CBSE Class 10 Mathematics Quadratic Equations Assignment Set 21

Exercise I: Check and Represent Quadratic Equations

Question 1. Check whether the following is a quadratic equation or not: \( (x + 1)^2 = 2(x - 3) \).
Answer:
\( (x + 1)^2 = x^2 + 2x + 1 \) and \( 2(x - 3) = 2x - 6 \).
\( x^2 + 2x + 1 = 2x - 6 \Rightarrow x^2 + 7 = 0 \), which is of the form \( ax^2 + bx + c = 0 \) with \( a = 1 \neq 0 \).
Hence it is a quadratic equation.
In simple words: Simplify both sides; if the highest power left is exactly 2 it is quadratic.

Teacher's Note:
1. Move every term to one side and simplify; the equation is quadratic only if the highest power of x that remains is 2 and its coefficient is not zero.
2. Expand the brackets carefully before deciding.

Question 2. Check whether the following is a quadratic equation or not: \( x^2 - 2x = (-2)(3 - x) \).
Answer:
\( (-2)(3 - x) = -6 + 2x \), so \( x^2 - 2x = -6 + 2x \Rightarrow x^2 - 4x + 6 = 0 \).
This is of the form \( ax^2 + bx + c = 0 \) with \( a = 1 \neq 0 \).
Hence it is a quadratic equation.
In simple words: Simplify both sides; if the highest power left is exactly 2 it is quadratic.

Teacher's Note:
1. Move every term to one side and simplify; the equation is quadratic only if the highest power of x that remains is 2 and its coefficient is not zero.
2. Expand the brackets carefully before deciding.

Question 3. Check whether the following is a quadratic equation or not: \( (x - 2)(x + 1) = (x - 1)(x + 3) \).
Answer:
L.H.S. \( = x^2 - x - 2 \) and R.H.S. \( = x^2 + 2x - 3 \).
\( x^2 - x - 2 = x^2 + 2x - 3 \Rightarrow -3x + 1 = 0 \), in which the \( x^2 \) terms cancel.
Hence it is not a quadratic equation (it is linear).
In simple words: Simplify both sides; if the highest power left is exactly 2 it is quadratic.

Teacher's Note:
1. Move every term to one side and simplify; the equation is quadratic only if the highest power of x that remains is 2 and its coefficient is not zero.
2. Expand the brackets carefully before deciding.

Question 4. Check whether the following is a quadratic equation or not: \( (x - 3)(2x + 1) = x(x + 5) \).
Answer:
L.H.S. \( = 2x^2 - 5x - 3 \) and R.H.S. \( = x^2 + 5x \).
\( 2x^2 - 5x - 3 = x^2 + 5x \Rightarrow x^2 - 10x - 3 = 0 \), which is of the form \( ax^2 + bx + c = 0 \) with \( a = 1 \neq 0 \).
Hence it is a quadratic equation.
In simple words: Simplify both sides; if the highest power left is exactly 2 it is quadratic.

Teacher's Note:
1. Move every term to one side and simplify; the equation is quadratic only if the highest power of x that remains is 2 and its coefficient is not zero.
2. Expand the brackets carefully before deciding.

Question 5. Check whether the following is a quadratic equation or not: \( (2x - 1)(x - 3) = (x + 5)(x - 1) \).
Answer:
L.H.S. \( = 2x^2 - 7x + 3 \) and R.H.S. \( = x^2 + 4x - 5 \).
\( 2x^2 - 7x + 3 = x^2 + 4x - 5 \Rightarrow x^2 - 11x + 8 = 0 \), which is of the form \( ax^2 + bx + c = 0 \).
Hence it is a quadratic equation.
In simple words: Simplify both sides; if the highest power left is exactly 2 it is quadratic.

Teacher's Note:
1. Move every term to one side and simplify; the equation is quadratic only if the highest power of x that remains is 2 and its coefficient is not zero.
2. Expand the brackets carefully before deciding.

Question 6. Check whether the following is a quadratic equation or not: \( x^2 + 3x + 1 = (x - 2)^2 \).
Answer:
\( (x - 2)^2 = x^2 - 4x + 4 \).
\( x^2 + 3x + 1 = x^2 - 4x + 4 \Rightarrow 7x - 3 = 0 \), in which the \( x^2 \) terms cancel.
Hence it is not a quadratic equation (it is linear).
In simple words: Simplify both sides; if the highest power left is exactly 2 it is quadratic.

Teacher's Note:
1. Move every term to one side and simplify; the equation is quadratic only if the highest power of x that remains is 2 and its coefficient is not zero.
2. Expand the brackets carefully before deciding.

Question 7. Check whether the following is a quadratic equation or not: \( (x + 2)^3 = 2x(x^2 - 1) \).
Answer:
\( (x + 2)^3 = x^3 + 6x^2 + 12x + 8 \) and \( 2x(x^2 - 1) = 2x^3 - 2x \).
\( x^3 + 6x^2 + 12x + 8 = 2x^3 - 2x \Rightarrow -x^3 + 6x^2 + 14x + 8 = 0 \), a polynomial of degree 3.
Hence it is not a quadratic equation (it is cubic).
In simple words: Simplify both sides; if the highest power left is exactly 2 it is quadratic.

Teacher's Note:
1. Move every term to one side and simplify; the equation is quadratic only if the highest power of x that remains is 2 and its coefficient is not zero.
2. Expand the brackets carefully before deciding.

Question 8. Check whether the following is a quadratic equation or not: \( x^3 - 4x^2 - x + 1 = (x - 2)^3 \).
Answer:
\( (x - 2)^3 = x^3 - 6x^2 + 12x - 8 \).
\( x^3 - 4x^2 - x + 1 = x^3 - 6x^2 + 12x - 8 \Rightarrow 2x^2 - 13x + 9 = 0 \); the \( x^3 \) terms cancel, leaving degree 2.
Hence it is a quadratic equation.
In simple words: Simplify both sides; if the highest power left is exactly 2 it is quadratic.

Teacher's Note:
1. Move every term to one side and simplify; the equation is quadratic only if the highest power of x that remains is 2 and its coefficient is not zero.
2. Expand the brackets carefully before deciding.

Question 9. Represent the following situation in the form of a quadratic equation: 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.
Answer:
Let the breadth of the plot be x m. Then the length \( = (2x + 1) \) m.
Area \( = \) length \( \times \) breadth: \( (2x + 1) \times x = 528 \Rightarrow 2x^2 + x = 528 \).
Hence the quadratic equation is \( 2x^2 + x - 528 = 0 \).
In simple words: Area = length times breadth, with length written as 2x + 1.

Teacher's Note:
1. Choose a variable for the unknown, express the other quantities in it and use the given condition to form one equation.
2. Bring the equation to the standard form \( ax^2 + bx + c = 0 \).

Question 10. Represent the following situation in the form of a quadratic equation: The product of two consecutive positive integers is 306. We need to find the integers.
Answer:
Let the first integer be x. Then the next consecutive positive integer is \( x + 1 \).
Product \( = 306 \Rightarrow x(x + 1) = 306 \Rightarrow x^2 + x = 306 \).
Hence the quadratic equation is \( x^2 + x - 306 = 0 \).
In simple words: Call the integers x and x + 1 and set their product equal to 306.

Teacher's Note:
1. Choose a variable for the unknown, express the other quantities in it and use the given condition to form one equation.
2. Bring the equation to the standard form \( ax^2 + bx + c = 0 \).

Question 11. Represent the following situation in the form of a quadratic equation: 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.
Answer:
Let the present age of Rohan be x years. Then his mother's age is \( (x + 26) \) years.
After 3 years, Rohan's age \( = x + 3 \) and his mother's age \( = x + 29 \).
Product of ages after 3 years is 360: \( (x + 3)(x + 29) = 360 \Rightarrow x^2 + 32x + 87 = 360 \).
Hence the quadratic equation is \( x^2 + 32x - 273 = 0 \).
In simple words: Add 3 years to both ages before multiplying.

Teacher's Note:
1. Choose a variable for the unknown, express the other quantities in it and use the given condition to form one equation.
2. Bring the equation to the standard form \( ax^2 + bx + c = 0 \).

Question 12. Represent the following situation in the form of a quadratic equation: 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.
Answer:
Let the original speed of the train be x km/h. Time taken \( = \frac{480}{x} \) hours.
With speed \( (x - 8) \) km/h the time taken is \( \left(\frac{480}{x} + 3\right) \) hours. Using distance \( = \) speed \( \times \) time: \( (x - 8)\left(\frac{480}{x} + 3\right) = 480 \).
\( 480 + 3x - \frac{3840}{x} - 24 = 480 \Rightarrow 3x - \frac{3840}{x} = 24 \Rightarrow 3x^2 - 24x - 3840 = 0 \).
Hence the quadratic equation is \( x^2 - 8x - 1280 = 0 \).
In simple words: Slower speed means more time; the distance stays 480 km.

Teacher's Note:
1. Choose a variable for the unknown, express the other quantities in it and use the given condition to form one equation.
2. Bring the equation to the standard form \( ax^2 + bx + c = 0 \).
3. Speed times time always gives the distance.

Exercise II: Factorisation Method and Word Problems

Question 13. Find the roots of the following quadratic equation by factorisation: \( x^2 - 3x - 10 = 0 \).
Answer:
Here \( ac = 1\times(-10) = -10 \). We need two numbers whose product is \( -10 \) and whose sum is \( -3 \): they are \( -5 \) and \( 2 \).
\( x^{2} - 3 x - 10 = 0 \Rightarrow x^2 - 5x + 2x - 10 = 0 \)
\( x(x - 5) + 2(x - 5) = 0 \Rightarrow (x + 2)(x - 5) = 0 \)
Equating each factor to zero: \( x = -2\ \text{or}\ x = 5 \).
In simple words: Break the middle term into two parts that multiply to give the product of the first and last terms.

Teacher's Note:
1. Split the middle term so that the two parts multiply to ac and add to b, then take out common factors.
2. Check each root by substituting it in the original equation.

Question 14. Find the roots of the following quadratic equation by factorisation: \( 2x^2 + x - 6 = 0 \).
Answer:
Here \( ac = 2\times(-6) = -12 \). We need two numbers whose product is \( -12 \) and whose sum is \( 1 \): they are \( -3 \) and \( 4 \).
\( 2 x^{2} + x - 6 = 0 \Rightarrow 2x^2 - 3x + 4x - 6 = 0 \)
\( x(2 x - 3) + 2(2 x - 3) = 0 \Rightarrow (x + 2)(2 x - 3) = 0 \)
Equating each factor to zero: \( x = -2\ \text{or}\ x = \frac{3}{2} \).
In simple words: Break the middle term into two parts that multiply to give the product of the first and last terms.

Teacher's Note:
1. Split the middle term so that the two parts multiply to ac and add to b, then take out common factors.
2. Check each root by substituting it in the original equation.

Question 15. Find the roots of the following quadratic equation by factorisation: \( \sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 \).
Answer:
Here \( ac = \sqrt{2} \times 5\sqrt{2} = 10 \); the middle term \( 7x = 5x + 2x \).
\( \sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} = 0 \Rightarrow x(\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5) = 0 \Rightarrow (\sqrt{2}x + 5)(x + \sqrt{2}) = 0 \).
\( \sqrt{2}x + 5 = 0 \Rightarrow x = -\frac{5}{\sqrt{2}} = -\frac{5\sqrt{2}}{2} \) and \( x + \sqrt{2} = 0 \Rightarrow x = -\sqrt{2} \).
Hence the roots are \( -\frac{5\sqrt{2}}{2} \) and \( -\sqrt{2} \).
In simple words: Multiply the first and last coefficients to get 10 and split 7 as 5 + 2.

Teacher's Note:
1. Split the middle term so that the two parts multiply to ac and add to b, then take out common factors.
2. Check each root by substituting it in the original equation.

Question 16. Find the roots of the following quadratic equation by factorisation: \( 2x^2 - x + \frac{1}{8} = 0 \).
Answer:
Multiplying by 8: \( 16x^2 - 8x + 1 = 0 \).
\( 16x^2 - 4x - 4x + 1 = 0 \Rightarrow 4x(4x - 1) - 1(4x - 1) = 0 \Rightarrow (4x - 1)^2 = 0 \).
Hence the roots are \( x = \frac{1}{4} \) and \( x = \frac{1}{4} \) (a repeated root).
In simple words: Clear the fraction first; the quadratic is a perfect square.

Teacher's Note:
1. Split the middle term so that the two parts multiply to ac and add to b, then take out common factors.
2. Check each root by substituting it in the original equation.

Question 17. Find the roots of the following quadratic equation by factorisation: \( 100x^2 - 20x + 1 = 0 \).
Answer:
Here \( ac = 100 \) and \( -20x = -10x - 10x \).
\( 100x^2 - 10x - 10x + 1 = 0 \Rightarrow 10x(10x - 1) - 1(10x - 1) = 0 \Rightarrow (10x - 1)^2 = 0 \).
Hence the roots are \( x = \frac{1}{10} \) and \( x = \frac{1}{10} \) (a repeated root).
In simple words: The left side is the perfect square \( (10x - 1)^2 \).

Teacher's Note:
1. Split the middle term so that the two parts multiply to ac and add to b, then take out common factors.
2. Check each root by substituting it in the original equation.

Question 18. Represent the following situation mathematically and solve it: John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had initially.
Answer:
Let John have x marbles initially. Then Jivanti has \( (45 - x) \) marbles.
After losing 5 marbles each: John has \( (x - 5) \) and Jivanti has \( (40 - x) \). Product \( = 124 \Rightarrow (x - 5)(40 - x) = 124 \).
\( 40x - x^2 - 200 + 5x = 124 \Rightarrow x^2 - 45x + 324 = 0 \Rightarrow x^2 - 36x - 9x + 324 = 0 \Rightarrow (x - 36)(x - 9) = 0 \).
So \( x = 36 \) or \( x = 9 \). If John had 36 marbles, Jivanti had 9; if John had 9, then Jivanti had 36.
Hence they had 36 and 9 marbles (in either order). Check: \( 31 \times 4 = 124 \).
In simple words: Initial marbles are x and 45 - x; subtract 5 from each before multiplying.

Teacher's Note:
1. Define the unknown, form the quadratic equation, factorise it and reject any root that does not suit the situation (such as a negative age or length).
2. Verify the answer in the original statement.

Question 19. Represent the following situation mathematically and solve it: A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs. 750. We would like to find out the number of toys produced on that day.
Answer:
Let the number of toys produced in a day be x. Then the cost of production of each toy is Rs. \( (55 - x) \).
Total cost \( = \) number of toys \( \times \) cost of each toy: \( x(55 - x) = 750 \Rightarrow x^2 - 55x + 750 = 0 \).
\( x^2 - 25x - 30x + 750 = 0 \Rightarrow (x - 25)(x - 30) = 0 \Rightarrow x = 25 \) or \( x = 30 \).
Hence the number of toys produced in a day is 25 or 30. Check: \( 25 \times 30 = 750 \).
In simple words: Total cost is toys times cost per toy, which gives x(55 - x) = 750.

Teacher's Note:
1. Define the unknown, form the quadratic equation, factorise it and reject any root that does not suit the situation (such as a negative age or length).
2. Verify the answer in the original statement.

Question 20. Find two numbers whose sum is 27 and product is 182.
Answer:
Let one number be x. Then the other number is \( 27 - x \). Product \( = 182 \Rightarrow x(27 - x) = 182 \Rightarrow x^2 - 27x + 182 = 0 \).
\( x^2 - 13x - 14x + 182 = 0 \Rightarrow (x - 13)(x - 14) = 0 \Rightarrow x = 13 \) or \( x = 14 \).
Hence the two numbers are 13 and 14.
In simple words: Write the second number as 27 minus the first and use the product.

Teacher's Note:
1. If one number is x, the other is (sum − x).
2. Check: 13 + 14 = 27 and 13 × 14 = 182.

Question 21. Find two consecutive positive integers, sum of whose squares is 365.
Answer:
Let the two consecutive positive integers be x and \( x + 1 \). Then \( x^2 + (x + 1)^2 = 365 \Rightarrow 2x^2 + 2x - 364 = 0 \Rightarrow x^2 + x - 182 = 0 \).
\( x^2 + 14x - 13x - 182 = 0 \Rightarrow (x + 14)(x - 13) = 0 \Rightarrow x = -14 \) or \( x = 13 \).
\( -14 \) is rejected because the integers are positive. So \( x = 13 \) and \( x + 1 = 14 \).
Hence the integers are 13 and 14. Check: 169 + 196 = 365.
In simple words: Take x and x + 1; reject the negative root.

Teacher's Note:
1. Positive integers rule out the root −14.
2. Check: 13² + 14² = 365.

Question 22. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Answer:
Let the base be x cm. Then the altitude is \( (x - 7) \) cm. By the Pythagoras theorem, \( x^2 + (x - 7)^2 = 13^2 \).
\( 2x^2 - 14x + 49 = 169 \Rightarrow x^2 - 7x - 60 = 0 \Rightarrow x^2 - 12x + 5x - 60 = 0 \Rightarrow (x - 12)(x + 5) = 0 \).
So \( x = 12 \) (since \( x = -5 \) is rejected, a length cannot be negative).
Hence the base is 12 cm and the altitude is 5 cm.
In simple words: Use the Pythagoras theorem with sides x, x - 7 and 13.

Teacher's Note:
1. Reject the negative root because a length cannot be negative.
2. Check: 12² + 5² = 144 + 25 = 169 = 13².

Question 23. 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 Rs. 90, find the number of articles produced and the cost of each article.
Answer:
Let the number of articles produced in a day be x. Then the cost of each article is Rs. \( (2x + 3) \).
Total cost \( = x(2x + 3) = 90 \Rightarrow 2x^2 + 3x - 90 = 0 \Rightarrow 2x^2 + 15x - 12x - 90 = 0 \Rightarrow (2x + 15)(x - 6) = 0 \).
\( x = -\frac{15}{2} \) or \( x = 6 \). The number of articles cannot be negative, so \( x = 6 \).
Cost of each article \( = 2 \times 6 + 3 = \) Rs. 15.
Hence 6 articles were produced and each cost Rs. 15.
In simple words: Cost per article is 2x + 3; multiply by x to get 90.

Teacher's Note:
1. Define the unknown, form the quadratic equation, factorise it and reject any root that does not suit the situation (such as a negative age or length).
2. Verify the answer in the original statement.