Chapter 4: Quadratic Equations
Exercise 4.1
Q1
Check whether the following are quadratic equations:
(i) \( (x + 1)^2 = 2(x - 3) \)
(ii) \( x^2 - 2x = (-2)(3 - x) \)
(iii) \( (x - 2)(x + 1) = (x - 1)(x + 3) \)
(iv) \( (x - 3)(2x + 1) = x(x + 5) \)
(v) \( (2x - 1)(x - 3) = (x + 5)(x - 1) \)
(vi) \( x^2 + 3x + 1 = (x - 2)^2 \)
(vii) \( (x + 2)^3 = 2x(x^2 - 1) \)
(viii) \( x^3 - 4x^2 - x + 1 = (x - 2)^3 \)
(i) \( (x + 1)^2 = 2(x - 3) \)
(ii) \( x^2 - 2x = (-2)(3 - x) \)
(iii) \( (x - 2)(x + 1) = (x - 1)(x + 3) \)
(iv) \( (x - 3)(2x + 1) = x(x + 5) \)
(v) \( (2x - 1)(x - 3) = (x + 5)(x - 1) \)
(vi) \( x^2 + 3x + 1 = (x - 2)^2 \)
(vii) \( (x + 2)^3 = 2x(x^2 - 1) \)
(viii) \( x^3 - 4x^2 - x + 1 = (x - 2)^3 \)
Solution:
(i) \( (x + 1)^2 = 2(x - 3) \)
\[ x^2 + 2x + 1 = 2x - 6 \]
\[ x^2 + 7 = 0 \]
It is of the form \( ax^2 + bx + c = 0 \). Yes, it is a quadratic equation.
(ii) \( x^2 - 2x = (-2)(3 - x) \)
\[ x^2 - 2x = -6 + 2x \]
\[ x^2 - 4x + 6 = 0 \]
It is of the form \( ax^2 + bx + c = 0 \). Yes, it is a quadratic equation.
(iii) \( (x - 2)(x + 1) = (x - 1)(x + 3) \)
\[ x^2 - x - 2 = x^2 + 2x - 3 \]
\[ -3x + 1 = 0 \]
It is a linear equation. No, it is not a quadratic equation.
(iv) \( (x - 3)(2x + 1) = x(x + 5) \)
\[ 2x^2 - 5x - 3 = x^2 + 5x \]
\[ x^2 - 10x - 3 = 0 \]
Yes, it is a quadratic equation.
(v) \( (2x - 1)(x - 3) = (x + 5)(x - 1) \)
\[ 2x^2 - 7x + 3 = x^2 + 4x - 5 \]
\[ x^2 - 11x + 8 = 0 \]
Yes, it is a quadratic equation.
(vi) \( x^2 + 3x + 1 = (x - 2)^2 \)
\[ x^2 + 3x + 1 = x^2 - 4x + 4 \]
\[ 7x - 3 = 0 \]
It is a linear equation. No, it is not a quadratic equation.
(vii) \( (x + 2)^3 = 2x(x^2 - 1) \)
\[ x^3 + 6x^2 + 12x + 8 = 2x^3 - 2x \]
\[ -x^3 + 6x^2 + 14x + 8 = 0 \]
It is a cubic equation. No, it is not a quadratic equation.
(viii) \( x^3 - 4x^2 - x + 1 = (x - 2)^3 \)
\[ x^3 - 4x^2 - x + 1 = x^3 - 6x^2 + 12x - 8 \]
\[ 2x^2 - 13x + 9 = 0 \]
Yes, it is a quadratic equation.
Q2
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.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
(iii) Rohan'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 Rohan's present age.
(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.
(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.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
(iii) Rohan'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 Rohan's present age.
(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.
Solution:
(i) Let the breadth be \( x \) m. Then length is \( 2x + 1 \) m.
Area = Length × Breadth = \( (2x + 1)x = 528 \)
\[ 2x^2 + x - 528 = 0 \]
(ii) Let the consecutive integers be \( x \) and \( x + 1 \).
Product = \( x(x + 1) = 306 \)
\[ x^2 + x - 306 = 0 \]
(iii) Let Rohan's age be \( x \). Mother's age is \( x + 26 \).
3 years from now: Rohan = \( x + 3 \), Mother = \( x + 26 + 3 = x + 29 \).
Product: \( (x + 3)(x + 29) = 360 \)
\[ x^2 + 29x + 3x + 87 = 360 \]
\[ x^2 + 32x - 273 = 0 \]
(iv) Let the speed of the train be \( x \) km/h.
Time taken = \( \frac{480}{x} \). New speed = \( x - 8 \). New time = \( \frac{480}{x - 8} \).
Difference in time is 3 hours: \( \frac{480}{x - 8} - \frac{480}{x} = 3 \)
\[ 480 \left( \frac{x - (x - 8)}{x(x - 8)} \right) = 3 \]
\[ 480 \times 8 = 3x(x - 8) \]
\[ 1280 = x^2 - 8x \]
\[ x^2 - 8x - 1280 = 0 \]
Exercise 4.2
Q1
Find the roots of the following quadratic equations by factorisation:
(i) \( x^2 - 3x - 10 = 0 \)
(ii) \( 2x^2 + x - 6 = 0 \)
(iii) \( \sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 \)
(iv) \( 2x^2 - x + \frac{1}{8} = 0 \)
(v) \( 100x^2 - 20x + 1 = 0 \)
(i) \( x^2 - 3x - 10 = 0 \)
(ii) \( 2x^2 + x - 6 = 0 \)
(iii) \( \sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 \)
(iv) \( 2x^2 - x + \frac{1}{8} = 0 \)
(v) \( 100x^2 - 20x + 1 = 0 \)
Solution:
(i) \( x^2 - 3x - 10 = 0 \)
\[ x^2 - 5x + 2x - 10 = 0 \]
\[ x(x - 5) + 2(x - 5) = 0 \]
\[ (x - 5)(x + 2) = 0 \Rightarrow x = 5, -2 \]
(ii) \( 2x^2 + x - 6 = 0 \)
\[ 2x^2 + 4x - 3x - 6 = 0 \]
\[ 2x(x + 2) - 3(x + 2) = 0 \]
\[ (2x - 3)(x + 2) = 0 \Rightarrow x = \frac{3}{2}, -2 \]
(iii) \( \sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 \)
\[ \sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} = 0 \]
\[ x(\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5) = 0 \]
\[ (\sqrt{2}x + 5)(x + \sqrt{2}) = 0 \Rightarrow x = -\frac{5}{\sqrt{2}}, -\sqrt{2} \]
(iv) \( 2x^2 - x + \frac{1}{8} = 0 \)
Multiply by 8: \( 16x^2 - 8x + 1 = 0 \)
\[ (4x - 1)^2 = 0 \Rightarrow x = \frac{1}{4}, \frac{1}{4} \]
(v) \( 100x^2 - 20x + 1 = 0 \)
\[ (10x - 1)^2 = 0 \Rightarrow x = \frac{1}{10}, \frac{1}{10} \]
Q2
Solve the problems given in Example 1.
(i) John and Jivanti have 45 marbles. Both lost 5 marbles each, product is 124.
(ii) Cottage industry toys production. Cost is 55 minus number of toys. Total cost 750.
(i) John and Jivanti have 45 marbles. Both lost 5 marbles each, product is 124.
(ii) Cottage industry toys production. Cost is 55 minus number of toys. Total cost 750.
Solution:
(i) Marbles Problem:
Equation derived: \( x^2 - 45x + 324 = 0 \)
\[ x^2 - 36x - 9x + 324 = 0 \]
\[ x(x - 36) - 9(x - 36) = 0 \Rightarrow (x - 36)(x - 9) = 0 \]
Answer: John and Jivanti had 36 and 9 marbles respectively.
(ii) Toys Problem:
Equation derived: \( x^2 - 55x + 750 = 0 \)
\[ x^2 - 30x - 25x + 750 = 0 \]
\[ x(x - 30) - 25(x - 30) = 0 \Rightarrow (x - 30)(x - 25) = 0 \]
Answer: Number of toys produced were 30 or 25.
Q3
Find two numbers whose sum is 27 and product is 182.
Solution:
Let the first number be \( x \). The second number is \( 27 - x \).
Product: \( x(27 - x) = 182 \)
\[ 27x - x^2 = 182 \Rightarrow x^2 - 27x + 182 = 0 \]
\[ x^2 - 13x - 14x + 182 = 0 \]
\[ (x - 13)(x - 14) = 0 \Rightarrow x = 13, 14 \]
Answer: The two numbers are 13 and 14.
Q4
Find two consecutive positive integers, sum of whose squares is 365.
Solution:
Let the integers be \( x \) and \( x + 1 \).
\[ x^2 + (x + 1)^2 = 365 \]
\[ 2x^2 + 2x + 1 = 365 \Rightarrow 2x^2 + 2x - 364 = 0 \]
Divide by 2: \( x^2 + x - 182 = 0 \)
\[ x^2 + 14x - 13x - 182 = 0 \]
\[ (x + 14)(x - 13) = 0 \Rightarrow x = 13, -14 \]
Since the integer is positive, \( x = 13 \).
Answer: The integers are 13 and 14.
Q5
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Solution:
Let the base be \( x \) cm. Altitude is \( x - 7 \) cm.
By Pythagoras theorem: \( x^2 + (x - 7)^2 = 13^2 \)
\[ x^2 + x^2 - 14x + 49 = 169 \]
\[ 2x^2 - 14x - 120 = 0 \Rightarrow x^2 - 7x - 60 = 0 \]
\[ x^2 - 12x + 5x - 60 = 0 \]
\[ (x - 12)(x + 5) = 0 \Rightarrow x = 12, -5 \]
Base cannot be negative, so \( x = 12 \).
Answer: Base = 12 cm, Altitude = 5 cm.
Q6
A cottage industry produces a certain number of pottery articles in a day. The cost of production of each article was 3 more than twice the number of articles. Total cost was ₹ 90. Find the number of articles and cost of each.
Solution:
Let the number of articles be \( x \). Cost per article = \( 2x + 3 \).
Total cost: \( x(2x + 3) = 90 \)
\[ 2x^2 + 3x - 90 = 0 \]
\[ 2x^2 + 15x - 12x - 90 = 0 \]
\[ x(2x + 15) - 6(2x + 15) = 0 \Rightarrow (x - 6)(2x + 15) = 0 \]
Since \( x \) (number of articles) must be positive, \( x = 6 \).
Cost = \( 2(6) + 3 = 15 \).
Answer: Number of articles = 6, Cost per article = ₹ 15.
Exercise 4.3
Q1
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) \( 2x^2 - 3x + 5 = 0 \)
(ii) \( 3x^2 - 4\sqrt{3}x + 4 = 0 \)
(iii) \( 2x^2 - 6x + 3 = 0 \)
(i) \( 2x^2 - 3x + 5 = 0 \)
(ii) \( 3x^2 - 4\sqrt{3}x + 4 = 0 \)
(iii) \( 2x^2 - 6x + 3 = 0 \)
Solution:
We use the discriminant \( D = b^2 - 4ac \).
(i) \( 2x^2 - 3x + 5 = 0 \)
\[ D = (-3)^2 - 4(2)(5) = 9 - 40 = -31 < 0 \]
Nature: No real roots.
(ii) \( 3x^2 - 4\sqrt{3}x + 4 = 0 \)
\[ D = (-4\sqrt{3})^2 - 4(3)(4) = 48 - 48 = 0 \]
Nature: Two equal real roots.
Roots: \( x = \frac{-b}{2a} = \frac{4\sqrt{3}}{6} = \frac{2\sqrt{3}}{3} \).
(iii) \( 2x^2 - 6x + 3 = 0 \)
\[ D = (-6)^2 - 4(2)(3) = 36 - 24 = 12 > 0 \]
Nature: Two distinct real roots.
Roots: \( x = \frac{-(-6) \pm \sqrt{12}}{4} = \frac{6 \pm 2\sqrt{3}}{4} = \frac{3 \pm \sqrt{3}}{2} \).
Q2
Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) \( 2x^2 + kx + 3 = 0 \)
(ii) \( kx(x - 2) + 6 = 0 \)
(i) \( 2x^2 + kx + 3 = 0 \)
(ii) \( kx(x - 2) + 6 = 0 \)
Solution:
For equal roots, discriminant \( D = 0 \).
(i) \( 2x^2 + kx + 3 = 0 \)
\[ b^2 - 4ac = k^2 - 4(2)(3) = 0 \]
\[ k^2 - 24 = 0 \Rightarrow k = \pm\sqrt{24} = \pm 2\sqrt{6} \]
(ii) \( kx(x - 2) + 6 = 0 \Rightarrow kx^2 - 2kx + 6 = 0 \)
\[ (-2k)^2 - 4(k)(6) = 0 \]
\[ 4k^2 - 24k = 0 \]
\[ 4k(k - 6) = 0 \]
Since \( k \neq 0 \) (quadratic term must exist), \( k = 6 \).
Q3
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.
Solution:
Let breadth be \( x \) m. Length = \( 2x \) m.
Area = \( x(2x) = 800 \Rightarrow 2x^2 = 800 \Rightarrow x^2 = 400 \).
\[ x = 20 \quad (\text{since breadth is positive}) \]
Discriminant of \( 2x^2 - 800 = 0 \) is positive, so it is possible.
Answer: Breadth = 20 m, Length = 40 m.
Q4
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.
Solution:
Let the age of one friend be \( x \). The other is \( 20 - x \).
Four years ago: Ages were \( x - 4 \) and \( 16 - x \).
Product: \( (x - 4)(16 - x) = 48 \)
\[ 16x - x^2 - 64 + 4x = 48 \]
\[ -x^2 + 20x - 112 = 0 \Rightarrow x^2 - 20x + 112 = 0 \]
Discriminant \( D = (-20)^2 - 4(1)(112) = 400 - 448 = -48 \).
Since \( D < 0 \), no real roots exist.
Answer: No, the situation is not possible.
Q5
Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth.
Solution:
Perimeter \( 2(l + b) = 80 \Rightarrow l + b = 40 \Rightarrow b = 40 - l \).
Area \( l \times b = 400 \Rightarrow l(40 - l) = 400 \).
\[ 40l - l^2 = 400 \Rightarrow l^2 - 40l + 400 = 0 \]
Discriminant \( D = (-40)^2 - 4(1)(400) = 1600 - 1600 = 0 \).
Since \( D = 0 \), roots are real and equal. It is possible.
\[ l = \frac{-(-40)}{2} = 20 \text{ m} \]
Breadth \( b = 40 - 20 = 20 \text{ m} \).
Answer: Length = 20 m, Breadth = 20 m (It is a square).