CBSE Class 10 Mathematics Quadratic Equations Worksheet Set 08

Short Answer Questions-

Question. In a flight of 600 km, an aircraft was slowed due to bad weather. Its average speed for the trip was reduced by 200 km/h and time of flight increased by 30 minutes. Find the original duration of flight. 
Answer: Sol. Let original speed of the aircraft be \( x \) km/h.
\(\therefore\) Time taken to cover a flight of 600 km = \( \frac{600}{x} \) h
When speed is reduced by 200 km/h
\(\therefore\) New speed = \( (x - 200) \) km/h
\(\therefore\) Time taken to cover 600 km = \( \frac{600}{x - 200} \) h
ATQ, \( \frac{600}{x - 200} - \frac{600}{x} = \frac{30}{60} \)
\( \implies \) \( 600 \left( \frac{1}{x - 200} - \frac{1}{x} \right) = \frac{1}{2} \)
\( \implies \) \( 600 \left( \frac{x - x + 200}{x(x - 200)} \right) = \frac{1}{2} \)
\( \implies \) \( 120000 \times 2 = x^2 - 200x \)
\( \implies \) \( x^2 - 200x - 240000 = 0 \)
\( x = \frac{200 \pm \sqrt{40000 + 960000}}{2} \)
\( \implies \) \( x = \frac{200 \pm \sqrt{1000000}}{2} = \frac{200 \pm 1000}{2} \)
\(\therefore\) \( x = \frac{200 + 1000}{2} \) (Negative sign neglected)
\( x = \frac{1200}{2} = 600 \implies x = 600 \) km/h
\(\therefore\) Original duration of flight = \( \frac{600}{x} = \frac{600}{600} \) h = 1 hour

Question. A two digit number is four times the sum of the digits. It is also equal to 3 times the product of digits. Find the number.
Answer: Sol. Let the ten's digit be \( x \) and unit's digit = \( y \).
\(\therefore\) Number = \( 10x + y \)
ATQ, \( 10x + y = 4(x + y) \)
\( \implies \) \( 6x = 3y \implies 2x = y \) ...(i)
Again \( 10x + y = 3xy \)
\( 10x + 2x = 3x(2x) \)
\( \implies \) \( 12x = 6x^2 \) [From equation (i)]
\( \implies \) \( 6x^2 - 12x = 0 \implies 6x(x - 2) = 0 \)
\( \implies \) \( x = 2 \) (rejecting \( x = 0 \))
From (i), \( 2x = y \implies y = 4 \)
\(\therefore\) The required number is \( 10x + y = 10 \times 2 + 4 = 24 \).

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. Find its usual speed. 
Answer: Sol. Given distance is 1500 km.
Usual speed = \( s \).
We know, speed = \( \frac{\text{distance}}{\text{time}} \)
\( \implies \) time = \( \frac{\text{distance}}{\text{speed}} \)
\( \implies \) From question, \( \frac{1500}{s} = \frac{1500}{s + 100} + \frac{1}{2} \) [half an hour late = 30 mins = 0.5 hr].
\( \frac{1500}{s} - \frac{1500}{s + 100} = \frac{1}{2} \)
\( \frac{1500(s + 100) - 1500s}{s(s + 100)} = \frac{1}{2} \)
\( \frac{150000}{s^2 + 100s} = \frac{1}{2} \)
Cross multiplying,
\( 300000 = s^2 + 100s \)
\( s^2 + 100s - 300000 = 0 \)
\( s^2 + 600s - 500s - 300000 = 0 \)
\( s(s + 600) - 500(s + 600) = 0 \)
\( (s - 500)(s + 600) = 0 \)
\( \implies \) \( s - 500 = 0 \) or \( s + 600 = 0 \)
\( \implies s = 500 \) km/h or \( s = -600 \) km/h.
But speed cannot be negative.
\( \implies \) The usual speed of the plane is 500 km/hr. 

Question. Solve the quadratic equation \( (x - 1)^2 - 5(x - 1) - 6 = 0 \).
Answer: Sol. We have,
\( (x - 1)^2 - 5(x - 1) - 6 = 0 \)
\( \implies \) \( y^2 - 5y - 6 = 0 \), where \( y = (x - 1) \)
\( \implies \) \( y^2 - 6y + y - 6 = 0 \)
\( \implies \) \( y(y - 6) + 1(y - 6) = 0 \)
\( \implies \) \( (y - 6)(y + 1) = 0 \)
\( \implies \) \( y = -1, 6 \)
\( \implies \) \( x - 1 = -1 \), or \( x - 1 = 6 \)
\( \implies \) \( x = 0, 7 \)
\(\therefore\) \( x = 0 \) or 7

Question. Solve for \( x: x^2 + 5x - (a^2 + a - 6) = 0 \) 
Answer: Sol. \( x^2 + 5x - (a^2 + a - 6) = 0 \)
\( x^2 + 5x - (a^2 + 3a - 2a - 6) = 0 \)
\( x^2 + 5x - [a(a + 3) - 2(a + 3)] = 0 \)
\( x^2 + 5x - (a - 2)(a + 3) = 0 \)
\(\therefore\) \( x^2 + (a + 3)x - (a - 2)x - (a - 2)(a + 3) = 0 \)
\( x[x + (a + 3)] - (a - 2)[x + (a + 3)] = 0 \)
\( [x + (a + 3)] [x - (a - 2)] = 0 \)
\(\therefore\) \( x = -(a + 3) \) or \( x = (a - 2) \)
\( \implies \) \( x = -(a + 3), (a - 2) \)
Alternative method
\( x^2 + 5x - (a^2 + a - 6) = 0 \)
\(\therefore\) \( x = \frac{-5 \pm \sqrt{5^2 - 4 \times 1 \times [-(a^2 + a - 6)]}}{2 \times 1} \)
\( = \frac{-5 \pm \sqrt{25 + 4a^2 + 4a - 24}}{2} = \frac{-5 \pm \sqrt{4a^2 + 4a + 1}}{2} \)
\( = \frac{-5 \pm \sqrt{(2a)^2 + 2.(2a).1 + 1^2}}{2} = \frac{-5 \pm \sqrt{(2a + 1)^2}}{2} \)
\( = \frac{-5 \pm (2a + 1)}{2} = \frac{-5 + 2a + 1}{2}, \frac{-5 - 2a - 1}{2} \)
\( = \frac{2a - 4}{2}, \frac{-2a - 6}{2} = (a - 2), -(a + 3) \)

Question. Using quadratic formula solve the following quadratic equation: \( p^2x^2 + (p^2 - q^2)x - q^2 = 0 \) 
Answer: Sol. We have, \( p^2x^2 + (p^2 - q^2)x - q^2 = 0 \)
Comparing this equation with \( ax^2 + bx + c = 0 \), we have
\( a = p^2, b = p^2 - q^2 \) and \( c = -q^2 \)
\(\therefore\) \( D = b^2 - 4ac = (p^2 - q^2)^2 - 4 \times p^2 \times (-q^2) \)
\( = (p^2 - q^2)^2 + 4p^2q^2 = (p^2 + q^2)^2 > 0 \)
So, the given equation has real roots given by
\( \alpha = \frac{-b + \sqrt{D}}{2a} = \frac{-(p^2 - q^2) + (p^2 + q^2)}{2p^2} = \frac{2q^2}{2p^2} = \frac{q^2}{p^2} \)
and \( \beta = \frac{-b - \sqrt{D}}{2a} = \frac{-(p^2 - q^2) - (p^2 + q^2)}{2p^2} = \frac{-2p^2}{2p^2} = -1 \)
Hence, roots are \( \frac{q^2}{p^2} \) and -1.

Question. Solve for \( x: \frac{1}{x+1} + \frac{2}{x+2} = \frac{4}{x+4}, x \neq -1, -2, -4 \) 
Answer: Sol. \( \frac{1}{x+1} + \frac{2}{x+2} = \frac{4}{x+4} \)
\( \implies \) \( \frac{x+2 + 2(x+1)}{(x+1)(x+2)} = \frac{4}{x+4} \)
\( \implies \) \( (x+4)(x+2+2x+2) = 4(x+1)(x+2) \)
\( \implies \) \( (x+4)(3x+4) = 4(x^2+3x+2) \)
\( \implies \) \( x^2 - 4x - 8 = 0 \)
\( \implies \) \( x = \frac{4 \pm \sqrt{16+32}}{2} = \frac{4 \pm 4\sqrt{3}}{2} = 2 \pm 2\sqrt{3} \)

Question. If the roots of the quadratic equation \( (a - b)x^2 + (b - c)x + (c - a) = 0 \) are equal, prove that \( 2a = b + c \). 
Answer: Sol. \( (a - b)x^2 + (b - c)x + (c - a) = 0 \)
The roots are equal, then \( D = 0 \)
Comparing eqn by \( ax^2 + bx + c = 0 \)
\( a = (a - b); b = (b - c); c = (c - a) \)
\( D = b^2 - 4ac \)
\( = (b - c)^2 - 4(a - b)(c - a) \)
Here, \( D = 0 \)
\( (b - c)^2 - 4(a - b)(c - a) = 0 \)
\( b^2 + c^2 - 2bc - 4(ac - a^2 - bc + ab) = 0 \)
\( b^2 + c^2 - 2bc - 4ac + 4a^2 + 4bc - 4ab = 0 \)
\( 4a^2 + b^2 + c^2 + 2bc - 4ab - 4ac = 0 \)
\( \implies \) \( (-2a + b + c)^2 = 0 \) \([a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = (a + b + c)^2]\)
\( -2a + b + c = 0 \)
\( b + c = 2a \)
Hence proved. 

Question. Find the values of \( k \), for which the quadratic equation \( (k + 4)x^2 + (k + 1)x + 1 = 0 \) has equal roots. 
Answer: Sol. For equal roots \( (k+1)^2 - 4(k+4) \times 1 = 0 \)
\( \implies \) \( k^2 - 2k - 15 = 0 \)
\( \implies \) \( (k+3)(k-5) = 0 \)
\( \implies \) \( k = -3, 5 \)

Question. Find the positive value(s) of \( k \) for which both quadratic equations \( x^2 + kx + 64 = 0 \) and \( x^2 - 8x + k = 0 \) will have real roots. 
Answer: Sol. (i) For \( x^2 + kx + 64 = 0 \) to have real roots
\( k^2 - 4(1)(64) \geq 0 \) i.e., \( k^2 - 256 \geq 0 \)
\( \implies \) \( (k - 16)(k + 16) \geq 0 \implies k \leq -16 \) or \( k \geq 16 \)
(ii) For \( x^2 - 8x + k = 0 \) to have real roots
\( (-8)^2 - 4(k) \geq 0 \) i.e., \( 64 - 4k \geq 0 \implies k \leq 16 \)
For (i) and (ii) to hold simultaneously \( k = 16 \).

Question. If the equation \( (1 + m^2)x^2 + 2mcx + c^2 - a^2 = 0 \) has equal roots, show that \( c^2 = a^2(1 + m^2) \). 
Answer: Sol. The given equation is \( (1 + m^2)x^2 + (2mc)x + (c^2 - a^2) = 0 \)
Here, \( A = 1 + m^2, B = 2mc \) and \( C = c^2 - a^2 \)
Since the given equation has equal roots, therefore \( D = 0 \implies B^2 - 4AC = 0 \).
\( \implies \) \( (2mc)^2 - 4(1 + m^2)(c^2 - a^2) = 0 \)
\( \implies \) \( 4m^2c^2 - 4(c^2 - a^2 + m^2c^2 - m^2a^2) = 0 \)
\( \implies \) \( m^2c^2 - c^2 + a^2 - m^2c^2 + m^2a^2 = 0 \) [Dividing throughout by 4]
\( \implies \) \( -c^2 + a^2(1 + m^2) = 0 \implies c^2 = a^2(1 + m^2) \)
Hence proved.

Question. If the roots of the equation \( (c^2 - ab)x^2 - 2(a^2 - bc)x + b^2 - ac = 0 \) in \( x \) are equal, then show that either \( a = 0 \) or \( a^3 + b^3 + c^3 = 3abc \).
Answer: Sol. For equal roots \( D = 0 \)
Therefore \( 4(a^2 - bc)^2 - 4(c^2 - ab)(b^2 - ac) = 0 \)
\( \implies \) \( 4[a^4 + b^2c^2 - 2a^2bc - b^2c^2 + ac^3 + ab^3 - a^2bc] = 0 \)
\( \implies \) \( a(a^3 + b^3 + c^3 - 3abc) = 0 \)
\( \implies \) Either \( a = 0 \) or \( a^3 + b^3 + c^3 = 3abc \)

Question. If the roots of the equation \( (a^2 + b^2)x^2 - 2(ac + bd)x + (c^2 + d^2) = 0 \) are equal, prove that \( \frac{a}{b} = \frac{c}{d} \). 
Answer: Sol. \( A = (a^2 + b^2), B = -2(ac + bd), C = (c^2 + d^2) \)
as roots are equal,
\( D = B^2 - 4AC = 0 \)
\( [-2(ac + bd)]^2 - 4(a^2 + b^2)(c^2 + d^2) = 0 \)
\( 4(a^2c^2 + 2abcd + b^2d^2) = 4(a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2) \)
\( 2abcd = a^2d^2 + b^2c^2 \)
\( 0 = a^2d^2 + b^2c^2 - 2abcd \)
\( 0 = (ad - bc)^2 \)
\( 0 = ad - bc \)
\( ad = bc \)
\( \implies \) \( \frac{a}{b} = \frac{c}{d} \)
Hence, proved. 

Long Answer Questions

Question. Solve for \( x: \frac{x-1}{2x+1} + \frac{2x+1}{x-1} = 2, where x \neq -\frac{1}{2}, 1 \).
Answer: Sol. Let \( \frac{x-1}{2x+1} \) be \( y \),
\( y + \frac{1}{y} = 2 \)
\( \frac{y^2 + 1}{y} = 2 \)
\( y^2 + 1 = 2y \)
\( y^2 - 2y + 1 = 0 \)
\( y^2 - y - y + 1 = 0 \)
\( y(y - 1) - 1(y - 1) = 0 \)
\( (y - 1)(y - 1) = 0 \)
\(\therefore\) \( y = 1 \) or 1.
Now, \( \frac{x-1}{2x+1} = 1 \) or \( \frac{x-1}{2x+1} = 1 \)
\( x - 1 = 2x + 1 \)
\( -2 = x \)
\(\therefore\) \( x = -2 \). 

Question. Solve the following quadratic equation:
\( 9x^2 - 9(a + b)x + [2a^2 + 5ab + 2b^2] = 0 \) 
Answer: Sol. Consider the equation \( 9x^2 - 9(a + b)x + [2a^2 + 5ab + 2b^2] = 0 \)
Now comparing with \( Ax^2 + Bx + C = 0 \), we get
\( A = 9, B = -9(a + b) \) and \( C = [2a^2 + 5ab + 2b^2] \)
Now discriminant,
\( D = B^2 - 4AC \)
\( = \{-9(a + b)\}^2 - 4 \times 9(2a^2 + 5ab + 2b^2) \)
\( = 9^2(a + b)^2 - 4 \times 9(2a^2 + 5ab + 2b^2) \)
\( = 9 \{9(a + b)^2 - 4(2a^2 + 5ab + 2b^2)\} \)
\( = 9 \{9a^2 + 9b^2 + 18ab - 8a^2 - 20ab - 8b^2\} \)
\( = 9 \{a^2 + b^2 - 2ab\} = 9(a - b)^2 \)
Now using the quadratic formula,
\( x = \frac{-B \pm \sqrt{D}}{2A} \), we get \( x = \frac{9(a + b) \pm \sqrt{9(a - b)^2}}{2 \times 9} \)
\( \implies \) \( x = \frac{9(a + b) \pm 3(a - b)}{2 \times 9} \)
\( \implies \) \( x = \frac{3(a + b) \pm (a - b)}{6} \)
\( \implies \) \( x = \frac{(3a + 3b) + (a - b)}{6} \) and \( x = \frac{(3a + 3b) - (a - b)}{6} \)
\( \implies \) \( x = \frac{(4a + 2b)}{6} \) and \( x = \frac{(2a + 4b)}{6} \)
\( \implies \) \( x = \frac{2a + b}{3} \) and \( x = \frac{a + 2b}{3} \) are required solutions.

Question. At \( t \) minutes past 2 pm, the time needed by the minute hand of clock to show 3 pm was found to be 3 minutes less than \( \frac{t^2}{4} \) minutes. Find \( t \). 
Answer: Sol. As we know that total time taken by the minute hand to run from 2 pm to 3 pm = 60 minutes.
According to the question
\( t + \left( \frac{t^2}{4} - 3 \right) = 60 \)
\( \implies \) \( 4t + t^2 - 12 = 240 \)
\( \implies \) \( t^2 + 4t - 252 = 0 \)
\( \implies \) \( t^2 + 18t - 14t - 252 = 0 \)
\( \implies \) \( t(t + 18) - 14(t + 18) = 0 \)
\( \implies \) \( (t + 18)(t - 14) = 0 \)
\( \implies \) \( t = 14 \) or \(-18\)
But time can not be negative.
\(\therefore\) \( t = 14 \) minutes

Question. Seven years ago Varun's age was five times the square of Swati's age. Three years hence, Swati's age will be two-fifth of Varun's age. Find their present ages.
Answer: Sol. Seven years ago, let Swati's age be \( x \) years. Then, seven years ago Varun's age was \( 5x^2 \) years.
\(\therefore\) Swati's present age = \( (x + 7) \) years
Varun's present age = \( (5x^2 + 7) \) years
Three years hence,
Swati's age = \( (x + 7 + 3) \) years = \( (x + 10) \) years
Varun's age \( (5x^2 + 7 + 3) \) years = \( (5x^2 + 10) \) years
According to the question,
\( x + 10 = \frac{2}{5} (5x^2 + 10) \)
\( \implies \) \( x + 10 = \frac{2}{5} \times 5 (x^2 + 2) \)
\( \implies \) \( x + 10 = 2x^2 + 4 \)
\( \implies \) \( 2x^2 - x - 6 = 0 \)
\( \implies \) \( 2x^2 - 4x + 3x - 6 = 0 \)
\( \implies \) \( 2x(x - 2) + 3(x - 2) = 0 \)
\( \implies \) \( (2x + 3)(x - 2) = 0 \)
\( \implies \) \( x - 2 = 0 \) or \( 2x + 3 = 0 \)
\( \implies \) \( x = 2 \) [\(\because\) \( 2x + 3 \neq 0 \) as \( x > 0 \)]
Hence, Swati's present age = \( (2 + 7) \) years = 9 years
and Varun's present age = \( (5 \times 2^2 + 7) \) years = 27 years

Question. \( A \) takes 6 days less than \( B \) to do a work. If both \( A \) and \( B \) working together can do it in 4 days, how many days will \( B \) take to finish it? 
Answer: Sol. Let \( B \) complete a work in \( x \) days.
Then \( A \) takes \( x - 6 \) days to complete it.
Together they complete it in 4 days.
According to work done per day,
\( \frac{1}{x - 6} + \frac{1}{x} = \frac{1}{4} \)
\( \frac{x + x - 6}{x(x - 6)} = \frac{1}{4} \)
\( 4(2x - 6) = x(x - 6) \)
\( 8x - 24 = x^2 - 6x \)
\(\therefore\) \( x^2 - 14x + 24 = 0 \)
\( x^2 - 12x - 2x + 24 = 0 \)
\( x(x - 12) - 2(x - 12) = 0 \)
\( (x - 2)(x - 12) = 0 \)
\(\therefore\) \( x = 2 \) or \( 12 \)
\( x = 2 \) is not possible because then \( x - 6 \) is \(-4\).
\(\therefore\) \( x = 12 \).
So, \( B \) takes 12 days to finish the work. 

Question. One-fourth of a herd of camels was seen in the forest. Twice the square root of the herd had gone to mountains and the remaining 15 camels were seen on the bank of a river. Find the total number of camels. 
Answer: Sol. Let \( x \) be the total number of camels.
Then, number of camels in the forest = \( \frac{x}{4} \)
number of camels on mountains = \( 2\sqrt{x} \)
and number of camels on the bank of river = 15
Thus, total number of camels = \( \frac{x}{4} + 2\sqrt{x} + 15 \)
Now by hypothesis, we have
\( \frac{x}{4} + 2\sqrt{x} + 15 = x \)
\( \implies \) \( 3x - 8\sqrt{x} - 60 = 0 \)
Let \( \sqrt{x} = y \), then \( x = y^2 \)
\( \implies \) \( 3y^2 - 8y - 60 = 0 \)
\( \implies \) \( 3y^2 - 18y + 10y - 60 = 0 \)
\( \implies \) \( 3y(y - 6) + 10(y - 6) = 0 \)
\( \implies \) \( (3y + 10)(y - 6) = 0 \)
\( \implies \) \( y = 6 \) or \( y = -\frac{10}{3} \)
Now, \( y = -\frac{10}{3} \implies x = \left( -\frac{10}{3} \right)^2 = \frac{100}{9} \) (\(\because\) \( x = y^2 \))
But, the number of camels cannot be a fraction.
\(\therefore\) \( y = 6 \)
\( \implies \) \( x = 6^2 = 36 \)
Hence, the number of camels = 36

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: Sol. Let the present age of Zeba be \( x \) years.
Age before 5 years = \( (x - 5) \) years
According to given condition,
\( \implies \) \( (x - 5)^2 = 5x + 11 \)
\( \implies \) \( x^2 + 25 - 10x = 5x + 11 \)
\( \implies \) \( x^2 - 10x - 5x + 25 - 11 = 0 \)
\( \implies \) \( x^2 - 15x + 14 = 0 \)
\( \implies \) \( x^2 - 14x - x + 14 = 0 \)
\( \implies \) \( x(x - 14) - 1(x - 14) = 0 \)
\( \implies \) \( (x - 1)(x - 14) = 0 \)
\( \implies \) \( x - 1 = 0 \) or \( x - 14 = 0 \)
\( x = 1 \) or \( x = 14 \)
But present age cannot be 1 year.
\(\therefore\) Present age of Zeba is 14 years.

Question. Rs. 9,000 were divided equally among a certain number of persons. Had there been 20 more persons, each would have got Rs. 160 less. Find the original number of persons. 
Answer: Sol. Let the original number of persons be \( x \).
\(\therefore\) Each person will get amount = Rs. \( \frac{9000}{x} \)
When 20 persons will increase
\(\therefore\) Each person will get = Rs. \( \frac{9000}{x + 20} \)
ATQ,
\( \frac{9000}{x} - \frac{9000}{x + 20} = 160 \)
\( \implies \) \( 9000 \left( \frac{1}{x} - \frac{1}{x + 20} \right) = 160 \)
\( \implies \) \( 9000 \left( \frac{x + 20 - x}{x(x + 20)} \right) = 160 \)
\( \implies \) \( 180000 = 160(x^2 + 20x) \)
\( \implies \) \( x^2 + 20x = 1125 \)
\( \implies \) \( x^2 + 20x - 1125 = 0 \)
\( \implies \) \( x^2 + 45x - 25x - 1125 = 0 \)
\( \implies \) \( x(x + 45) - 25(x + 45) = 0 \)
\( \implies \) \( (x - 25)(x + 45) = 0 \)
\( \implies \) \( x - 25 = 0 \) (but \( x + 45 = 0 \implies x \neq -45 \))
\( \implies \) \( x = 25 \)
\(\therefore\) Original number of persons is 25.

Question. Two taps running together can fill a tank in \( 3\frac{1}{13} \) hours. If one tap takes 3 hours more than the other to fill the tank, then how much time will each tap take to fill the tank? 
Answer: Sol. Let, time taken by faster tap to fill the tank be \( x \) hours.
Therefore, time taken by slower tap to fill the tank = \( (x + 3) \) hours
Since the faster tap takes \( x \) hours to fill the tank.
\(\therefore\) Portion of the tank filled by the faster tap in one hour = \( \frac{1}{x} \)
Portion of the tank filled by the slower tap in one hour = \( \frac{1}{x + 3} \)
Portion of the tank filled by the two tap together in one hour = \( \frac{1}{\frac{40}{13}} = \frac{13}{40} \)
According to question,
\( \frac{1}{x} + \frac{1}{x + 3} = \frac{13}{40} \)
\( \implies \) \( \frac{x + 3 + x}{x(x + 3)} = \frac{13}{40} \)
\( \implies \) \( 40(2x + 3) = 13x(x + 3) \)
\( \implies \) \( 80x + 120 = 13x^2 + 39x \)
\( \implies \) \( 13x^2 - 41x - 120 = 0 \)
\( \implies \) \( 13x^2 - 65x + 24x - 120 = 0 \)
\( \implies \) \( 13x(x - 5) + 24(x - 5) = 0 \)
\( \implies \) \( (x - 5)(13x + 24) = 0 \)
Either \( x - 5 = 0 \) or \( 13x + 24 = 0 \)
\( \implies \) \( x = 5 \) or \( x = -\frac{24}{13} \)
\( \implies \) \( x = 5 \) [\(\because\) \( x \) cannot be negative]
Hence, time taken by faster tap to fill the tank = \( x = 5 \) hours
and time taken by slower tap = \( x + 3 = 5 + 3 = 8 \) hours.

Question. A two digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.
Answer: Sol. Let the digit at tens place be \( x \).
Then, digit at unit place = \( \frac{18}{x} \)
\(\therefore\) Number = \( 10x + \frac{18}{x} \)
and number obtained by interchanging the digits = \( 10 \times \frac{18}{x} + x \)
According to question,
\( (10x + \frac{18}{x}) - 63 = 10 \times \frac{18}{x} + x \implies (10x + \frac{18}{x}) - (10 \times \frac{18}{x} + x) = 63 \)
\( \implies \) \( 10x + \frac{18}{x} - \frac{180}{x} - x = 63 \implies 9x - \frac{162}{x} - 63 = 0 \)
\( \implies \) \( 9x^2 - 63x - 162 = 0 \implies x^2 - 7x - 18 = 0 \)
\( \implies \) \( x^2 - 9x + 2x - 18 = 0 \implies x(x - 9) + 2(x - 9) = 0 \)
\( \implies \) \( (x - 9)(x + 2) = 0 \implies x = 9 \) or \( x = -2 \)
\( x = 9 \) [\(\because\) a digit can never be negative]
Hence, the required number = \( 10 \times 9 + \frac{18}{9} = 92 \).

Case Study-based Questions

Read the following and answer any four questions from (i) to (v).
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 is the quadratic equation in terms of 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. What is the speed of Raj's car?
(a) 20 km/hour
(b) 15 km/hour
(c) 25 km/hour
(d) 10 km/hour
Answer: (a) 20 km/hour

Question. How much time Ajay took to travel 400 km?
(a) 20 hours
(b) 40 hours
(c) 25 hours
(d) 16 hours
Answer: (d) 16 hours

Question. What is the speed of Ajay's car?
(a) 15 km/h
(b) 20 km/h
(c) 25 km/h
(d) 30 km/h
Answer: (c) 25 km/h

Sol. We have speed of Raj's Car = \( x \) km/h
Therefore, speed of Ajay's Car = \( (x + 5) \) km/h
Now, time taken to complete the journey by Raj = \( \frac{400}{x} \)
and time taken to complete the journey by Ajay = \( \frac{400}{x + 5} \)
According to question, Raj took 4 hours more than Ajay.
\(\therefore\) \( \frac{400}{x} - \frac{400}{x + 5} = 4 \implies 400 \left( \frac{x + 5 - x}{x(x + 5)} \right) = 4 \)
\( \implies \) \( 500 = x(x + 5) = x^2 + 5x \)
\( \implies \) \( x^2 + 5x - 500 = 0 \)
\( \implies \) \( x^2 + 25x - 20x - 500 = 0 \)
\( \implies \) \( x(x + 25) - 20(x + 25) = 0 \)
\( \implies \) \( (x + 25)(x - 20) = 0 \)
\( \implies \) \( x = 20, x \neq -25 \) (Because speed can never be negative)
\( \implies \) \( x = 20 \) km/h
(i) Distance covered by Ajay's car in two hours = Speed \(\times\) Time = \( (x + 5) \times 2 = 2(x + 5) \) km. Option (a) is correct.
(ii) We get the quadratic equation \( x^2 + 5x - 500 = 0 \) for the speed of Raj's car. Option (c) is correct.
(iii) Speed of Raj's car = \( x \) km/h = 20 km/h. Option (a) is correct.
(iv) Time taken by Ajay to travel 400 km = \( \frac{400}{x + 5} = \frac{400}{20 + 5} = \frac{400}{25} = 16 \) hours. Option (d) is correct.
(v) Speed of Ajay's car = \( (x + 5) \) km/h = \( 20 + 5 = 25 \) km/h. Option (c) is correct.

The speed of a motor boat is 20 km/h. For covering the distance of 15 km the boat took 1 hour more for upstream than downstream. Based on above information answer the following questions.

Question. What is the speed of current?
Answer: Sol. We have downstream speed of motor boat = \( (20 + x) \) km/h
and upstream speed of motor boat = \( (20 - x) \) km/h
where \( x \) km/h is the speed of the stream.
We have,
\( \frac{15}{20 - x} - \frac{15}{20 + x} = 1 \)
\( \frac{15(20 + x - 20 + x)}{(20 - x)(20 + x)} = 1 \)
\( \implies \) \( 15 \times 2x = 400 - x^2 \)
\( \implies \) \( x^2 + 30x - 400 = 0 \)
\( \implies \) \( x^2 + 40x - 10x - 400 = 0 \)
\( \implies \) \( x(x + 40) - 10(x + 40) = 0 \)
\( \implies \) \( (x + 40)(x - 10) = 0 \)
\( \implies \) \( x = 10 \) [\(\because\) speed can never be negative, \( x \neq -40 \)]
\(\therefore\) \( x = 10 \) km/h

Question. Which is the correct quadratic equation for the speed of the current?
Answer: The quadratic equation \( x^2 + 30x - 400 = 0 \) is for the speed of the current (Stream).

Question. How much time boat took in downstream?
Answer: Time taken by boat in downstream = \( \frac{15}{20 + x} \) hour = \( \frac{15}{20 + 10} = \frac{15}{30} = \frac{1}{2} \) hour = 30 minutes.

SECTION A

Choose and write the correct option in the following questions.

Question. A student solved a quadratic equation and obtains the roots as \( -4 \) and 3. Part of the student's work to verify the root is shown: \( (-4)^2 + 2(-4) - 9 = 0 \). Based on the student's work, which of these is correct? 
(a) The student calculated the roots of the equation that can be obtained by adding 1 to the equation that the student solved.
(b) The student calculated the roots of the equation that can be obtained by adding \( -1 \) to the equation that the student solved.
(c) The student calculated the roots correctly.
(d) The student calculated the roots correctly but should replace \( 2(-4) \) in his work with \( 2(3) \).
Answer: (a) The student calculated the roots of the equation that can be obtained by adding 1 to the equation that the student solved.

Question. The roots of the quadratic equation \( x^2 - 9x + 20 = 0 \) are
(a) \( -4, 5 \)
(b) \( -4, -5 \)
(c) \( 4, 5 \)
(d) \( 4, -5 \)
Answer: (c) 4, 5

Question. The smallest positive value of \( k \) for which the equation \( x^2 + kx + 9 = 0 \) has real roots is
(a) \( -6 \)
(b) 6
(c) 36
(d) 3
Answer: (b) 6

Solve the following questions.

Question. Show that \( x = -2 \) is a solution of \( 3x^2 + 13x + 14 = 0 \).
Answer: Substituting \( x = -2 \) in \( 3x^2 + 13x + 14 = 0 \):
\( 3(-2)^2 + 13(-2) + 14 = 3(4) - 26 + 14 = 12 - 26 + 14 = 0 \).
Since LHS = RHS, \( x = -2 \) is a solution.

Question. Find the value of \( k \) for which the roots of the equation \( 3x^2 - 10x + k = 0 \) are reciprocal of each other. 
Answer: \( k = 3 \)

SECTION B

Question. Solve the following questions.

Question. What are the roots of the equation \( 4x^2 - 2x - 20 = x^2 + 9x? \)
Answer: \( -\frac{4}{3}, 5 \)

Question. Solve for \( x: \frac{x + 3}{x + 2} = \frac{3x - 7}{2x - 3}, x \neq -2, \frac{3}{2} \) 
Answer: \( 5, -1 \)

Question. Determine the condition for one root of the quadratic equation \( ax^2 + bx + c = 0 \) to be thrice the other.
Answer: \( 3b^2 = 16ac \)

Question. Find the value of \( k \) such that the equation \( (k - 12)x^2 + 2(k - 12)x + 2 = 0 \) has equal roots. 
Answer: \( k = 14 \)

Solve the following questions.

Question. Solve for \( x: 3\left( \frac{3x - 1}{2x + 3} \right) - 2\left( \frac{2x + 3}{3x - 1} \right) = 5; x \neq \frac{1}{3}, -\frac{3}{2} \) 
Answer: \( 0, -7 \)

Question. Solve for \( x: \frac{1}{a + b + x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x}; a + b + x \neq 0, a, b, x \neq 0 \) 
Answer: \( x = -a, -b \)

Question. The difference of two natural numbers is 3 and the difference of their reciprocals is \( \frac{3}{28} \). Find the numbers.
Answer: \( 7 \) and \( 4 \)

Question. Find the value of \( c \) for which the quadratic equation \( 4x^2 - 2(c + 1)x + (c + 1) = 0 \) has equal roots, which are real. 
Answer: \( c = -1, 3 \)

Solve the following questions.

Question. The sum of the areas of two squares is \( 640 \text{ m}^2 \). If the difference of their perimeters is \( 64 \text{ m} \), find the sides of the squares. 
Answer: \( 24 \text{ m} \) and \( 8 \text{ m} \)

Question. Two water taps together can fill a tank in 9 hours 36 minutes. The tap of larger diameter takes 8 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank. 
Answer: \( 24 \text{ hours}, 16 \text{ hours} \)

Question. A motorboat whose speed is 24 km/h in still water takes 1 hour more to go 32 km upstream than to return downstream to the same spot. Find the speed of the stream. 
Answer: \( 8 \text{ km/h} \)