CBSE Class 10 Mathematics Quadratic Equations VBQs Set G

Topics Covered

• 1. Quadratic Equations
• 2. Solution of a Quadratic Equation by Factorisation
• 3. Solution of a Quadratic Equation by Quadratic Formula
• 4. Nature of Roots

1. Quadratic Equations

The second degree polynomial equations is commonly known as quadratic equation, i.e., if \( p(x) \) is a quadratic polynomial, then \( p(x) = 0 \) is called a quadratic equation. The general form of quadratic equation in the variable \( x \) is \( ax^2 + bx + c = 0 \), where \( a, b, c \) are real numbers and \( a \neq 0 \).

For example, \( 2x^2 + x - 150 = 0 \), \( 3x^2 - 2x + 5 = 0 \), \( 4x - 3x^2 + 2 = 0 \) are quadratic equations.

Question. Check whether the following are quadratic equations: (a) \( x(x + 2) - 3 = x(x + 4) \)
Answer: Since \( x(x + 2) - 3 = x(x + 4) \Rightarrow x^2 + 2x - 3 = x^2 + 4x \Rightarrow 2x + 3 = 0 \). This is linear equation not a quadratic equation.

Question. Check whether the following are quadratic equations: (b) \( (x + 2)^3 = x^3 - 4x^2 + 2 \)
Answer: \( (x + 2)^3 = x^3 - 4x^2 + 2 \Rightarrow x^3 + 6x^2 + 12x + 8 = x^3 - 4x^2 + 2 \Rightarrow 10x^2 + 12x + 6 = 0 \Rightarrow 5x^2 + 6x + 3 = 0 \). This is a quadratic equation.

Question. Represent the following situation in the form a quadratic equation: Abdul and Sneha together have 30 oranges. Both of them ate 3 oranges each and the product of the number of oranges they have now is 120. We would like to find out how many oranges they had initially.
Answer: Let the number of number of left oranges with Abdul and Sneha be \( x \) and \( y \) respectively. Then, \( x + y = 30 \Rightarrow y = 30 - x \). The number of oranges left with both Abdul and Sneha are \( x - 3 \) and \( y - 3 \) respectively. The product of number of left oranges = 120 \( \Rightarrow (x - 3) (y - 3) = 120 \Rightarrow (x - 3) (30 - x - 3) = 120 \Rightarrow (x - 3) (27 - x) = 120 \Rightarrow 27x - x^2 - 81 + 3x = 120 \Rightarrow x^2 - 30x + 201 = 0 \).

Question. Represent the following situation in the form a quadratic equation: The area of a rectangular plot is 428 m\(^2\). The length of the plot (in metres) is two more than twice its breadth. We need to find the length and breadth of the plot.
Answer: Let the breadth of the plot be \( x \). Then the length of the plot = \( 2x + 2 \). Since, area of the plot = 428 m\(^2\) (Given) \( \therefore x(2x + 2) = 428 \Rightarrow 2x^2 + 2x - 428 = 0 \Rightarrow x^2 + x - 214 = 0 \).

Roots of a Quadratic Equation
A real number \( \alpha \) is called a root of the quadratic equation \( ax^2 + bx + c = 0 \), \( a \neq 0 \) if \( a\alpha^2 + b\alpha + c = 0 \). In other words, \( x = \alpha \) is a root or solution of the quadratic equation \( ax^2 + bx + c = 0 \) as it satisfies it.

Question. Determine whether 3 is a root of the equation \( \sqrt{x^2 - 4x + 3} + \sqrt{x^2 - 9} = \sqrt{4x^2 - 14x + 6} \)
Answer: L.H.S. = \( \sqrt{3^2 - 12 + 3} + \sqrt{9 - 9} = 0 + 0 = 0 \). R.H.S. = \( \sqrt{4(3)^2 - 14(3) + 6} = \sqrt{36 - 42 + 6} = 0 \). Since L.H.S. = R.H.S., 3 is the root of the given quadratic equation.

Question. If \( \frac{1}{2} \) is a root of the equation \( x^2 + kx - \frac{5}{4} = 0 \), then find the value of \( k \).
Answer: It is given that \( \frac{1}{2} \) is a root of quadratic equation. \( \therefore (\frac{1}{2})^2 + k(\frac{1}{2}) - \frac{5}{4} = 0 \Rightarrow \frac{1}{4} + \frac{k}{2} - \frac{5}{4} = 0 \Rightarrow \frac{1 + 2k - 5}{4} = 0 \Rightarrow 2k - 4 = 0 \Rightarrow k = 2 \).

Exercise 

Question. Which of the following is a quadratic equation?
(a) \( x^2 + 2x + 1 = (4 - x)^2 + 3 \)
(b) \( -2x^2 = (5 - x) (2x - \frac{2}{5}) \)
(c) \( (k + 1)x^2 + \frac{3}{2}x = 7 \) (where \( k = -1 \))
(d) \( x^3 - x^2 = (x - 1)^3 \)
Answer: (d) \( x^3 - x^2 = (x - 1)^3 \)

Question. Which of the following equations has 2 as a root?
(a) \( x^2 - 4x + 5 = 0 \)
(b) \( x^2 + 3x - 12 = 0 \)
(c) \( 2x^2 - 7x + 6 = 0 \)
(d) \( 3x^2 - 6x - 2 = 0 \)
Answer: (c) \( 2x^2 - 7x + 6 = 0 \)

Question. The degree of quadratic equation is
(a) 0
(b) 1
(c) 2
(d) 5
Answer: (c) 2

Question. Assertion (A): The equation \( x^2 + 3x + 1 = (x - 2)^2 \) is a quadratic equation.
Reason (R): Any equation of the form \( ax^2 + bx + c = 0 \) where \( a \neq 0 \), is a quadratic equation.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer: (d) Assertion (A) is false but reason (R) is true.

Question. Assertion (A): \( (2x - 1)^2 - 4x^2 + 5 = 0 \) is not a quadratic equation.
Reason (R): \( x = 0, 3 \) are the roots of the equation \( 2x^2 - 6x = 0 \).
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer: (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).

Question. If \( x = -\frac{1}{2} \) is a solution of the quadratic equation \( 3x^2 + 2kx - 3 = 0 \), find the value of \( k \).
Answer: \( k = -2\frac{1}{4} \)

Question. Find the value of \( k \) for which \( x = \sqrt{3} \) is a solution of the equation \( kx^2 + \sqrt{3}x - 4 = 0 \).
Answer: \( k = \frac{1}{3} \)

Question. If \( x = \frac{2}{3} \) and \( x = -3 \) are roots of the quadratic equations \( ax^2 + 7x + b = 0 \), find the values of \( a \) and \( b \).
Answer: \( a = 3, b = -6 \)

Question. Show that \( x = -2 \) is a solution of the equation \( 3x^2 + 13x + 14 = 0 \).
Answer: L.H.S. = \( 3(-2)^2 + 13(-2) + 14 = 12 - 26 + 14 = 0 \). Since L.H.S. = R.H.S., \( x = -2 \) is a solution.

Question. Represent the following situation in the form of a quadratic equation: John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they have now is 124. We would like to find out how many marbles they had to start with.
Answer: \( x^2 - 45x + 324 = 0 \)

Question. Represent the following situation in the form of a quadratic equation: 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 ₹ 750. We would like to find out the number of toys produced on that day.
Answer: \( x^2 - 55x + 750 = 0 \)

Raj and Ajay are very close friends. Both the families decide to go to Ranikhet by their own cars. Raj’s car travels at a speed of \( x \) km/h while Ajay’s car travels 5 km/h faster than Raj’s car. Raj took 4 hours more than Ajay to complete the journey of 400 km.

Question. What will be the distance covered by Ajay’s car in two hours?
(a) \( 2(x + 5) \) km
(b) \( (x - 5) \) km
(c) \( 2(x + 10) \) km
(d) \( (2x + 5) \) km
Answer: (a) \( 2(x + 5) \) km

Question. Which of the following quadratic equations describes the speed of Raj’s car?
(a) \( x^2 - 5x - 500 = 0 \)
(b) \( x^2 + 4x - 400 = 0 \)
(c) \( x^2 + 5x - 500 = 0 \)
(d) \( x^2 - 4x + 400 = 0 \)
Answer: (c) \( x^2 + 5x - 500 = 0 \)

Question. The roots of the quadratic equation which describe the speed of Raj’s car are
(a) 15, – 20
(b) 20, – 15
(c) 20, – 25
(d) 25, – 25
Answer: (c) 20, – 25

Question. Which of the following quadratic equations has 2 as a root?
(a) \( x^2 - 4x + 5 = 0 \)
(b) \( x^2 + 3x - 12 = 0 \)
(c) \( 2x^2 - 7x + 6 = 0 \)
(d) \( 3x^2 - 6x - 2 = 0 \)
Answer: (c) \( 2x^2 - 7x + 6 = 0 \)

Question. The positive root of \( \sqrt{3x^2 + 6} = 9 \) is
(a) 5
(b) –5
(c) 3
(d) –3
Answer: (a) 5

2. SOLUTION OF A QUADRATIC EQUATION BY FACTORISATION

To find the solution of a quadratic equation by factorisation method, we first express the given equation as product of two linear factors by splitting the middle term. By equating each factor to zero, we get possible solutions/roots of the given quadratic equation. Let the given quadratic equation be \( ax^2 + bx + c = 0 \). Let the quadratic polynomial \( ax^2 + bx + c \) be expressed as the product of two linear factors say \( (px + q) \) and \( (rx + s) \) where, \( p, q, r, s \) are real numbers such that \( p \neq 0, r \neq 0 \).

Then \( ax^2 + bx + c = 0 \Rightarrow (px + q)(rx + s) = 0 \Rightarrow \) Either \( (px + q) = 0 \) or \( (rx + s) = 0 \Rightarrow x = -\frac{q}{p} \) or \( x = -\frac{s}{r} \).

Question. Solve the quadratic equation \( 7x^2 = 8 - 10x \) by the factorisation method.
Answer: \( x = \frac{4}{7}, x = -2 \)

Question. Solve the quadratic equation \( x(x + 9) = 52 \) by the factorisation method.
Answer: \( x = -13, x = 4 \)

Question. Solve the quadratic equation \( 3(x^2 - 4) = 5x \) by the factorisation method.
Answer: \( x = -\frac{4}{3}, x = 3 \)

Question. Solve the quadratic equation \( 3x^2 - 2\sqrt{6}x + 2 = 0 \) by the factorisation method.
Answer: \( x = \frac{\sqrt{2}}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}} \)

Question. Solve the equation \( \frac{4}{x} - 3 = \frac{5}{2x + 3} \); \( x \neq 0, -\frac{3}{2} \) for \( x \).
Answer: \( x = 1 \) or \( x = -2 \)

Question. Solve for \( x \): \( \frac{16}{x} - 1 = \frac{15}{x + 1} \); \( x \neq 0, -1 \).
Answer: \( x = \pm 4 \)

Question. The sum of the squares of two consecutive odd numbers is 394. Find the numbers.
Answer: The required numbers are 13 and 15.

Question. The difference of two natural numbers is 5 and the difference of their reciprocals is \( \frac{1}{10} \). Find the numbers.
Answer: The required numbers are 10 and 5.

Question. If Zeba was younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than five times her actual age. What is her age now?
Answer: Present age of Zeba is 14 years.

Question. Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back at the same point in 4 hours 30 minutes. Find the speed of the stream.
Answer: Speed of stream = 5 km/hr

Question. A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time, it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.
Answer: Speed of plane = 500 km/hr

Question. Solve for \( x \): \( \frac{1}{x+1} + \frac{3}{5x+1} = \frac{5}{x+4} \), \( x \neq -1, -\frac{1}{5}, -4 \).
Answer: \( x = -\frac{11}{17}, x = 1 \)