8 km/h, so time = 48/8 = 6 hours. (Check: at 6 km/h the journey takes 8 hours, 2 more ✓.)
Let Abdul Hadi take x days alone; Abdullah takes x + 10 days. Their combined rate: 1/x + 1/(x + 10) = 1/12. Multiply by 12x(x + 10): 12(x + 10) + 12x = x(x + 10), so 24x + 120 = x² + 10x, i.e. x² − 14x − 120 = (x − 20)(x + 6) = 0. Reject x = −6; Abdul Hadi alone takes 20 days (Abdullah 30 days; check: 1/20 + 1/30 = 5/60 = 1/12 ✓).
Safe means d(s) ≤ 50: 0.02s² + 0.1s − 50 ≤ 0. Multiply by 50: s² + 5s − 2500 ≤ 0. Roots: s = [−5 ± √(25 + 10000)]/2 = [−5 ± √10025]/2. The positive root is (−5 + √10025)/2 ≈ 47.56. Since speed s ≥ 0 and the parabola opens upward, the inequality holds between the roots, so the safe range is 0 ≤ s ≤ (−5 + √10025)/2 km/h, about 47.6 km/h.
“At least 40 metres” means h(t) ≥ 40: −5t² + 20t + 30 ≥ 40, so −5t² + 20t − 10 ≥ 0. Divide by −5 (inequality flips): t² − 4t + 2 ≤ 0. Roots: t = [4 ± √(16 − 8)]/2 = 2 ± √2 ≈ 0.59 and 3.41. The parabola t² − 4t + 2 opens upward, so it is ≤ 0 between its roots: the rocket is at least 40 m up during [2 − √2, 2 + √2] seconds, about 0.59 s to 3.41 s after launch.
اس باب میں کیا ہے؟ Quadratic functions کا پورا نظریہ: parabola کا sketch اور analysis, completing تھے square سے maximum/minimum, domain محدود کر کے inverse, absolute value والی equations اور inequalities, دوسری قسم کی equations جو quadratic میں بدل جاتی ہیں (rational aur radical), اور آخر میں اصل زندگی کے مسائل.
Revision guidance: vertex form اور completing تھے square کو ہاتھ سے تین دفعہ لکھ کر پکا کرو — Exercise 3.1 Q1 اسی پر ہے. Inverse میں سب سے بڑا پھندا branch کا چناؤ ہے: ہمیشہ یہ پوچھو کے inverse کا output f کے domain میں ہے یا نہیں. Absolute value اور radical سوالوں میں جانچ (checking) کوئی extra step نہیں, وہ جواب کا حصہ ہے — بنا جانچ کے extraneous root کو جواب میں لکھنا غلطی ہے. Exercise: 📝 Official Exercise Questions — with Answers
اس باب میں دو درجی تفاعل کا خاکہ اور تجزیہ، مربع مکمل کر کے زیادہ سے زیادہ اور کم سے کم قدر، معکوس، مطلق قدر والی مساواتیں اور نامساواتیں، دو درجی میں تبدیل ہونے والی مساواتیں، اور عملی مسائل شامل ہیں۔
نظرثانی کی رہنمائی: ورٹیکس صورت اور مربع مکمل کرنے کا طریقہ اچھی طرح یاد کریں، معکوس میں شاخ کا انتخاب ڈومین سے کریں، اور مطلق قدر و جذری مساواتوں میں ہر جواب کی جانچ ضرور کریں۔
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