Exercise 4.1
1. Check whether the following are quadratic equations:
(i) \((x+1)^2 = 2(x-3)\)
Step 1: Expand both sides
\((x+1)^2 = x^2 + 2x + 1\)
\(2(x-3) = 2x - 6\)
Step 2: Write the equation
\(x^2 + 2x + 1 = 2x - 6\)
Step 3: Simplify
\(x^2 + 2x + 1 - 2x + 6 = 0\)
\(x^2 + 7 = 0\)
Conclusion: This is a quadratic equation since the highest power of x is 2.
(ii) \(x^2 - 2x = (-2)(3-x)\)
Step 1: Expand the right side
\((-2)(3-x) = -6 + 2x\)
Step 2: Write the equation
\(x^2 - 2x = -6 + 2x\)
Step 3: Simplify
\(x^2 - 2x - 2x + 6 = 0\)
\(x^2 - 4x + 6 = 0\)
Conclusion: This is a quadratic equation since the highest power of x is 2.
(iii) \((x-2)(x+1) = (x-1)(x+3)\)
Step 1: Expand both sides
\((x-2)(x+1) = x^2 + x - 2x - 2 = x^2 - x - 2\)
\((x-1)(x+3) = x^2 + 3x - x - 3 = x^2 + 2x - 3\)
Step 2: Write the equation
\(x^2 - x - 2 = x^2 + 2x - 3\)
Step 3: Simplify
\(x^2 - x - 2 - x^2 - 2x + 3 = 0\)
\(-3x + 1 = 0\)
Conclusion: This is NOT a quadratic equation since the highest power of x is 1.
(iv) \((x-3)(2x+1) = x(x+5)\)
Step 1: Expand both sides
\((x-3)(2x+1) = 2x^2 + x - 6x - 3 = 2x^2 - 5x - 3\)
\(x(x+5) = x^2 + 5x\)
Step 2: Write the equation
\(2x^2 - 5x - 3 = x^2 + 5x\)
Step 3: Simplify
\(2x^2 - 5x - 3 - x^2 - 5x = 0\)
\(x^2 - 10x - 3 = 0\)
Conclusion: This is a quadratic equation since the highest power of x is 2.
(v) \((2x-1)(x-3) = (x+5)(x-1)\)
Step 1: Expand both sides
\((2x-1)(x-3) = 2x^2 - 6x - x + 3 = 2x^2 - 7x + 3\)
\((x+5)(x-1) = x^2 - x + 5x - 5 = x^2 + 4x - 5\)
Step 2: Write the equation
\(2x^2 - 7x + 3 = x^2 + 4x - 5\)
Step 3: Simplify
\(2x^2 - 7x + 3 - x^2 - 4x + 5 = 0\)
\(x^2 - 11x + 8 = 0\)
Conclusion: This is a quadratic equation since the highest power of x is 2.
(vi) \(x^2 + 3x + 1 = (x-2)^2\)
Step 1: Expand the right side
\((x-2)^2 = x^2 - 4x + 4\)
Step 2: Write the equation
\(x^2 + 3x + 1 = x^2 - 4x + 4\)
Step 3: Simplify
\(x^2 + 3x + 1 - x^2 + 4x - 4 = 0\)
\(7x - 3 = 0\)
Conclusion: This is NOT a quadratic equation since the highest power of x is 1.
(vii) \((x+2)^3 = 2x(x^2-1)\)
Step 1: Expand both sides
\((x+2)^3 = x^3 + 6x^2 + 12x + 8\)
\(2x(x^2-1) = 2x^3 - 2x\)
Step 2: Write the equation
\(x^3 + 6x^2 + 12x + 8 = 2x^3 - 2x\)
Step 3: Simplify
\(x^3 + 6x^2 + 12x + 8 - 2x^3 + 2x = 0\)
\(-x^3 + 6x^2 + 14x + 8 = 0\)
Conclusion: This is NOT a quadratic equation since the highest power of x is 3.
(viii) \(x^3 - 4x^2 - x + 1 = (x-2)^3\)
Step 1: Expand the right side
\((x-2)^3 = x^3 - 6x^2 + 12x - 8\)
Step 2: Write the equation
\(x^3 - 4x^2 - x + 1 = x^3 - 6x^2 + 12x - 8\)
Step 3: Simplify
\(x^3 - 4x^2 - x + 1 - x^3 + 6x^2 - 12x + 8 = 0\)
\(2x^2 - 13x + 9 = 0\)
Conclusion: This is a quadratic equation since the highest power of x is 2.
2. Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Step 1: Let breadth = x m
Step 2: Then length = (2x + 1) m
Step 3: Area = length × breadth = x(2x + 1) = 528
Step 4: So the quadratic equation is: \(2x^2 + x - 528 = 0\)
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
Step 1: Let the first integer = x
Step 2: Then the next consecutive integer = x + 1
Step 3: Their product = x(x + 1) = 306
Step 4: So the quadratic equation is: \(x^2 + x - 306 = 0\)
(iii) Ram's mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Ram's present age.
Step 1: Let Ram's present age = x years
Step 2: Then his mother's present age = (x + 26) years
Step 3: After 3 years: Ram's age = (x + 3) years, Mother's age = (x + 29) years
Step 4: Product of their ages after 3 years = (x + 3)(x + 29) = 360
Step 5: So the quadratic equation is: \(x^2 + 32x + 87 - 360 = 0\) or \(x^2 + 32x - 273 = 0\)
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Step 1: Let the actual speed = x km/h
Step 2: Then actual time = \(\frac{480}{x}\) hours
Step 3: Reduced speed = (x - 8) km/h
Step 4: Time at reduced speed = \(\frac{480}{x-8}\) hours
Step 5: According to the problem: \(\frac{480}{x-8} - \frac{480}{x} = 3\)
Step 6: Multiply both sides by x(x-8): \(480x - 480(x-8) = 3x(x-8)\)
Step 7: Simplify: \(3840 = 3x^2 - 24x\)
Step 8: So the quadratic equation is: \(3x^2 - 24x - 3840 = 0\) or \(x^2 - 8x - 1280 = 0\)
Exercise 4.2
1. Find the roots of the following quadratic equations by factorisation:
(i) \(x^2 - 3x - 10 = 0\)
Step 1: Factorize the quadratic expression
\(x^2 - 3x - 10 = x^2 - 5x + 2x - 10\)
\(= x(x - 5) + 2(x - 5)\)
\(= (x - 5)(x + 2)\)
Step 2: Set each factor to zero
\(x - 5 = 0\) ⇒ \(x = 5\)
\(x + 2 = 0\) ⇒ \(x = -2\)
Conclusion: The roots are 5 and -2.
(ii) \(2x^2 + x - 6 = 0\)
Step 1: Factorize the quadratic expression
\(2x^2 + x - 6 = 2x^2 + 4x - 3x - 6\)
\(= 2x(x + 2) - 3(x + 2)\)
\(= (2x - 3)(x + 2)\)
Step 2: Set each factor to zero
\(2x - 3 = 0\) ⇒ \(x = \frac{3}{2}\)
\(x + 2 = 0\) ⇒ \(x = -2\)
Conclusion: The roots are \(\frac{3}{2}\) and -2.
(iii) \(\sqrt{2} x^2 + 7x + 5\sqrt{2} = 0\)
Step 1: Factorize the quadratic expression
\(\sqrt{2} x^2 + 7x + 5\sqrt{2} = \sqrt{2} x^2 + 2x + 5x + 5\sqrt{2}\)
\(= \sqrt{2}x(x + \sqrt{2}) + 5(x + \sqrt{2})\)
\(= (\sqrt{2}x + 5)(x + \sqrt{2})\)
Step 2: Set each factor to zero
\(\sqrt{2}x + 5 = 0\) ⇒ \(x = -\frac{5}{\sqrt{2}} = -\frac{5\sqrt{2}}{2}\)
\(x + \sqrt{2} = 0\) ⇒ \(x = -\sqrt{2}\)
Conclusion: The roots are \(-\frac{5\sqrt{2}}{2}\) and \(-\sqrt{2}\).
(iv) \(2x^2 - x + \frac{1}{8} = 0\)
Step 1: Multiply the entire equation by 8 to eliminate fractions
\(16x^2 - 8x + 1 = 0\)
Step 2: Factorize
\(16x^2 - 8x + 1 = 16x^2 - 4x - 4x + 1\)
\(= 4x(4x - 1) - 1(4x - 1)\)
\(= (4x - 1)(4x - 1) = (4x - 1)^2\)
Step 3: Set the factor to zero
\(4x - 1 = 0\) ⇒ \(x = \frac{1}{4}\)
Conclusion: The equation has equal roots: \(\frac{1}{4}\).
(v) \(100x^2 - 20x + 1 = 0\)
Step 1: Factorize
\(100x^2 - 20x + 1 = 100x^2 - 10x - 10x + 1\)
\(= 10x(10x - 1) - 1(10x - 1)\)
\(= (10x - 1)(10x - 1) = (10x - 1)^2\)
Step 2: Set the factor to zero
\(10x - 1 = 0\) ⇒ \(x = \frac{1}{10}\)
Conclusion: The equation has equal roots: \(\frac{1}{10}\).
(vi) \(2x^2 - 7x + 6 = 0\)
Step 1: Factorize
\(2x^2 - 7x + 6 = 2x^2 - 4x - 3x + 6\)
\(= 2x(x - 2) - 3(x - 2)\)
\(= (2x - 3)(x - 2)\)
Step 2: Set each factor to zero
\(2x - 3 = 0\) ⇒ \(x = \frac{3}{2}\)
\(x - 2 = 0\) ⇒ \(x = 2\)
Conclusion: The roots are \(\frac{3}{2}\) and 2.
(vii) \(x^2 - 10x - 96 = 0\)
Step 1: Factorize
\(x^2 - 10x - 96 = x^2 - 16x + 6x - 96\)
\(= x(x - 16) + 6(x - 16)\)
\(= (x - 16)(x + 6)\)
Step 2: Set each factor to zero
\(x - 16 = 0\) ⇒ \(x = 16\)
\(x + 6 = 0\) ⇒ \(x = -6\)
Conclusion: The roots are 16 and -6.
(viii) \(\sqrt{3}x^2 + 10x + 7\sqrt{3} = 0\)
Step 1: Factorize
\(\sqrt{3}x^2 + 10x + 7\sqrt{3} = \sqrt{3}x^2 + 7x + 3x + 7\sqrt{3}\)
\(= \sqrt{3}x(x + \sqrt{3}) + 7(x + \sqrt{3})\)
\(= (\sqrt{3}x + 7)(x + \sqrt{3})\)
Step 2: Set each factor to zero
\(\sqrt{3}x + 7 = 0\) ⇒ \(x = -\frac{7}{\sqrt{3}} = -\frac{7\sqrt{3}}{3}\)
\(x + \sqrt{3} = 0\) ⇒ \(x = -\sqrt{3}\)
Conclusion: The roots are \(-\frac{7\sqrt{3}}{3}\) and \(-\sqrt{3}\).
(ix) \(x^2 + 2\sqrt{2}x + 2 = 0\)
Step 1: Factorize
\(x^2 + 2\sqrt{2}x + 2 = x^2 + \sqrt{2}x + \sqrt{2}x + 2\)
\(= x(x + \sqrt{2}) + \sqrt{2}(x + \sqrt{2})\)
\(= (x + \sqrt{2})(x + \sqrt{2}) = (x + \sqrt{2})^2\)
Step 2: Set the factor to zero
\(x + \sqrt{2} = 0\) ⇒ \(x = -\sqrt{2}\)
Conclusion: The equation has equal roots: \(-\sqrt{2}\).
(x) \(14x + 5 - 3x^2 = 0\)
Step 1: Rewrite in standard form
\(-3x^2 + 14x + 5 = 0\) or \(3x^2 - 14x - 5 = 0\)
Step 2: Factorize
\(3x^2 - 14x - 5 = 3x^2 - 15x + x - 5\)
\(= 3x(x - 5) + 1(x - 5)\)
\(= (3x + 1)(x - 5)\)
Step 3: Set each factor to zero
\(3x + 1 = 0\) ⇒ \(x = -\frac{1}{3}\)
\(x - 5 = 0\) ⇒ \(x = 5\)
Conclusion: The roots are \(-\frac{1}{3}\) and 5.
2. Solve the problems given in Example 1.
Example 1 problems would be solved similarly using factorization method as shown above.
3. Find two numbers whose sum is 27 and product is 182.
Step 1: Let the numbers be x and y
Step 2: Given: x + y = 27 and xy = 182
Step 3: The quadratic equation with roots x and y is: \(t^2 - (x+y)t + xy = 0\)
\(t^2 - 27t + 182 = 0\)
Step 4: Factorize
\(t^2 - 27t + 182 = t^2 - 13t - 14t + 182\)
\(= t(t - 13) - 14(t - 13)\)
\(= (t - 13)(t - 14)\)
Step 5: The numbers are 13 and 14.
4. Find two consecutive positive integers, sum of whose squares is 365.
Step 1: Let the integers be x and x+1
Step 2: Given: \(x^2 + (x+1)^2 = 365\)
Step 3: Expand: \(x^2 + x^2 + 2x + 1 = 365\)
\(2x^2 + 2x + 1 - 365 = 0\)
\(2x^2 + 2x - 364 = 0\)
\(x^2 + x - 182 = 0\) (dividing by 2)
Step 4: Factorize
\(x^2 + x - 182 = x^2 + 14x - 13x - 182\)
\(= x(x + 14) - 13(x + 14)\)
\(= (x - 13)(x + 14)\)
Step 5: Since we need positive integers, x = 13
Conclusion: The integers are 13 and 14.
5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Step 1: Let base = x cm, then altitude = (x - 7) cm
Step 2: Using Pythagoras theorem: \(x^2 + (x-7)^2 = 13^2\)
Step 3: Expand: \(x^2 + x^2 - 14x + 49 = 169\)
\(2x^2 - 14x + 49 - 169 = 0\)
\(2x^2 - 14x - 120 = 0\)
\(x^2 - 7x - 60 = 0\) (dividing by 2)
Step 4: Factorize
\(x^2 - 7x - 60 = x^2 - 12x + 5x - 60\)
\(= x(x - 12) + 5(x - 12)\)
\(= (x + 5)(x - 12)\)
Step 5: Since length cannot be negative, x = 12
Conclusion: Base = 12 cm, Altitude = 5 cm
6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹90, find the number of articles produced and the cost of each article.
Step 1: Let number of articles = x
Step 2: Then cost of each article = (2x + 3) rupees
Step 3: Total cost = x(2x + 3) = 90
\(2x^2 + 3x - 90 = 0\)
Step 4: Factorize
\(2x^2 + 3x - 90 = 2x^2 + 15x - 12x - 90\)
\(= x(2x + 15) - 6(2x + 15)\)
\(= (x - 6)(2x + 15)\)
Step 5: Since number of articles cannot be negative, x = 6
Conclusion: Number of articles = 6, Cost of each article = 2×6 + 3 = ₹15
Exercise 4.3
1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) \(2x^2 - 7x + 3 = 0\)
Step 1: Divide by coefficient of x²
\(x^2 - \frac{7}{2}x + \frac{3}{2} = 0\)
Step 2: Move constant term to RHS
\(x^2 - \frac{7}{2}x = -\frac{3}{2}\)
Step 3: Add square of half of coefficient of x to both sides
Half of \(\frac{7}{2}\) is \(\frac{7}{4}\), square is \(\frac{49}{16}\)
\(x^2 - \frac{7}{2}x + \frac{49}{16} = -\frac{3}{2} + \frac{49}{16}\)
\(\left(x - \frac{7}{4}\right)^2 = -\frac{24}{16} + \frac{49}{16} = \frac{25}{16}\)
Step 4: Take square root
\(x - \frac{7}{4} = \pm \frac{5}{4}\)
Step 5: Solve for x
\(x = \frac{7}{4} \pm \frac{5}{4}\)
\(x = \frac{7+5}{4} = 3\) or \(x = \frac{7-5}{4} = \frac{1}{2}\)
Conclusion: The roots are 3 and \(\frac{1}{2}\).
(ii) \(2x^2 + x - 4 = 0\)
Step 1: Divide by coefficient of x²
\(x^2 + \frac{1}{2}x - 2 = 0\)
Step 2: Move constant term to RHS
\(x^2 + \frac{1}{2}x = 2\)
Step 3: Add square of half of coefficient of x to both sides
Half of \(\frac{1}{2}\) is \(\frac{1}{4}\), square is \(\frac{1}{16}\)
\(x^2 + \frac{1}{2}x + \frac{1}{16} = 2 + \frac{1}{16} = \frac{33}{16}\)
\(\left(x + \frac{1}{4}\right)^2 = \frac{33}{16}\)
Step 4: Take square root
\(x + \frac{1}{4} = \pm \frac{\sqrt{33}}{4}\)
Step 5: Solve for x
\(x = -\frac{1}{4} \pm \frac{\sqrt{33}}{4} = \frac{-1 \pm \sqrt{33}}{4}\)
Conclusion: The roots are \(\frac{-1 + \sqrt{33}}{4}\) and \(\frac{-1 - \sqrt{33}}{4}\).
(iii) \(4x^2 + 4\sqrt{3}x + 3 = 0\)
Step 1: Divide by coefficient of x²
\(x^2 + \sqrt{3}x + \frac{3}{4} = 0\)
Step 2: Move constant term to RHS
\(x^2 + \sqrt{3}x = -\frac{3}{4}\)
Step 3: Add square of half of coefficient of x to both sides
Half of \(\sqrt{3}\) is \(\frac{\sqrt{3}}{2}\), square is \(\frac{3}{4}\)
\(x^2 + \sqrt{3}x + \frac{3}{4} = -\frac{3}{4} + \frac{3}{4} = 0\)
\(\left(x + \frac{\sqrt{3}}{2}\right)^2 = 0\)
Step 4: Take square root
\(x + \frac{\sqrt{3}}{2} = 0\)
Step 5: Solve for x
\(x = -\frac{\sqrt{3}}{2}\)
Conclusion: The equation has equal roots: \(-\frac{\sqrt{3}}{2}\).
(iv) \(2x^2 + x + 4 = 0\)
Step 1: Divide by coefficient of x²
\(x^2 + \frac{1}{2}x + 2 = 0\)
Step 2: Move constant term to RHS
\(x^2 + \frac{1}{2}x = -2\)
Step 3: Add square of half of coefficient of x to both sides
Half of \(\frac{1}{2}\) is \(\frac{1}{4}\), square is \(\frac{1}{16}\)
\(x^2 + \frac{1}{2}x + \frac{1}{16} = -2 + \frac{1}{16} = -\frac{31}{16}\)
\(\left(x + \frac{1}{4}\right)^2 = -\frac{31}{16}\)
Conclusion: Since the square of a real number cannot be negative, this equation has no real roots.
2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.
The quadratic formula is: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
(i) \(2x^2 - 7x + 3 = 0\)
Step 1: Identify coefficients: a = 2, b = -7, c = 3
Step 2: Calculate discriminant: D = b² - 4ac = (-7)² - 4×2×3 = 49 - 24 = 25
Step 3: Apply formula: \(x = \frac{-(-7) \pm \sqrt{25}}{2×2} = \frac{7 \pm 5}{4}\)
Step 4: \(x = \frac{7+5}{4} = 3\) or \(x = \frac{7-5}{4} = \frac{1}{2}\)
(ii) \(2x^2 + x - 4 = 0\)
Step 1: Identify coefficients: a = 2, b = 1, c = -4
Step 2: Calculate discriminant: D = 1² - 4×2×(-4) = 1 + 32 = 33
Step 3: Apply formula: \(x = \frac{-1 \pm \sqrt{33}}{4}\)
(iii) \(4x^2 + 4\sqrt{3}x + 3 = 0\)
Step 1: Identify coefficients: a = 4, b = 4√3, c = 3
Step 2: Calculate discriminant: D = (4√3)² - 4×4×3 = 48 - 48 = 0
Step 3: Apply formula: \(x = \frac{-4\sqrt{3} \pm 0}{8} = -\frac{\sqrt{3}}{2}\)
(iv) \(2x^2 + x + 4 = 0\)
Step 1: Identify coefficients: a = 2, b = 1, c = 4
Step 2: Calculate discriminant: D = 1² - 4×2×4 = 1 - 32 = -31
Step 3: Since D < 0, there are no real roots.
3. Find the roots of the following equations:
(i) \(x - \frac{1}{x} = 3, x \neq 0\)
Step 1: Multiply both sides by x
\(x^2 - 1 = 3x\)
\(x^2 - 3x - 1 = 0\)
Step 2: Use quadratic formula: a = 1, b = -3, c = -1
D = (-3)² - 4×1×(-1) = 9 + 4 = 13
\(x = \frac{3 \pm \sqrt{13}}{2}\)
(ii) \(\frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30}, x \neq -4, 7\)
Step 1: Combine the fractions on LHS
\(\frac{(x-7) - (x+4)}{(x+4)(x-7)} = \frac{11}{30}\)
\(\frac{-11}{(x+4)(x-7)} = \frac{11}{30}\)
Step 2: Cross multiply
\(-11 × 30 = 11(x+4)(x-7)\)
\(-330 = 11(x^2 - 3x - 28)\)
\(-30 = x^2 - 3x - 28\)
\(x^2 - 3x + 2 = 0\)
Step 3: Factorize
\((x-1)(x-2) = 0\)
x = 1 or x = 2
4. The sum of the reciprocals of Rehman's ages, (in years) 3 years ago and 5 years from now is \(\frac{1}{3}\). Find his present age.
Step 1: Let present age = x years
Step 2: Age 3 years ago = (x-3) years
Step 3: Age 5 years from now = (x+5) years
Step 4: According to the problem:
\(\frac{1}{x-3} + \frac{1}{x+5} = \frac{1}{3}\)
Step 5: Combine LHS
\(\frac{(x+5)+(x-3)}{(x-3)(x+5)} = \frac{1}{3}\)
\(\frac{2x+2}{x^2+2x-15} = \frac{1}{3}\)
Step 6: Cross multiply
\(3(2x+2) = x^2+2x-15\)
\(6x+6 = x^2+2x-15\)
\(x^2 - 4x - 21 = 0\)
Step 7: Factorize
\((x-7)(x+3) = 0\)
Step 8: Since age cannot be negative, x = 7
Conclusion: Rehman's present age is 7 years.
5. In a class test, the sum of Shefali's marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
Step 1: Let marks in Mathematics = x
Step 2: Then marks in English = 30 - x
Step 3: According to the problem:
\((x+2)(30-x-3) = 210\)
\((x+2)(27-x) = 210\)
Step 4: Expand
\(27x - x^2 + 54 - 2x = 210\)
\(-x^2 + 25x + 54 - 210 = 0\)
\(-x^2 + 25x - 156 = 0\)
\(x^2 - 25x + 156 = 0\)
Step 5: Factorize
\((x-12)(x-13) = 0\)
Step 6: x = 12 or x = 13
Conclusion: If Mathematics = 12, English = 18; If Mathematics = 13, English = 17
6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Step 1: Let shorter side = x metres
Step 2: Then longer side = (x+30) metres
Step 3: Diagonal = (x+60) metres
Step 4: Using Pythagoras theorem:
\(x^2 + (x+30)^2 = (x+60)^2\)
Step 5: Expand
\(x^2 + x^2 + 60x + 900 = x^2 + 120x + 3600\)
\(2x^2 + 60x + 900 = x^2 + 120x + 3600\)
\(x^2 - 60x - 2700 = 0\)
Step 6: Use quadratic formula: a = 1, b = -60, c = -2700
D = (-60)² - 4×1×(-2700) = 3600 + 10800 = 14400
\(\sqrt{D} = 120\)
\(x = \frac{60 \pm 120}{2}\)
\(x = \frac{60+120}{2} = 90\) or \(x = \frac{60-120}{2} = -30\)
Step 7: Since length cannot be negative, x = 90
Conclusion: Shorter side = 90 m, Longer side = 120 m
7. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Step 1: Let larger number = x, smaller number = y
Step 2: According to the problem:
\(x^2 - y^2 = 180\) ...(1)
\(y^2 = 8x\) ...(2)
Step 3: Substitute (2) in (1)
\(x^2 - 8x = 180\)
\(x^2 - 8x - 180 = 0\)
Step 4: Factorize
\((x-18)(x+10) = 0\)
x = 18 or x = -10
Step 5: If x = 18, then y² = 8×18 = 144, so y = ±12
Step 6: If x = -10, then y² = 8×(-10) = -80 (not possible)
Conclusion: The numbers are 18 and 12, or 18 and -12
8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Step 1: Let actual speed = x km/h
Step 2: Then actual time = \(\frac{360}{x}\) hours
Step 3: Increased speed = (x+5) km/h
Step 4: Time at increased speed = \(\frac{360}{x+5}\) hours
Step 5: According to the problem:
\(\frac{360}{x} - \frac{360}{x+5} = 1\)
Step 6: Multiply both sides by x(x+5)
\(360(x+5) - 360x = x(x+5)\)
\(1800 = x^2 + 5x\)
\(x^2 + 5x - 1800 = 0\)
Step 7: Factorize
\((x+45)(x-40) = 0\)
Step 8: Since speed cannot be negative, x = 40
Conclusion: The speed of the train is 40 km/h.
9. Two water taps together can fill a tank in \(9\frac{3}{8}\) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Step 1: Let time taken by smaller tap = x hours
Step 2: Then time taken by larger tap = (x-10) hours
Step 3: Part filled by smaller tap in 1 hour = \(\frac{1}{x}\)
Step 4: Part filled by larger tap in 1 hour = \(\frac{1}{x-10}\)
Step 5: Together they fill \(\frac{1}{x} + \frac{1}{x-10}\) in 1 hour
Step 6: Time taken together = \(\frac{75}{8}\) hours
\(\frac{1}{x} + \frac{1}{x-10} = \frac{8}{75}\)
Step 7: Combine LHS
\(\frac{x-10+x}{x(x-10)} = \frac{8}{75}\)
\(\frac{2x-10}{x^2-10x} = \frac{8}{75}\)
Step 8: Cross multiply
\(75(2x-10) = 8(x^2-10x)\)
\(150x - 750 = 8x^2 - 80x\)
\(8x^2 - 230x + 750 = 0\)
\(4x^2 - 115x + 375 = 0\)
Step 9: Use quadratic formula: a = 4, b = -115, c = 375
D = (-115)² - 4×4×375 = 13225 - 6000 = 7225
\(\sqrt{D} = 85\)
\(x = \frac{115 \pm 85}{8}\)
\(x = \frac{115+85}{8} = 25\) or \(x = \frac{115-85}{8} = 3.75\)
Step 10: If x = 3.75, then x-10 = -6.25 (not possible)
Conclusion: Smaller tap takes 25 hours, larger tap takes 15 hours.
10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.
Step 1: Let speed of passenger train = x km/h
Step 2: Then speed of express train = (x+11) km/h
Step 3: Time taken by passenger train = \(\frac{132}{x}\) hours
Step 4: Time taken by express train = \(\frac{132}{x+11}\) hours
Step 5: According to the problem:
\(\frac{132}{x} - \frac{132}{x+11} = 1\)
Step 6: Multiply both sides by x(x+11)
\(132(x+11) - 132x = x(x+11)\)
\(1452 = x^2 + 11x\)
\(x^2 + 11x - 1452 = 0\)
Step 7: Factorize
\((x+44)(x-33) = 0\)
Step 8: Since speed cannot be negative, x = 33
Conclusion: Passenger train speed = 33 km/h, Express train speed = 44 km/h.
11. Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.
Step 1: Let sides of squares be x and y metres, with x > y
Step 2: According to the problem:
\(x^2 + y^2 = 468\) ...(1)
\(4x - 4y = 24\) ⇒ \(x - y = 6\) ...(2)
Step 3: From (2), x = y + 6
Step 4: Substitute in (1)
\((y+6)^2 + y^2 = 468\)
\(y^2 + 12y + 36 + y^2 = 468\)
\(2y^2 + 12y + 36 - 468 = 0\)
\(2y^2 + 12y - 432 = 0\)
\(y^2 + 6y - 216 = 0\)
Step 5: Factorize
\((y+18)(y-12) = 0\)
Step 6: Since length cannot be negative, y = 12
Step 7: Then x = 12 + 6 = 18
Conclusion: Sides of the squares are 18 m and 12 m.
Exercise 4.4
1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) \(2x^2 - 3x + 5 = 0\)
Step 1: Identify coefficients: a = 2, b = -3, c = 5
Step 2: Calculate discriminant: D = b² - 4ac = (-3)² - 4×2×5 = 9 - 40 = -31
Step 3: Since D < 0, the equation has no real roots.
Conclusion: The equation has no real roots.
(ii) \(3x^2 - 4\sqrt{3}x + 4 = 0\)
Step 1: Identify coefficients: a = 3, b = -4√3, c = 4
Step 2: Calculate discriminant: D = b² - 4ac = (-4√3)² - 4×3×4 = 48 - 48 = 0
Step 3: Since D = 0, the equation has equal real roots.
Step 4: The root is: \(x = \frac{-b}{2a} = \frac{4\sqrt{3}}{6} = \frac{2\sqrt{3}}{3}\)
Conclusion: The equation has equal roots: \(\frac{2\sqrt{3}}{3}\).
(iii) \(2x^2 - 6x + 3 = 0\)
Step 1: Identify coefficients: a = 2, b = -6, c = 3
Step 2: Calculate discriminant: D = b² - 4ac = (-6)² - 4×2×3 = 36 - 24 = 12
Step 3: Since D > 0, the equation has two distinct real roots.
Step 4: The roots are: \(x = \frac{-b \pm \sqrt{D}}{2a} = \frac{6 \pm \sqrt{12}}{4} = \frac{6 \pm 2\sqrt{3}}{4} = \frac{3 \pm \sqrt{3}}{2}\)
Conclusion: The roots are \(\frac{3 + \sqrt{3}}{2}\) and \(\frac{3 - \sqrt{3}}{2}\).
2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) \(2x^2 + kx + 3 = 0\)
Step 1: For equal roots, discriminant D = 0
Step 2: D = b² - 4ac = k² - 4×2×3 = k² - 24 = 0
Step 3: k² = 24 ⇒ k = ±2√6
(ii) \(kx(x-2) + 6 = 0\)
Step 1: Rewrite in standard form: kx² - 2kx + 6 = 0
Step 2: For equal roots, discriminant D = 0
Step 3: D = b² - 4ac = (-2k)² - 4×k×6 = 4k² - 24k = 0
Step 4: 4k(k - 6) = 0 ⇒ k = 0 or k = 6
Step 5: If k = 0, the equation becomes 6 = 0, which is not quadratic
Conclusion: k = 6
3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.
Step 1: Let breadth = x m, then length = 2x m
Step 2: Area = length × breadth = 2x × x = 2x² = 800
Step 3: x² = 400 ⇒ x = 20 (since length cannot be negative)
Step 4: Length = 2×20 = 40 m
Conclusion: Yes, it is possible. Length = 40 m, Breadth = 20 m.
4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Step 1: Let present ages be x and (20-x) years
Step 2: Four years ago, their ages were (x-4) and (16-x) years
Step 3: According to the problem: (x-4)(16-x) = 48
Step 4: Expand: 16x - x² - 64 + 4x = 48
-x² + 20x - 64 - 48 = 0
-x² + 20x - 112 = 0
x² - 20x + 112 = 0
Step 5: Discriminant D = (-20)² - 4×1×112 = 400 - 448 = -48
Step 6: Since D < 0, the equation has no real roots
Conclusion: The situation is not possible.
5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth.
Step 1: Let length = l m, breadth = b m
Step 2: Perimeter = 2(l+b) = 80 ⇒ l+b = 40 ⇒ b = 40-l
Step 3: Area = l×b = l(40-l) = 400
40l - l² = 400
l² - 40l + 400 = 0
Step 4: Discriminant D = (-40)² - 4×1×400 = 1600 - 1600 = 0
Step 5: Since D = 0, the equation has equal roots
l = \frac{40}{2} = 20
Step 6: b = 40 - 20 = 20
Conclusion: Yes, it is possible. The park is a square with length = breadth = 20 m.