Welcome dear students! Today we are going to learn about Quadratic Equations from Class 10 Maths.
In Chapter 2, you studied different types of polynomials. One type was the quadratic polynomial of the form ax² + bx + c, where a ≠ 0. When we equate this polynomial to zero, we get a quadratic equation. Quadratic equations arise in many real-life situations. For instance, suppose a charity trust builds a prayer hall with a carpet area of 300 square metres, and its length is one metre more than twice its breadth. Let the breadth be x metres. The length is (2x + 1) metres. Figure 4.1 shows a rectangle with breadth x and length 2x + 1. The area is (2x + 1) times x, which equals 2x² + x square metres. Given the area is 300 square metres, we write 2x² + x = 300, which rearranges to 2x² + x – 300 = 0. This is a quadratic equation. Historically, Babylonians solved equations like x² – px + q = 0. Euclid used geometric methods. Indian mathematicians like Brahmagupta solved ax² + bx = c. Sridharacharya derived the quadratic formula by completing the square. Al-Khwarizmi and Abraham bar Hiyya Ha-Nasi also made significant contributions.
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Moving to Section 4.2, a quadratic equation in the variable x is an equation of the form ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0. Examples include 2x² + x – 300 = 0, 2x² – 3x + 1 = 0, 4x – 3x² + 2 = 0, and 1 – x² + 300 = 0. Any equation p(x) = 0 where p(x) is a polynomial of degree 2 is quadratic. Writing terms in descending order of degrees gives the standard form: ax² + bx + c = 0, a ≠ 0. Let us examine Example 1. Part (i): John and Jivanti together have 45 marbles. Both lose 5 marbles each. The product of their remaining marbles is 124. Let John have x marbles. Jivanti has 45 – x. After losing 5, John has x – 5, and Jivanti has 45 – x – 5, which is 40 – x. Their product is (x – 5)(40 – x). Expanding gives 40x – x² – 200 + 5x, which simplifies to –x² + 45x – 200. Setting this equal to 124 gives –x² + 45x – 324 = 0, or x² – 45x + 324 = 0. Part (ii): A cottage industry produces x toys daily. Cost per toy is 55 – x rupees. Total cost is 750 rupees. The equation is x(55 – x) = 750, which simplifies to 55x – x² = 750, or x² – 55x + 750 = 0.
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Example 2 asks us to check if equations are quadratic. Part (i): (x – 2)² + 1 = 2x – 3. Expanding the left side gives x² – 4x + 4 + 1 = x² – 4x + 5. The equation becomes x² – 4x + 5 = 2x – 3, which simplifies to x² – 6x + 8 = 0. This is of the form ax² + bx + c = 0, so it is quadratic. Part (ii): x(x + 1) + 8 = (x + 2)(x – 2). Expanding gives x² + x + 8 = x² – 4. Subtracting x² from both sides yields x + 12 = 0. This is linear, not quadratic. Part (iii): x(2x + 3) = x² + 1. Expanding the left side gives 2x² + 3x = x² + 1. Rearranging gives x² + 3x – 1 = 0. This is quadratic. Part (iv): (x + 2)³ = x³ – 4. Expanding the left side gives x³ + 6x² + 12x + 8 = x³ – 4. Subtracting x³ gives 6x² + 12x + 12 = 0, which divides by 6 to give x² + 2x + 2 = 0. This is quadratic. A crucial remark here: Be careful! In part (ii) above, the given equation appears to be a quadratic equation, but it is not a quadratic equation. In part (iv) above, the given equation appears to be a cubic equation, an equation of degree 3, and not a quadratic equation. But it turns out to be a quadratic equation. As you can see, often we need to simplify the given equation before deciding whether it is quadratic or not.
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Now, Exercise 4.1. Question 1: Check if quadratic. (i) (x + 1)² = 2(x – 3) expands to x² + 2x + 1 = 2x – 6, simplifying to x² + 7 = 0. Quadratic. (ii) x² – 2x = –2(3 – x) becomes x² – 2x = –6 + 2x, simplifying to x² – 4x + 6 = 0. Quadratic. (iii) (x – 2)(x + 1) = (x – 1)(x + 3) expands to x² – x – 2 = x² + 2x – 3, simplifying to –3x + 1 = 0. Not quadratic. (iv) (x – 3)(2x + 1) = x(x + 5) expands to 2x² – 5x – 3 = x² + 5x, simplifying to x² – 10x – 3 = 0. Quadratic. (v) (2x – 1)(x – 3) = (x + 5)(x – 1) expands to 2x² – 7x + 3 = x² + 4x – 5, simplifying to x² – 11x + 8 = 0. Quadratic. (vi) x² + 3x + 1 = (x – 2)² expands to x² + 3x + 1 = x² – 4x + 4, simplifying to 7x – 3 = 0. Not quadratic. (vii) (x + 2)³ = 2x(x² – 1) expands to x³ + 6x² + 12x + 8 = 2x³ – 2x, rearranging to x³ – 6x² – 14x – 8 = 0. Not quadratic. (viii) x³ – 4x² – x + 1 = (x – 2)³ expands to x³ – 4x² – x + 1 = x³ – 6x² + 12x – 8, simplifying to 2x² – 13x + 9 = 0. Quadratic.
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Question 2: Represent situations as quadratic equations. (i) Area 528 m², length is one more than twice breadth x. Equation: x(2x + 1) = 528, which gives 2x² + x – 528 = 0. (ii) Product of consecutive positive integers is 306. Let integers be x and x + 1. Equation: x(x + 1) = 306, which gives x² + x – 306 = 0. (iii) Rohan’s present age is x. Mother is x + 26. In 3 years, ages are x + 3 and x + 29. Product: (x + 3)(x + 29) = 360. Expanding gives x² + 32x + 87 = 360, which simplifies to x² + 32x – 273 = 0. (iv) A train travels 480 km at uniform speed x km/h. Time taken is 480/x hours. If speed is x – 8 km/h, time is 480/(x – 8) hours. The difference in time is 3 hours: 480/(x – 8) – 480/x = 3. Multiplying through by x(x – 8) gives 480x – 480(x – 8) = 3x(x – 8). Simplifying the left side: 480x – 480x + 3840 = 3x² – 24x. So, 3840 = 3x² – 24x. Rearranging gives 3x² – 24x – 3840 = 0. Dividing by 3 gives x² – 8x – 1280 = 0. To find the speed, we solve this equation. We factorise x² – 8x – 1280. We look for two numbers multiplying to –1280 and adding to –8. They are –40 and 32. So, x² – 40x + 32x – 1280 = 0. Grouping terms: x(x – 40) + 32(x – 40) = 0. This gives (x – 40)(x + 32) = 0. So x = 40 or x = –32. Since speed cannot be negative, the speed of the train is 40 km/h.
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Section 4.3 covers solving by factorisation. Consider 2x² – 3x + 1 = 0. Substituting x = 1 gives 2(1)² – 3(1) + 1 = 0. Thus, 1 is a root. A real number α is a root of ax² + bx + c = 0 if aα² + bα + c = 0. Roots and zeroes are identical. A quadratic has at most two roots. Example 3: Solve 2x² – 5x + 3 = 0. Split –5x into –2x – 3x. 2x² – 2x – 3x + 3 = 0. Factor by grouping: 2x(x – 1) – 3(x – 1) = 0. This gives (2x – 3)(x – 1) = 0. So 2x – 3 = 0 gives x = 3/2, and x – 1 = 0 gives x = 1. Roots are 1 and 3/2. Example 4: Solve 6x² – x – 2 = 0. Split –x into 3x – 4x. 6x² + 3x – 4x – 2 = 0. Grouping: 3x(2x + 1) – 2(2x + 1) = 0. This gives (3x – 2)(2x + 1) = 0. So 3x – 2 = 0 gives x = 2/3, and 2x + 1 = 0 gives x = –1/2. Roots are 2/3 and –1/2.
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Example 5: Solve √3 x² – 2√6 x + 2√3 = 0. Split middle term: √3 x² – √6 x – √6 x + 2√3 = 0. Factor by grouping: √3 x(x – √2) – √6(x – √2) = 0. This gives (√3 x – √6)(x – √2) = 0. So √3 x – √6 = 0 gives x = √6/√3 = √2, and x – √2 = 0 gives x = √2. The root √2 is repeated twice. Example 6: Find prayer hall dimensions from 2x² + x – 300 = 0. Split x into –24x + 25x. 2x² – 24x + 25x – 300 = 0. Grouping: 2x(x – 12) + 25(x – 12) = 0. This gives (x – 12)(2x + 25) = 0. So x = 12 or x = –12.5. Breadth cannot be negative, so x = 12 m. Length = 2(12) + 1 = 25 m.
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Exercise 4.2 Question 1: (i) x² – 3x – 10 = 0. Factors of –10 summing to –3 are –5 and 2. So (x – 5)(x + 2) = 0. Roots: 5, –2. (ii) 2x² + x – 6 = 0. Split x into 4x – 3x. 2x² + 4x – 3x – 6 = 0. 2x(x + 2) – 3(x + 2) = 0. (2x – 3)(x + 2) = 0. Roots: 3/2, –2. (iii) √2 x² + 7x + 5√2 = 0. Split 7x into 5x + 2x. √2 x² + 5x + 2x + 5√2 = 0. x(√2 x + 5) + √2(√2 x + 5) = 0. (x + √2)(√2 x + 5) = 0. Roots: –√2, –5/√2. (iv) 2x² – x + 1/8 = 0. Multiply by 8: 16x² – 8x + 1 = 0. This is (4x – 1)² = 0. Roots: 1/4 and 1/4, a repeated root. (v) 100x² – 20x + 1 = 0. This is (10x – 1)² = 0. Roots: 1/10 and 1/10, a repeated root.
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Exercise 4.2 Question 2: Solve Example 1 problems. Marbles: x² – 45x + 324 = 0. Factors of 324 summing to –45 are –9 and –36. (x – 9)(x – 36) = 0. x = 9 or 36. Therefore, if John had 9 marbles, Jivanti had 36. If John had 36 marbles, Jivanti had 9. Toys: x² – 55x + 750 = 0. Factors of 750 summing to –55 are –25 and –30. (x – 25)(x – 30) = 0. x = 25 or 30. Therefore, if 25 toys were produced, the cost per toy was 30 rupees. If 30 toys were produced, the cost per toy was 25 rupees. Question 3: Sum 27, product 182. x(27 – x) = 182 gives x² – 27x + 182 = 0. Factors of 182 summing to –27 are –13 and –14. (x – 13)(x – 14) = 0. Numbers are 13 and 14. Question 4: Consecutive integers, sum of squares 365. x² + (x + 1)² = 365. Expanding: x² + x² + 2x + 1 = 365. 2x² + 2x – 364 = 0. Divide by 2: x² + x – 182 = 0. Factors of –182 summing to 1 are 14 and –13. (x + 14)(x – 13) = 0. Positive integer is 13. Numbers: 13, 14.
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Question 5: Right triangle altitude 7 cm less than base. Hypotenuse 13 cm. Let base be x. Altitude is x – 7. By Pythagoras: x² + (x – 7)² = 13². Expanding: x² + x² – 14x + 49 = 169. 2x² – 14x – 120 = 0. Divide by 2: x² – 7x – 60 = 0. Factors of –60 summing to –7 are –12 and 5. (x – 12)(x + 5) = 0. x = 12 or –5. Base is 12 cm. Altitude is 12 – 7 = 5 cm. Question 6: Cost per article = 2x + 3. Total cost x(2x + 3) = 90. 2x² + 3x – 90 = 0. Split 3x into 15x – 12x. 2x² + 15x – 12x – 90 = 0. x(2x + 15) – 6(2x + 15) = 0. (x – 6)(2x + 15) = 0. x = 6 or –15/2. Number of articles is 6. Cost per article = 2(6) + 3 = 15 rupees.
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Section 4.4 covers the nature of roots. The roots of ax² + bx + c = 0 are given by the quadratic formula: x = [–b ± √(b² – 4ac)] / 2a. If b² – 4ac > 0, we get two distinct real roots. If b² – 4ac = 0, we get two equal real roots, both equal to –b/2a. If b² – 4ac < 0, there are no real roots. The expression b² – 4ac is called the discriminant. Example 7: 2x² – 4x + 3 = 0. a = 2, b = –4, c = 3. Discriminant D = (–4)² – 4(2)(3) = 16 – 24 = –8. Since D < 0, there are no real roots. Example 8: Circular park diameter 13 m. Gates A and B are diametrically opposite. Pole P is on the boundary. Let BP = x. Difference in distances AP – BP = 7, so AP = x + 7. Figure 4.2 shows a circle with diameter AB and point P on the circumference. Angle APB is 90 degrees because it is subtended by the diameter. By Pythagoras theorem: AP² + PB² = AB². So (x + 7)² + x² = 13². Expanding: x² + 14x + 49 + x² = 169. 2x² + 14x – 120 = 0. Divide by 2: x² + 7x – 60 = 0. Discriminant D = 7² – 4(1)(–60) = 49 + 240 = 289. D > 0, so real roots exist. Using quadratic formula: x = [–7 ± √289] / 2 = [–7 ± 17] / 2. So x = 5 or x = –12. Distance must be positive, so x = 5 m. BP = 5 m, AP = 12 m.
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Example 9: 3x² – 2x + 1/3 = 0. a = 3, b = –2, c = 1/3. D = (–2)² – 4(3)(1/3) = 4 – 4 = 0. Since D = 0, there are two equal real roots. Roots are –b/2a = 2/6 = 1/3. So roots are 1/3 and 1/3. Now, Exercise 4.3. Question 1: Find nature of roots and find them if real. (i) 2x² – 3x + 5 = 0. a = 2, b = –3, c = 5. D = (–3)² – 4(2)(5) = 9 – 40 = –31. D < 0, so no real roots exist. (ii) 3x² – 4√3 x + 4 = 0. a = 3, b = –4√3, c = 4. D = (–4√3)² – 4(3)(4) = 48 – 48 = 0. D = 0, so two equal real roots exist. Roots are –b/2a = 4√3 / 6 = 2√3 / 3. (iii) 2x² – 6x + 3 = 0. a = 2, b = –6, c = 3. D = (–6)² – 4(2)(3) = 36 – 24 = 12. D > 0, so two distinct real roots exist. Using formula: x = [6 ± √12] / 4. √12 simplifies to 2√3. So x = [6 ± 2√3] / 4 = [3 ± √3] / 2. Roots are (3 + √3)/2 and (3 – √3)/2.
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Question 2: Find k for equal roots. (i) 2x² + kx + 3 = 0. a = 2, b = k, c = 3. For equal roots, D = 0. So k² – 4(2)(3) = 0. k² – 24 = 0. k² = 24. k = ±√24 = ±2√6. (ii) kx(x – 2) + 6 = 0. Expanding gives kx² – 2kx + 6 = 0. a = k, b = –2k, c = 6. D = (–2k)² – 4(k)(6) = 4k² – 24k. For equal roots, 4k² – 24k = 0. 4k(k – 6) = 0. So k = 0 or k = 6. But if k = 0, the equation becomes 6 = 0, which is not quadratic. Therefore, k = 6. Question 3: Rectangular mango grove, length is twice breadth. Area is 800 m². Let breadth be x. Length is 2x. Area = x(2x) = 2x². So 2x² = 800, which rearranges to 2x² – 800 = 0. Here a = 2, b = 0, c = –800. We verify using the discriminant: D = b² – 4ac = 0² – 4(2)(–800) = 6400. Since D > 0, real roots exist. Solving using the quadratic formula: x = [0 ± √6400] / 4 = ±80 / 4 = ±20. Since breadth cannot be negative, x = 20. Breadth is 20 m, length is 40 m. Yes, it is possible.
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Question 4: Sum of two friends' ages is 20. Four years ago, product of ages was 48. Let present ages be x and 20 – x. Four years ago: x – 4 and 16 – x. Product: (x – 4)(16 – x) = 48. Expanding: 16x – x² – 64 + 4x = 48. –x² + 20x – 64 = 48. Rearranging: x² – 20x + 112 = 0. Discriminant D = (–20)² – 4(1)(112) = 400 – 448 = –48. Since D < 0, there are no real roots. Therefore, this situation is not possible. Question 5: Rectangular park perimeter 80 m, area 400 m². Let length be x, breadth be y. Perimeter 2(x + y) = 80, so x + y = 40, meaning y = 40 – x. Area xy = 400. Substituting y: x(40 – x) = 400. 40x – x² = 400. x² – 40x + 400 = 0. Discriminant D = (–40)² – 4(1)(400) = 1600 – 1600 = 0. D = 0, so equal roots exist. Root x = 40 / 2 = 20. So length is 20 m, breadth is 20 m. Yes, it is possible.
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Section 4.5 Summary: 1. A quadratic equation in x is of the form ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0. 2. A real number α is a root if aα² + bα + c = 0. Zeroes of the polynomial and roots of the equation are the same. 3. If we can factorise ax² + bx + c into two linear factors, we find roots by setting each factor to zero. 4. Quadratic formula: x = [–b ± √(b² – 4ac)] / 2a, provided b² – 4ac ≥ 0. 5. Nature of roots: If b² – 4ac > 0, two distinct real roots. If b² – 4ac = 0, two equal real roots. If b² – 4ac < 0, no real roots.
Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]