CBSE Class 10 Mathematics Quadratic Equations VBQs Set N

The velocity of a motorboat is 20 km/hr. For protecting the space of 15 km the boat took 1 hour extra for upstream than downstream.

Question. Let velocity of the stream be \( x \) km/hr, then velocity of the motorboat is upstream shall be :
(a) 20 km/hr
(b) \( (20 + x) \) km/hr
(c) \( (20 - x) \) km/hr
(d) 2 km/hr
Answer: (c) \( (20 - x) \) km/hr

Question. What is the relation between velocity, distance and time?
(a) velocity \( = \frac{\text{distance}}{\text{time}} \)
(b) distance \( = \frac{\text{velocity}}{\text{time}} \)
(c) time = velocity \( \times \) distance
(d) velocity = distance \( \times \) time
Answer: (b) distance \( = \frac{\text{velocity}}{\text{time}} \)

Question. Which is the correct quadratic equation for the given situation?
(a) \( x^2 + 30x - 200 = 0 \)
(b) \( x^2 + 20x - 400 = 0 \)
(c) \( x^2 + 30x - 400 = 0 \)
(d) \( x^2 - 20x - 400 = 0 \)
Answer: (c) \( x^2 + 30x - 400 = 0 \)
Explanation :
Speed of motorboat = 20 km/hr
Let speed of stream = \( x \) km/hr
Speed of boat in upstream = \( (20 - x) \) km/hr
Speed of boat in downstream = \( (20 + x) \) km/hr
According to question
\( \frac{15}{20 - x} - \frac{15}{20 + x} = 1 \)
\( \frac{15(20 + x) - 15(20 - x)}{400 - x^2} = 1 \)
\( \frac{300 + 15x - 300 + 15x}{400 - x^2} = 1 \)
\( 30x = 400 - x^2 \)
\( x^2 + 30x - 400 = 0 \)

Question. What is the velocity of the motorboat in the still water?
(a) 20 km/hour
(b) 10 km/hour
(c) 15 km/hour
(d) 25 km/hour
Answer: (b) 10 km/hour
Explanation :
\( x^2 + 30x - 400 = 0 \)
\( \Rightarrow x^2 + 40x - 10x - 400 = 0 \)
\( \Rightarrow x(x + 40) - 10(x + 40) = 0 \)
\( \Rightarrow (x + 40)(x - 10) = 0 \)
\( x = - 40, 10 \)
(Speed cannot be negative)
\( x = 10 \) km/hr

Question. How much time a boat took in downstream to cover the distance :
(a) 90 minutes
(b) 15 minutes
(c) 30 minutes
(d) 45 minutes
Answer: (c) 30 minutes
Explanation :
Speed of boat = \( (20 + 10) \) km/hr
= 30 km/hr (downstream)
Time \( = \frac{\text{Distance}}{\text{Speed}} \)
Time \( = \frac{15}{30} \)
Time \( = \frac{1}{2} \text{ hour} \)
Time = 30 minute

Point A and B representing Chandigarh and Kurukshetra respectively are almost 90 km apart from each other on the highway. A car starts from Chandigarh and another from Kurukshetra at the same time. If these cars go in the same direction they meet in 9 hours and if these cars go in opposite direction, they meet in 9/7 hours. Let X and Y be two cars starting from points A and B respectively and their speed be \( x \) km/hr and \( y \) km/hr.

Question. When both cars move in the same direction, then the situation can be represented algebraically as :
(a) \( x - y = 10 \)
(b) \( x + y = 10 \)
(c) \( x + y = 9 \)
(d) \( x - y = 9 \)
Answer: (a) \( x - y = 10 \)
Explanation :
Suppose two cars meet at point Q. Then, distance travelled by car X = AQ and distance travelled by car Y = BQ. It is given that two cars meet in 9 hours.
\( \therefore \) Distance travelled by car X in 9 hours = 9x km \( \Rightarrow \) AQ = 9x
Distance travelled by car Y in 9 hours = 9y km \( \Rightarrow \) BQ = 9y
Clearly, AQ – BQ = AB
\( \Rightarrow 9x - 9y = 90 \)
\( \Rightarrow x - y = 10 \)

Question. When both cars move in opposite direction, then the situation can be represented algebraically as
(a) \( x - y = 70 \)
(b) \( x + y = 90 \)
(c) \( x + y = 70 \)
(d) \( x + y = 10 \)
Answer: (c) \( x + y = 70 \)
Explanation :
Suppose two cars meet at point P. Then, Distance travelled by car X = AP and distance travelled by car Y = BP. In this case, two cars meet in \( \frac{9}{7} \) hours.
\( \therefore \) Distance travelled by car X in 9/7 hours \( = \frac{9}{7}x \) km \( \Rightarrow \) AP \( = \frac{9}{7}x \) km
Distance travelled by car Y in \( \frac{9}{7} \) hours is \( \frac{9}{7}y \) km \( \Rightarrow \) BP \( = \frac{9}{7}y \)
Clearly, AP + BP = AB
\( \Rightarrow \frac{9}{7}x + \frac{9}{7}y = 90 \)
\( \Rightarrow \frac{9}{7}(x + y) = 90 \)
\( \Rightarrow x + y = 70 \)

Question. Speed of car X is :
(a) 30 km/hr
(b) 40 km/hr
(c) 50 km/hr
(d) 60 km/hr
Answer: (b) 40 km/hr
Explanation :
We have \( x - y = 10 \) and \( x + y = 70 \)
On solving both the equations, we get
\( x = 40 \) km/hr
Hence, speed of car X is 40 km/hr.

Question. Speed of car Y is
(a) 50 km/hr
(b) 40 km/hr
(c) 30 km/hr
(d) 60 km/hr
Answer: (c) 30 km/hr
Explanation :
We have \( x - y = 10 \)
\( \Rightarrow 40 - y = 10 \)
\( \Rightarrow y = 30 \)
Hence, speed of car y is 30 km/hr.

Question. If the speed of car X and car Y, each is increased by 10 km/hr, and cars are moving in opposite direction, then after how much time they will meet?
(a) 5 hrs
(b) 4 hrs
(c) 2 hrs
(d) 1 hr
Answer: (d) 1 hr
Explanation :
New speed of car X = 50 km/hr
New speed of car Y = 40 km/hr
Then, \( t = \frac{90}{50 + 40} = 1 \) hour

(c) If \( \alpha \) is a root of the given quadratic equation, then its other root is \( -\alpha \).
(d) All of the options
Answer: (d) All of the options
Explanation :
\( x(x - 1) (x + 7) = x(6x - 9) \)
\( \Rightarrow x^3 + 6x^2 - 7x = 6x^2 - 9x \)
\( \Rightarrow x^3 + 2x = 0 \)
is not a quadratic equation.

Question. Which of the following is not a method of finding solutions of the given quadratic equation ?
(a) Factorisation method
(b) Completing the square method
(c) Formula method
(d) None of the options
Answer: (d) None of the options

Rahul and Aryan are good friends. They decided go to Panipat by their own vehicles. Rahul’s automotive travels at a velocity of \( x \) km/h whereas Aryan‘s automotive travels 5 km/h faster than Rahul’s automotive. Rahul took 4 hours more than Aryan to finish the journey of 400 km.

Question. What would be the distance covered by Aryan’s automotive 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
Explanation :
\( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \)
Distance = Speed \( \times \) Time
\( D = (x + 5) \times 2 \)

Question. Which of the given quadratic equation describe the velocity of Rahul’s automotive?
(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) \( x^2 + 5x - 500 = 0 \)
Explanation :
Speed of Rahul’s car = \( x \) km/hr
Speed of Aryan’s car = \( (x + 5) \) km/hr
Total distance = 400 km
Rahul took 4 hour more than Aryan
Time taken by Aryan‘s car \( = \frac{400}{x + 5} \)
Time taken by Rahul’s car \( = (\frac{400}{x} + 5) + 4 \) hr
Distance = Speed \( \times \) Time
\( 400 = x \times [(\frac{400}{x} + 5) + 4] \)
\( x^2 + 5x - 500 = 0 \)

Question. What is the velocity of Rahul’s automotive?
(a) 20 km/hour
(b) 15 km/hour
(c) 25 km/hour
(d) 10 km/hour
Answer: (a) 20 km/hour
Explanation :
The equation formed is \( (x^2 + 5x - 500) = 0 \)
\( \Rightarrow x^2 + 25x - 20x - 500 = 0 \)
\( \Rightarrow x(x + 25) - 20(x + 25) = 0 \)
\( \Rightarrow (x + 25)(x - 20) = 0 \)
\( x = - 25, 20 \)
(Speed cannot be negative)
\( x = 20 \) km/hr

Question. How much time Aryan took to complete the journey of 400 km?
(a) 20 hours
(b) 40 hours
(c) 25 hours
(d) 16 hours
Answer: (d) 16 hours
Explanation :
Rahul car’s speed = 20 km/hr
Aryan car’s speed = \( 20 + 5 = 25 \) km/hr
\( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \)
\( \Rightarrow t = \frac{D}{S} \)
\( \Rightarrow t = \frac{400}{25} \)
\( t = 16 \)

Question. What is the relation between velocity, distance and time?
(a) Velocity \( = \frac{\text{Distance}}{\text{Time}} \)
(b) Distance \( = \frac{\text{Velocity}}{\text{Time}} \)
(c) Time = Velocity \( \times \) Distance
(d) Velocity = Distance \( \times \) Time
Answer: (b) Distance \( = \frac{\text{Velocity}}{\text{Time}} \)

The velocity of a motorboat is 20 km/hr. For protecting the space of 15 km the boat took 1 hour extra for upstream than downstream.

Question. Let velocity of the stream be \( x \) km/hr, then velocity of the motorboat is upstream shall be :
(a) 20 km/hr
(b) \( (20 + x) \) km/hr
(c) \( (20 - x) \) km/hr
(d) 2 km/hr
Answer: (c) \( (20 - x) \) km/hr

Question. What is the relation between velocity, distance and time?
(a) velocity \( = \frac{\text{distance}}{\text{time}} \)
(b) distance \( = \frac{\text{velocity}}{\text{time}} \)
(c) time = velocity \( \times \) distance
(d) velocity = distance \( \times \) time
Answer: (b) distance \( = \frac{\text{velocity}}{\text{time}} \)

Question. Which is the correct quadratic equation for the given situation?
(a) \( x^2 + 30x - 200 = 0 \)
(b) \( x^2 + 20x - 400 = 0 \)
(c) \( x^2 + 30x - 400 = 0 \)
(d) \( x^2 - 20x - 400 = 0 \)
Answer: (c) \( x^2 + 30x - 400 = 0 \)
Explanation :
Speed of motorboat = 20 km/hr
Let speed of stream = \( x \) km/hr
Speed of boat in upstream = \( (20 - x) \) km/hr
Speed of boat in downstream = \( (20 + x) \) km/hr
According to question
\( \frac{15}{20 - x} + 1 = \frac{15}{20 - x} \)
\( x^2 + 30x - 400 = 0 \)

Question. What is the velocity of the motorboat in the still water?
(a) 20 km/hour
(b) 10 km/hour
(c) 15 km/hour
(d) 25 km/hour
Answer: (b) 10 km/hour
Explanation :
\( x^2 + 30x - 400 = 0 \)
\( \Rightarrow x^2 + 40x - 10x - 400 = 0 \)
\( \Rightarrow x(x + 40) - 10(x + 40) = 0 \)
\( \Rightarrow (x + 40)(x - 10) = 0 \)
\( x = - 40, 10 \)
(Speed cannot be negative)
\( x = 10 \) km/hr

Question. How much time a boat took in downstream to cover the distance :
(a) 90 minutes
(b) 15 minutes
(c) 30 minutes
(d) 45 minutes
Answer: (c) 30 minutes
Explanation :
Speed of boat \( = (20 + 10) \) km/hr \( = 30 \) km/hr (downstream)
\( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \)
\( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \)
Time \( = \frac{15}{30} = \frac{1}{2} \text{ hour} \)
Time = 30 minute

Point A and B representing Chandigarh and Kurukshetra respectively are almost 90 km apart from each other on the highway. A car starts from Chandigarh and another from Kurukshetra at the same time. If these cars go in the same direction they meet in 9 hours and if these cars go in opposite direction, they meet in 9/7 hours. Let X and Y be two cars starting from points A and B respectively and their speed be \( x \) km/hr and \( y \) km/hr.

Question. When both cars move in the same direction, then the situation can be represented algebraically as :
(a) \( x - y = 10 \)
(b) \( x + y = 10 \)
(c) \( x + y = 9 \)
(d) \( x - y = 9 \)
Answer: (a) \( x - y = 10 \)
Explanation :
Suppose two cars meet at point Q.
Then, distance travelled by car X = AQ and distance travelled by car Y = BQ.
It is given that two cars meet in 9 hours.
\( \therefore \) Distance travelled by car X in 9 hours \( = 9x \) km \( \Rightarrow \) AQ \( = 9x \)
Distance travelled by car Y in 9 hours \( = 9y \) km \( \Rightarrow \) BQ \( = 9y \)
Clearly, AQ – BQ = AB
\( \Rightarrow 9x - 9y = 90 \)
\( \Rightarrow x - y = 10 \)

Question. When both cars move in opposite direction, then the situation can be represented algebraically as
(a) \( x - y = 70 \)
(b) \( x + y = 90 \)
(c) \( x + y = 70 \)
(d) \( x + y = 10 \)
Answer: (c) \( x + y = 70 \)
Explanation :
Suppose two cars meet at point P.
Then, Distance travelled by car X = AP and distance travelled by car Y = BP.
In this case, two cars meet in \( \frac{9}{7} \) hours.
\( \therefore \) Distance travelled by car X in 9/7 hours \( = \frac{9}{7}x \) km
\( \therefore \text{AP} = \frac{9}{7}x \) km
Distance travelled by car Y in \( \frac{9}{7} \) hours is \( \frac{9}{7}y \) km
\( \Rightarrow \text{BP} = \frac{9}{7}y \)
Clearly, AP + BP = AB
\( \Rightarrow \frac{9}{7}x + \frac{9}{7}y = 90 \)
\( \Rightarrow \frac{9}{7}(x + y) = 90 \)
\( \Rightarrow x + y = 70 \)

Question. Speed of car X is :
(a) 30 km/hr
(b) 40 km/hr
(c) 50 km/hr
(d) 60 km/hr
Answer: (b) 40 km/hr
Explanation :
We have \( x - y = 10 \) and \( x + y = 70 \)
On solving both the equations, we get \( x = 40 \) km/hr
Hence, speed of car X is 40 km/hr.

Question. Speed of car Y is
(a) 50 km/hr
(b) 40 km/hr
(c) 30 km/hr
(d) 60 km/hr
Answer: (c) 30 km/hr
Explanation :
We have \( x - y = 10 \)
\( \Rightarrow 40 - y = 10 \)
\( \Rightarrow y = 30 \)
Hence, speed of car y is 30 km/hr.

Question. If the speed of car X and car Y, each is increased by 10 km/hr, and cars are moving in opposite direction, then after how much time they will meet?
(a) 5 hrs
(b) 4 hrs
(c) 2 hrs
(d) 1 hr
Answer: (d) 1 hr
Explanation :
New speed of car X = 50 km/hr
New speed of car Y = 40 km/hr
Then, \( t = \frac{90}{50 + 40} = 1 \) hour

Question. A quadratic equation can be defined as an equation of degree 2. This means that the highest exponent of the polynomial in it is 2. The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \), where \( a \), \( b \) and \( c \) are real numbers are \( a \neq 0 \). Every quadratic equation has two roots depending of the nature of its discriminant, \( D = b^2 - 4ac \). Based on the above information, answer the following questions.

Question. (i) Which of the following quadratic equation have no real roots?
(a) \( - 4x^2 + 7x - 4 = 0 \)
(b) \( - 4x^2 + 7x - 2 = 0 \)
(c) \( - 2x^2 + 5x - 2 = 0 \)
(d) \( 3x^2 + 6x + 2 = 0 \)
Answer: (a) \( - 4x^2 + 7x - 4 = 0 \)
Explanation :
To have no real roots, Discriminant \( (D = b^2 - 4ac) \) should be \( < 0 \).
(a) \( D = 7^2 - 4(- 4)(- 4) = 49 - 64 = - 15 < 0 \)
(b) \( D = 7^2 - 4(- 4)(- 2) = 49 - 32 = 17 > 0 \)
(c) \( D = 5^2 - 4(- 2)(- 2) = 25 - 16 = 9 > 0 \)
(d) \( D = 6^2 - 4(3)(2) = 36 - 24 = 12 > 0 \)

Question. (ii) Which of the following quadratic equation have rational roots?
(a) \( x^2 + x - 1 = 0 \)
(b) \( x^2 - 5x + 6 = 0 \)
(c) \( 4x^2 - 3x - 2 = 0 \)
(d) \( 6x^2 - x + 11 = 0 \)
Answer: (b) \( x^2 - 5x + 6 = 0 \)
Explanation :
To have rational roots Discriminant \( (D = b^2 - 4ac) \) should be \( > 0 \) and also a perfect square.
(a) \( D = 1^2 - 4(1)(- 1) = 1 + 4 = 5 \), which is not a perfect square.
(b) \( D = (- 5)^2 - 4(1)(6) = 25 - 24 = 1 \), which is a perfect square.
(c) \( D = (- 3)^2 - 4(4)(- 2) = 9 + 32 = 41 \), which is not a perfect square.
(d) \( D = (- 1)^2 - 4(6)(11) = 1 - 264 = - 263 \), which is not a perfect square.

Question. (iii) Which of the following quadratic equation have irrational roots?
(a) \( 3x^2 + 2x + 2 = 0 \)
(b) \( 4x^2 - 7x + 3 = 0 \)
(c) \( 6x^2 - 3x - 5 = 0 \)
(d) \( 2x^2 + 3x - 2 = 0 \)
Answer: (c) \( 6x^2 - 3x - 5 = 0 \)
Explanation :
To have irrational roots, Discriminant \( (D = b^2 - 4ac) \) should be \( > 0 \) but not a perfect square.
(a) \( D = 2^2 - 4(3)(2) = 4 - 24 = - 20 < 0 \)
(b) \( D = (- 7)^2 - 4(4)(3) = 49 - 48 = 1 > 0 \) and a perfect square.
(c) \( D = (- 3)^2 - 4(6)(- 5) = 9 + 120 = 129 > 0 \) and not a perfect square.
(d) \( D = 3^2 - 4(2)(- 2) = 9 + 16 = 25 > 0 \) and a perfect square.

Question. (iv) Which of the following quadratic equations have equal roots?
(a) \( x^2 - 3x + 4 = 0 \)
(b) \( 2x^2 - 2x + 1 = 0 \)
(c) \( 5x^2 - 10x + 1 = 0 \)
(d) \( 9x^2 + 6x + 1 = 0 \)
Answer: (d) \( 9x^2 + 6x + 1 = 0 \)
Explanation :
To have equal roots, Discriminant \( (D = b^2 - 4ac) \) should be \( = 0 \).
(a) \( D = (- 3)^2 - 4(1)(4) = 9 - 16 = - 7 < 0 \)
(b) \( D = (- 2)^2 - 4(2)(1) = 4 - 8 = - 4 < 0 \)
(c) \( D = (- 10)^2 - 4(5)(1) = 100 - 20 = 80 > 0 \)
(d) \( D = 6^2 - 4(9)(1) = 36 - 36 = 0 \)

Question. (v) Which of the following quadratic equations has two distinct equal roots?
(a) \( x^2 + 3x + 1 = 0 \)
(b) \( - x^2 + 3x - 3 = 0 \)
(c) \( 4x^2 + 8x + 4 = 0 \)
(d) \( 3x^2 + 6x + 4 = 0 \)
Answer: (a) \( x^2 + 3x + 1 = 0 \)
Explanation :
To have two distinct real roots, Discriminant \( (D = b^2 - 4ac) \) should be \( > 0 \).
(a) \( D = 3^2 - 4(1)(1) = 9 - 4 = 5 > 0 \)
(b) \( D = 3^2 - 4(- 1)(- 3) = 9 - 12 = - 3 < 0 \)
(c) \( D = 8^2 - 4(4)(4) = 64 - 64 = 0 \)
(d) \( D = 6^2 - 4(3)(4) = 36 - 48 = - 12 < 0 \)

In our daily life we use quadratic formula as for calculating areas, determining a product‘s profit or formulating the speed of an object and many more. Based on the above information, answer the following questions :

Question. (i) If the roots of the quadratic equation are 2, – 3, then its equation is :
(a) \( x^2 - 2x + 3 = 0 \)
(b) \( x^2 + x - 6 = 0 \)
(c) \( 2x^2 - 3x + 1 = 0 \)
(d) \( x^2 - 6x - 1 = 0 \)
Answer: (b) \( x^2 + x - 6 = 0 \)
Explanation :
Roots of the quadratic equation are 2 and – 3.
\( \therefore \) The required quadratic equation is
\( \Rightarrow (x - 2)(x + 3) = 0 \)
\( \Rightarrow x^2 + x - 6 = 0 \)

Question. (ii) If one root of the quadratic equation \( 2x^2 + kx + 1 = 0 \) is \( -1/2 \), then \( k = \)
(a) 3
(b) – 5
(c) – 3
(d) 5
Answer: (a) 3
Explanation :
We have, \( 2x^2 + kx + 1 = 0 \)
Since, \( -1/2 \) is the root of the equation, so it will satisfy the given equation.
\( \therefore 2(-\frac{1}{2})^2 + k(-\frac{1}{2}) + 1 = 0 \)
\( \Rightarrow 1 - k + 2 = 0 \)
\( \Rightarrow k = 3 \)

Question. (iii) Which of the following quadratic equation has equal and opposite roots?
(a) \( x^2 - 4 = 0 \)
(b) \( 16x^2 - 9 = 0 \)
(c) \( 3x^2 + 5x - 5 = 0 \)
(d) Both (a) and (b)
Answer: (d) Both (a) and (b)
Explanation :
If the roots of the quadratic equations are opposites to each other then coefficient of \( x \) (sum of roots) is 0.
So, both (a) and (b) have the coefficient of \( x = 0 \).

Question. (iv) Which of the following quadratic equation can be represented as \( (x - 2)^2 + 19 = 0 \)?
(a) \( x^2 + 4x - 15 = 0 \)
(b) \( x^2 - 4x + 15 = 0 \)
(c) \( x^2 - 4x + 23 = 0 \)
(d) \( x^2 + 4x + 23 = 0 \)
Answer: (c) \( x^2 - 4x + 23 = 0 \)
Explanation :
The given equation is \( (x - 2)^2 + 19 = 0 \)
\( \Rightarrow x^2 - 4x + 4 + 19 = 0 \)
\( \Rightarrow x^2 - 4x + 23 = 0 \)

Question. (v) If one root of a quadratic equation is \( 1 + 5\sqrt{7} \)
(a) \( 1 + 5\sqrt{7} \)
(b) \( 1 - 5\sqrt{7} \)
(c) \( - 1 + 5\sqrt{7} \)
(d) \( - 1 - 5\sqrt{7} \)
Answer: (b) \( 1 - 5\sqrt{7} \)
Explanation :
If one root of a quadratic equation is irrational, then its other root is also irrational and also its conjugate i.e., if one root is \( p + (\sqrt{q}) \) then its other root is \( p - (\sqrt{q}) \)

Quadratic equations started around 3000 B.C. with the Babylonians. They were one of the world’s first civilisation, and came up with some great ideas like agriculture, irrigation and writting. There were many reasons why Babylonians needed to solve quadratic equations. For example, to know what amount of crop you can grow on the square field. Based on the above information, represent the following questions in the form of quadratic equation.

Question. (i) The sum of squares of two consecutive integers is 650 :
(a) \( x^2 + 2x - 650 = 0 \)
(b) \( 2x^2 + 2x - 649 = 0 \)
(c) \( x^2 - 2x - 650 = 0 \)
(d) \( 2x^2 + 6x - 550 = 0 \)
Answer: (b) \( 2x^2 + 2x - 649 = 0 \)
Explanation :
Let two consecutive integers be \( x \), \( x + 1 \).
Given, \( x^2 + (x + 1)^2 = 650 \)
\( \Rightarrow 2x^2 + 2x + 1 - 650 = 0 \)
\( \Rightarrow 2x^2 + 2x - 649 = 0 \)

Question. (ii) The sum of two numbers is 15 and the sum of their reciprocals is \( \frac{3}{10} \):
(a) \( x^2 + 10x - 150 = 0 \)
(b) \( 15x^2 - x + 150 = 0 \)
(c) \( x^2 - 15x + 50 = 0 \)
(d) \( 3x^2 - 10x + 15 = 0 \)
Answer: (c) \( x^2 - 15x + 50 = 0 \)
Explanation :
Let the two numbers be \( x \) and \( 15 - x \).
Given, \( \frac{1}{x} + \frac{1}{15 - x} = \frac{3}{10} \)
\( \Rightarrow 10(15 - x + x) = 3x(15 - x) \)
\( \Rightarrow 50 = 15x - x^2 \)
\( \Rightarrow x^2 - 15x + 50 = 0 \)

Question. (iii) Two numbers differe by 3 and their product is 504 :
(a) \( 3x^2 - 504 = 0 \)
(b) \( x^2 - 504x + 3 = 0 \)
(c) \( 504x^2 + 3 = x \)
(d) \( x^2 + 3x - 504 = 0 \)
Answer: (d) \( x^2 + 3x - 504 = 0 \)
Explanation :
Let the numbers be \( x \) and \( x + 3 \).
Given, \( x(x + 3) = 504 \)
\( \Rightarrow x^2 + 3x - 504 = 0 \)

Question. (iv) A natural number whose square diminished by 84 is thrice of 8 more of given number :
(a) \( x^2 + 8x - 84 = 0 \)
(b) \( 3x^2 - 84x + 3 = 0 \)
(c) \( x^2 - 3x - 108 = 0 \)
(d) \( x^2 - 11x + 60 = 0 \)
Answer: (c) \( x^2 - 3x - 108 = 0 \)
Explanation :
Let the number be \( x \).
According to question,
\( x^2 - 84 = 3(x + 8) \)
\( \Rightarrow x^2 - 84 = 3x + 24 \)
\( \Rightarrow x^2 - 3x - 108 = 0 \)

Question. (v) A natural number when increased by 12, equals 160 times its reciprocal :
(a) \( x^2 - 12x + 160 = 0 \)
(b) \( x^2 - 160x + 12 = 0 \)
(c) \( 12x^2 - x - 160 = 0 \)
(d) \( x^2 + 12x - 160 = 0 \)
Answer: (d) \( x^2 + 12x - 160 = 0 \)
Explanation :
Let the number be \( x \).
According to question,
\( x + 12 = \frac{160}{x} \)
\( \Rightarrow x^2 + 12x - 160 = 0 \)

Amit is preparing for his upcoming semester exam. For this, he has to practice the chapter of quadratic equtaions. So, he started with factorization method. Let two linear factors of \( ax^2 + bx + c \) be \( (px + q) \) and \( (rx + s) \).
\( \therefore ax^2 + bx + c = (px + q)(rx + s) = prx^2 + (ps + qr) x + qs \). Now, factorize each of the following quadratic equations and find their roots.

Question. (i) \( 6x^2 + x - 2 = 0 \)
(a) 1, 6
(b) \( \frac{1}{2}, \frac{-2}{3} \)
(c) \( \frac{1}{3}, \frac{-1}{2} \)
(d) \( \frac{3}{2}, - 2 \)
Answer: (b) \( \frac{1}{2}, \frac{-2}{3} \)
Explanation :
We have \( 6x^2 + x - 2 = 0 \)
\( \Rightarrow 6x^2 - 3x + 4x - 2 = 0 \)
\( \Rightarrow (3x + 2)(2x - 1) = 0 \)
\( \Rightarrow x = \frac{1}{2}, \frac{-2}{3} \)

Question. (ii) \( 2x^2 + x - 300 = 0 \)
(a) 30, \( \frac{21}{5} \)
(b) 60, \( \frac{-2}{5} \)
(c) 12, \( \frac{-25}{2} \)
(d) None of the options
Answer: (c) 12, \( \frac{-25}{2} \)
Explanation :
\( 2x^2 + x - 300 = 0 \)
\( \Rightarrow 2x^2 - 24x + 25x - 300 = 0 \)
\( \Rightarrow (x - 12)(2x + 25) = 0 \)
\( \Rightarrow x = 12, \frac{-25}{2} \)

Question. (iii) \( x^2 - 8x + 16 = 0 \)
(a) 3, 3
(b) 3, – 3
(c) 4, – 4
(d) 4, 4
Answer: (d) 4, 4
Explanation :
\( x^2 - 8x + 16 = 0 \)
\( \Rightarrow (x - 4)^2 = 0 \)
\( \Rightarrow (x - 4)(x - 4) = 0 \)
\( \Rightarrow x = 4, 4 \)

Question. (iv) \( 6x^2 - 13x + 5 = 0 \)
(a) 2, \( \frac{3}{5} \)
(b) – 2, \( \frac{5}{3} \)
(c) \( \frac{1}{2}, \frac{-3}{5} \)
(d) \( \frac{1}{2}, \frac{5}{3} \)
Answer: (d) \( \frac{1}{2}, \frac{5}{3} \)
Explanation :
\( 6x^2 - 13x + 5 = 0 \)
\( \Rightarrow 6x^2 - 3x - 10x + 5 = 0 \)
\( \Rightarrow (2x - 1)(3x - 5) = 0 \)
\( \Rightarrow x = \frac{1}{2}, \frac{5}{3} \)

Question. (v) \( 100x^2 - 20x + 1 = 0 \)
(a) \( \frac{1}{10}, \frac{1}{10} \)
(b) – 10, – 10
(c) \( -10, \frac{1}{10} \)
(d) \( \frac{-1}{10}, \frac{-1}{10} \)
Answer: (a) \( \frac{1}{10}, \frac{1}{10} \)
Explanation :
\( 100x^2 - 20x + 1 = 0 \)
\( \Rightarrow (10x - 1)^2 = 0 \)
\( \Rightarrow x = \frac{1}{10}, \frac{1}{10} \)