CBSE Class 10 Mathematics Quadratic Equations VBQs Set E

Question. Find the values(s) of \( k \) for which the equation \( x^2 + 5kx + 16 = 0 \) has real and equal roots.
Answer: For roots to be real and equal, \( b^2 - 4ac = 0 \)
\( (5k)^2 - 4 \times 1 \times 16 = 0 \)
\( 25k^2 = 64 \)
\( k = \pm \frac{8}{5} \).

Question. Find the value(s) of \( k \) if the quadratic equation \( 3x^2 - k\sqrt{3}x + 4 = 0 \) has real roots.
Answer: If discriminant of quadratic equation is equal to zero, or more than zero, then roots are real.
Given, \( 3x^2 - k\sqrt{3}x + 4 = 0 \)
Comparing with \( ax^2 + bx + c = 0 \), we get \( a = 3, b = -k\sqrt{3} \) and \( c = 4 \)
Since, Discriminant, \( D = b^2 - 4ac \)
and for real roots \( b^2 - 4ac \ge 0 \)
\( (-k\sqrt{3})^2 - 4 \times 3 \times 4 \ge 0 \)
\( 3k^2 - 48 \ge 0 \)
\( 3k^2 \ge 48 \)
\( k^2 \ge 16 \)
\( (k - 4)(k + 4) \ge 0 \)
\( k \le -4 \) and \( k \ge 4 \).

Short Answer Type Questions-I

Question. Solve for \( x \): \( \sqrt{2x + 9} + x = 13 \).
Answer: \( \sqrt{2x + 9} + x = 13 \)
\( \sqrt{2x + 9} = 13 - x \)
Squaring both sides:
\( 2x + 9 = (13 - x)^2 \)
\( 2x + 9 = 169 + x^2 - 26x \)
\( x^2 - 28x + 160 = 0 \)
\( (x - 8)(x - 20) = 0 \)
Either \( x = 8 \) or \( x = 20 \).
However, \( x = 20 \) does not satisfy the original equation, so \( x = 8 \).

Question. If \( x = \frac{2}{3} \) and \( x = -3 \) are roots of the quadratic equation \( ax^2 + 7x + b = 0 \), find the values of \( a \) and \( b \).
Answer: Substituting \( x = \frac{2}{3} \) in \( ax^2 + 7x + b = 0 \)
\( \frac{4}{9}a + \frac{14}{3} + b = 0 \)
\( 4a + 42 + 9b = 0 \)
\( 4a + 9b = -42 \) ...(i)
Again, substituting \( x = -3 \) in \( ax^2 + 7x + b = 0 \)
\( 9a - 21 + b = 0 \)
\( 9a + b = 21 \) ...(ii)
Solving (i) and (ii), we get \( a = 3 \) and \( b = -6 \).

Question. Solve the following quadratic equation for \( x \): \( 4x^2 - 4a^2x + (a^4 - b^4) = 0 \).
Answer: Given, \( 4x^2 - 4a^2x + (a^4 - b^4) = 0 \)
Comparing with \( Ax^2 + Bx + C = 0 \), we get
Here, \( A = 4, B = -4a^2 \) and \( C = (a^4 - b^4) \)
Since, \( x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \)
Then, \( x = \frac{-(-4a^2) \pm \sqrt{(-4a^2)^2 - 4 \times 4 \times (a^4 - b^4)}}{2 \times 4} \)
\( x = \frac{4a^2 \pm \sqrt{16a^4 - 16a^4 + 16b^4}}{8} \)
\( x = \frac{4a^2 \pm \sqrt{16b^4}}{8} \)
\( x = \frac{4a^2 \pm 4b^2}{8} \)
\( x = \frac{a^2 \pm b^2}{2} \)
\( \therefore x = \frac{a^2 + b^2}{2} \) or \( x = \frac{a^2 - b^2}{2} \).

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: Let unit's digit and ten's digit of the two digit number be \( x \) and \( y \) respectively
\( \therefore \) Number is \( 10y + x \)
According to question,
\( 10y + x = 4(y + x) \)
\( \Rightarrow 10y + x = 4y + 4x \)
\( \Rightarrow 10y - 4y = 4x - x \)
\( \Rightarrow 6y = 3x \)
\( \Rightarrow 2y = x \) ...(i)
Also, \( 10y + x = 3xy \) ...(ii)
From eq (i), \( 10y + 2y = 3(2y)y \)
\( \Rightarrow 12y = 6y^2 \)
\( \Rightarrow 6y^2 - 12y = 0 \)
\( \Rightarrow 6y(y - 2) = 0 \)
\( \Rightarrow y = 0 \) or \( y = 2 \)
Rejecting \( y = 0 \) as tens digit should not be zero for a two digit number.
\( \Rightarrow y = 2 \)
\( \Rightarrow x = 4 \)
\( \therefore \) Required number \( = 10y + x = 10 \times 2 + 4 = 24 \).

Question. In a cricket match, Harbhajan took three wickets less than twice the number of wickets taken by Zahir. The product of the number of wickets taken by these two is 20. Represent the above situation in the form of quadratic equation.
Answer: Let the number of wickets taken by Zahir be \( x \).
Then, the number of wickets taken by Harbhajan \( = 2x - 3 \)
According to question, \( x(2x - 3) = 20 \)
\( 2x^2 - 3x = 20 \)
\( \therefore \) Required quadratic equation is, \( 2x^2 - 3x - 20 = 0 \).

Question. For what value of \( k \), the given quadratic equation \( kx^2 - 6x - 1 = 0 \) has no real roots?
Answer: Given quadratic equation \( \Rightarrow kx^2 - 6x - 1 = 0 \)
where \( a = k, b = -6, c = -1 \)
For no real roots (imaginary roots), discriminant must be less than 0.
That is, \( D < 0 \)
\( \Rightarrow b^2 - 4ac < 0 \)
\( \Rightarrow (-6)^2 - 4(k)(-1) < 0 \)
\( \Rightarrow 36 + 4k < 0 \)
\( 4k < -36 \Rightarrow k < -9 \)
\( k \) should be less than -9.

Question. Find the value of \( k \) for which the roots of the quadratic equation \( 2x^2 + kx + 8 = 0 \) will have the equal roots?
Answer: For equal roots, \( D = 0 \)
\( \therefore b^2 - 4ac = 0 \)
\( \Rightarrow k^2 = 4 \times 2 \times 8 \)
\( k^2 = 64 \)
\( k = \pm\sqrt{64} \)
\( k = \pm 8 \).

Question. If 2 is a root of the equation \( x^2 + kx + 12 = 0 \) and the equation \( x^2 + kx + q = 0 \) has equal roots, find the value of \( q \).
Answer: Since, 2 is the root of \( x^2 + kx + 12 = 0 \)
\( (2)^2 + 2k + 12 = 0 \)
\( 2k + 16 = 0 \)
\( k = -8 \)
Putting, \( k = -8 \) in \( x^2 + kx + q = 0 \)
\( \Rightarrow x^2 - 8x + q = 0 \)
For equal roots, \( (-8)^2 - 4(1)q = 0 \)
\( 64 - 4q = 0 \)
\( 4q = 64 \)
\( q = 16 \).

Question. Find \( k \) so that the quadratic equation \( (k + 1)x^2 - 2(k + 1)x + 1 = 0 \) has equal roots.
Answer: Since, \( (k + 1)x^2 - 2(k + 1)x + 1 = 0 \) has equal roots.
\( D = 0 \)
i.e., \( b^2 = 4ac \)
\( \Rightarrow 4(k + 1)^2 = 4(k + 1) \)
\( \Rightarrow k^2 + 2k + 1 = k + 1 \)
\( \Rightarrow k^2 + k = 0 \)
\( \Rightarrow k(k + 1) = 0 \)
\( \Rightarrow k = 0 \) or -1
\( k = -1 \) does not satisfy the equation.
So, \( k = 0 \).

Short Answer Type Questions-II

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: Let original speed of flight be \( x \) km/h, then according to question,
\( \frac{600}{x - 200} - \frac{600}{x} = 30 \text{ minutes} \)
\( \left[ \because \text{Time} = \frac{\text{Distance}}{\text{Speed}} \right] \)
\( \Rightarrow 600 \left[ \frac{1}{x - 200} - \frac{1}{x} \right] = \frac{30}{60} \text{ hour} \)
\( \Rightarrow \frac{x - x + 200}{x(x - 200)} = \frac{1}{2 \times 600} \)
\( \Rightarrow \frac{200}{x^2 - 200x} = \frac{1}{1200} \)
\( \Rightarrow x^2 - 200x = 240000 \)
\( \Rightarrow x^2 - 200x - 240000 = 0 \)
Here, \( a = 1, b = -200 \) and \( c = -240000 \)
\( \therefore x = \frac{200 \pm \sqrt{40000 + 960000}}{2 \times 1} \)
\( = \frac{200 \pm \sqrt{1000000}}{2} \)
\( = \frac{200 \pm 1000}{2} \)
\( = \frac{200 + 1000}{2}, \frac{200 - 1000}{2} \)
\( = 600, -400 \)
Since, speed cannot be negative, therefore original speed \( = 600 \text{ km/h.} \)
and original distance \( = 600 \text{ km} \)
\( \therefore \text{Time} = \frac{\text{original distance}}{\text{original speed}} = \frac{600 \text{ km}}{600 \text{ km/hr.}} = 1 \text{ h} \)
Hence, the original duration of flight is 1 h.

Question. A train covers 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. Find the original speed of the train.
Answer: Let the speed of the train \( = x \text{ km/h} \)
Total distance covered by the train \( = 480 \text{ km} \)
\( \therefore \text{Time taken to cover the 480 km distance} = \frac{480}{x} \text{ h} \)
If the speed has decreased 8 km/h, i.e., \( (x - 8) \text{ km/h} \)
Then, time taken to cover the 480 km distance \( = \frac{480}{x - 8} \text{ h} \)
According to question,
\( \frac{480}{x - 8} - \frac{480}{x} = 3 \)
\( \Rightarrow 480 \left[ \frac{x - x + 8}{x(x - 8)} \right] = 3 \)
\( \Rightarrow \frac{8}{x^2 - 8x} = \frac{3}{480} = \frac{1}{160} \)
\( \Rightarrow x^2 - 8x - 1280 = 0 \)
Comparing it with \( ax^2 + bx + c = 0 \), we get \( a = 1, b = -8 \) and \( c = -1280 \)
\( x = \frac{8 \pm \sqrt{64 + 4 \times 1280}}{2 \times 1} = \frac{8 \pm \sqrt{5184}}{2} = \frac{8 \pm 72}{2} \)
\( = \frac{8 + 72}{2}, \frac{8 - 72}{2} = \frac{80}{2}, \frac{-64}{2} = 40, -32 \)
Since, negative speed cannot be possible. Hence, the original speed of the train \( = 40 \text{ km/h.} \)

Question. Solve for \( x \): \( \frac{1}{x + 4} - \frac{1}{x + 7} = \frac{11}{30} \), \( x \neq -4, -7 \).
Answer: Given, \( \frac{1}{x + 4} - \frac{1}{x + 7} = \frac{11}{30} \)
\( \Rightarrow \frac{x + 7 - x - 4}{(x + 4)(x + 7)} = \frac{11}{30} \)
\( \Rightarrow \frac{3}{x^2 + 4x + 7x + 28} = \frac{11}{30} \)
\( \Rightarrow \frac{3}{x^2 + 11x + 28} = \frac{11}{30} \)
\( \Rightarrow 11x^2 + 121x + 308 = 90 \)
\( \Rightarrow 11x^2 + 121x + 218 = 0 \)
Comparing it with \( ax^2 + bx + c = 0 \), we get \( a = 11, b = 121 \) and \( c = 218 \)
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-121 \pm \sqrt{14641 - 9592}}{22} \)
\( \Rightarrow x = \frac{-121 \pm \sqrt{5049}}{22} = \frac{-121 \pm 71.06}{22} \)
\( \Rightarrow x = \frac{-121 + 71.06}{22}, \frac{-121 - 71.06}{22} = \frac{-49.94}{22}, \frac{-192.06}{22} \)
\( \Rightarrow x = -2.27, -8.73 \).

Question. A fast train takes 3 hours less than a slow train for a journey of 600 km. If the speed of the slow train is 10 km/h less than that of the fast train, find the speed of each train.
Answer: Total distance of the journey \( = 600 \text{ km} \)
Let speed of fast train \( = x \text{ km/h} \),
then speed of slow train \( = (x - 10) \text{ km/h} \)
According to question, \( \frac{600}{x - 10} - \frac{600}{x} = 3 \)
\( \left[ \because \text{Time} = \frac{\text{Distance}}{\text{Speed}} \right] \)
\( \Rightarrow 600 \left[ \frac{x - x + 10}{x(x - 10)} \right] = 3 \)
\( \Rightarrow \frac{6000}{x^2 - 10x} = 3 \)
\( \Rightarrow x^2 - 10x - 2000 = 0 \)
\( \Rightarrow x^2 - 50x + 40x - 2000 = 0 \)
\( \Rightarrow x(x - 50) + 40(x - 50) = 0 \)
\( \Rightarrow (x - 50)(x + 40) = 0 \)
Either, \( x = 50 \) or \( x = -40 \)
As speed can not be negative. So, the speed of fast train \( = 50 \text{ km/h} \), and the speed of slow train \( = 50 - 10 = 40 \text{ km/h} \).

Question. Solve for \( x \): \( x^2 + 5x - (a^2 + a - 6) = 0 \)
Answer: Given, \( x^2 + 5x - (a^2 + a - 6) = 0 \)
Then, \( x = \frac{-5 \pm \sqrt{25 + 4(a^2 + a - 6)}}{2} \)
\( \left[ \because x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \right] \)
\( = \frac{-5 \pm (2a + 1)}{2} = \frac{2a - 4}{2} \) or \( \frac{-2a - 6}{2} \)
Thus, \( x = a - 2 \) or \( x = -(a + 3) \)
Alternate Solution:
\( x^2 + 5x - (a^2 + a - 6) = 0 \)
\( \Rightarrow x^2 + 5x - (a^2 + 3a - 2a - 6) = 0 \)
\( \Rightarrow x^2 + 5x - [a(a + 3) - 2(a + 3)] = 0 \)
\( \Rightarrow x^2 + 5x - (a + 3)(a - 2) = 0 \)
\( \Rightarrow x^2 + [(a + 3) - (a - 2)]x - (a + 3)(a - 2) = 0 \)
\( \Rightarrow x^2 + (a + 3)x - (a - 2)x - (a + 3)(a - 2) = 0 \)
\( \Rightarrow x[x + (a + 3)] - (a - 2) [x + (a + 3)] = 0 \)
\( \Rightarrow [x + (a + 3)] [x - (a - 2)] = 0 \)
\( \Rightarrow x = -(a + 3) \) or \( x = a - 2 \)
Hence, roots of given equation are \( -(a + 3) \) and \( (a - 2) \).

Question. Divide 27 into two parts such that the sum of their reciprocals is \( \frac{3}{20} \).
Answer: Let two parts be \( x \) and \( 27 - x \).
\( \therefore \frac{1}{x} + \frac{1}{27 - x} = \frac{3}{20} \)
\( \Rightarrow \frac{27 - x + x}{x(27 - x)} = \frac{3}{20} \)
\( \Rightarrow x^2 - 27x + 180 = 0 \)
\( \Rightarrow (x - 15)(x - 12) = 0 \)
\( \Rightarrow x = 12 \) or \( 15 \).
The two parts are 12 and 15.

Question. A plane left 30 minutes late then its scheduled time and in order to reach of 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: Let usual speed of the plane be \( x \) km/hr.
\( \therefore \frac{1500}{x} - \frac{1500}{x + 100} = \frac{30}{60} \)
\( \Rightarrow x^2 + 100x - 300000 = 0 \)
\( \Rightarrow x^2 + 600x - 500x - 300000 = 0 \)
\( \Rightarrow (x + 600)(x - 500) = 0 \)
\( x = -600 \) or \( x = 500 \)
(Rejecting negative value)
Speed of plane \( = 500 \text{ km/h} \)

Question. Solve for \( x \): \( \frac{x + 1}{x - 1} + \frac{x - 2}{x + 2} = 4 - \frac{2x + 3}{x - 2} \); where \( x \neq 1, -2, 2 \)
Answer: Here, \( \frac{x^2 + 3x + 2 + x^2 - 3x + 2}{x^2 + x - 2} = \frac{4x - 8 - 2x - 3}{x - 2} \)
\( \Rightarrow (2x^2 + 4)(x - 2) = (2x - 11)(x^2 + x - 2) \)
\( \Rightarrow 5x^2 + 19x - 30 = 0 \)
\( \Rightarrow (5x - 6)(x + 5) = 0 \)
\( \Rightarrow x = -5 \) or \( \frac{6}{5} \)

Question. Solve the following quadratic equation for \( x \): \( x^2 + \left( \frac{a}{a + b} + \frac{a + b}{a} \right)x + 1 = 0 \).
Answer: Here, \( x^2 + \frac{a}{a + b}x + \frac{a + b}{a}x + 1 = 0 \)
\( \Rightarrow x \left( x + \frac{a}{a + b} \right) + \frac{a + b}{a} \left( x + \frac{a}{a + b} \right) = 0 \)
\( \Rightarrow \left( x + \frac{a}{a + b} \right) \left( x + \frac{a + b}{a} \right) = 0 \)
\( \Rightarrow x = \frac{-a}{a + b} \) or \( x = \frac{-(a + b)}{a} \).

Question. Solve the following quadratic equation for \( x \): \( 9x^2 - 9(a + b)x + 2a^2 + 5ab + 2b^2 = 0 \)
Answer: Given, \( 9x^2 - 9(a + b)x + 2a^2 + 5ab + 2b^2 = 0 \)
First, we solve, \( 2a^2 + 5ab + 2b^2 = 2a^2 + 4ab + ab + 2b^2 \)
\( = 2a(a + 2b) + b(a + 2b) = (a + 2b)(2a + b) \)
Hence, the equation becomes \( 9x^2 - 9(a + b)x + (a + 2b)(2a + b) = 0 \)
\( \Rightarrow 9x^2 - 3[3a + 3b]x + (a + 2b)(2a + b) = 0 \)
\( \Rightarrow 9x^2 - 3[(a + 2b) + (2a + b)]x + (a + 2b)(2a + b) = 0 \)
\( \Rightarrow 3x[3x - (a + 2b)] - (2a + b) [3x - (a + 2b)] = 0 \)
\( \Rightarrow [3x - (a + 2b)][3x - (2a + b)] = 0 \)
\( \Rightarrow x = \frac{a + 2b}{3} \) or \( x = \frac{2a + b}{3} \)
Hence, the roots are \( \frac{a + 2b}{3}, \frac{2a + b}{3} \).

Question. Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60. Find the numbers.
Answer: Let the three consecutive natural numbers be \( x \), \( x + 1 \) and \( x + 2 \).
\( \therefore (x + 1)^2 = (x + 2)^2 - (x)^2 + 60 \)
\( \Rightarrow x^2 + 2x + 1 = x^2 + 4x + 4 - x^2 + 60 \)
\( \Rightarrow x^2 - 2x - 63 = 0 \)
\( \Rightarrow (x - 9)(x + 7) = 0 \)
Thus, \( x = 9 \) or \( x = -7 \)
Rejecting -7, we get \( x = 9 \). Hence, three numbers are 9, 10 and 11.

Question. P and Q are centres of circles of radii 9 cm and 2 cm respectively. PQ = 17 cm. R is the centre of the circle of radius \( x \) cm which touches given circles externally. Given that angle PRQ is 90°. Write an equation in \( x \) and solve it.
Answer: In right \( \Delta PQR \), by Pythagoras theorem \( PQ^2 = PR^2 + RQ^2 \)
\( \Rightarrow 17^2 = (x + 9)^2 + (x + 2)^2 \)
\( 289 = x^2 + 18x + 81 + x^2 + 4x + 4 \)
\( \Rightarrow 2x^2 + 22x - 204 = 0 \)
\( \Rightarrow x^2 + 11x - 102 = 0 \)
\( \Rightarrow (x + 17)(x - 6) = 0 \)
\( \Rightarrow x = 6 \) or \( x = -17 \) (as \( x \) can't be negative)
Thus, \( x = 6 \text{ cm} \).

Question. Solve the quadratic equation \( (x - 1)^2 - 5(x - 1) - 6 = 0 \)
Answer: \( (x - 1)^2 - 5(x - 1) - 6 = 0 \)
\( \Rightarrow x^2 - 2x + 1 - 5x + 5 - 6 = 0 \)
\( \Rightarrow x^2 - 7x = 0 \)
\( \Rightarrow x(x - 7) = 0 \)
\( \therefore x = 0 \) or 7.

Question. Write all the values of \( p \) for which the quadratic equation \( x^2 + px + 16 = 0 \) has equal roots. Find the roots of the equation so obtained.
Answer: \( x^2 + px + 16 = 0 \) have equal roots if \( D = p^2 - 4(16)(1) = 0 \)
\( p^2 = 64 \Rightarrow p = \pm 8 \)
when \( p = 8, x^2 + 8x + 16 = 0 \Rightarrow (x + 4)^2 = 0 \Rightarrow x = -4, -4 \)
when \( p = -8, x^2 - 8x + 16 = 0 \Rightarrow (x - 4)^2 = 0 \Rightarrow x = 4, 4 \)
Hence, the required roots are either -4, -4 or 4, 4.

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: Given \( A = (a^2 + b^2), B = -2(ac + bd), C = (c^2 + d^2) \).
As roots are equal, \( D = B^2 - 4AC = 0 \).
\( [-2(ac + bd)]^2 = 4(a^2 + b^2)(c^2 + d^2) \)
\( 4(a^2c^2 + 2abcd + b^2d^2) = 4(a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2) \)
\( 2abcd = a^2d^2 + b^2c^2 \)
\( a^2d^2 + b^2c^2 - 2abcd = 0 \)
\( (ad - bc)^2 = 0 \Rightarrow ad - bc = 0 \Rightarrow ad = bc \)
\( \Rightarrow \frac{a}{b} = \frac{c}{d} \). Hence, proved.

Question. If \( ad \neq bc \), then prove that the equation \( (a^2 + b^2)x^2 + 2(ac + bd)x + (c^2 + d^2) = 0 \) has no real roots.
Answer: \( D = B^2 - 4AC = [2(ac + bd)]^2 - 4(a^2 + b^2)(c^2 + d^2) \)
\( = 4[a^2c^2 + 2abcd + b^2d^2 - a^2c^2 - a^2d^2 - b^2c^2 - b^2d^2] \)
\( = 4[2abcd - a^2d^2 - b^2c^2] = -4[a^2d^2 + b^2c^2 - 2abcd] = -4(ad - bc)^2 \)
Since \( ad \neq bc \), therefore \( (ad - bc)^2 > 0 \).
Thus \( D < 0 \). Hence, the equation has no real roots.

Question. If 2 is a root of the quadratic equation \( 3x^2 + px - 8 = 0 \) and the quadratic equation \( 4x^2 - 2px + k = 0 \) has equal roots, find \( k \).
Answer: Putting \( x = 2 \) in \( 3x^2 + px - 8 = 0 \), we get \( 12 + 2p - 8 = 0 \Rightarrow p = -2 \).
The second equation is \( 4x^2 + 4x + k = 0 \).
For equal roots, \( D = b^2 - 4ac = 0 \).
\( (4)^2 - 4(4)(k) = 0 \Rightarrow 16 - 16k = 0 \Rightarrow k = 1 \).

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: Comparing given equation with \( Ax^2 + Bx + C = 0 \), we have \( A = a - b, B = b - c, C = c - a \).
Since roots are equal, \( D = B^2 - 4AC = 0 \).
\( (b - c)^2 - 4(a - b)(c - a) = 0 \)
\( b^2 + c^2 - 2bc - 4(ac - a^2 - bc + ab) = 0 \)
\( b^2 + c^2 - 2bc - 4ac + 4a^2 + 4bc - 4ab = 0 \)
\( 4a^2 + b^2 + c^2 - 4ab + 2bc - 4ac = 0 \)
\( (-2a + b + c)^2 = 0 \)
\( -2a + b + c = 0 \Rightarrow 2a = b + c \). Hence, proved.

Long Answer Type Questions

(5 Marks Each)

Question. Solve the following equation: \( \frac{1}{x} - \frac{1}{x - 2} = 3 \), \( x \neq 0, 2 \)
Answer: \( \frac{1}{x} - \frac{1}{x - 2} = 3 \Rightarrow \frac{x - 2 - x}{x(x - 2)} = 3 \)
\( \Rightarrow -2 = 3x^2 - 6x \Rightarrow 3x^2 - 6x + 2 = 0 \)
Comparing with \( ax^2 + bx + c = 0 \), we get \( a = 3, b = -6, c = 2 \).
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(2)}}{2(3)} \)
\( = \frac{6 \pm \sqrt{36 - 24}}{6} = \frac{6 \pm \sqrt{12}}{6} = \frac{6 \pm 2\sqrt{3}}{6} = \frac{3 \pm \sqrt{3}}{3} \).
Hence, \( x = \frac{3 + \sqrt{3}}{3}, \frac{3 - \sqrt{3}}{3} \).

Question. A train covers a distance of 360 km at a uniform speed. Had the speed been 5 km/h more, it would have taken 48 minutes less for the journey find the original speed of the train.
Answer: Let original speed of the train be \( x \) km/h. \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \).
Time taken for 360 km, \( t_1 = \frac{360}{x} \).
When speed increased by 5 km/h, \( t_2 = \frac{360}{x + 5} \).
Given \( t_1 - t_2 = 48 \text{ minutes} = \frac{48}{60} = \frac{4}{5} \text{ h.} \)
\( \Rightarrow \frac{360}{x} - \frac{360}{x + 5} = \frac{4}{5} \Rightarrow 360 \left( \frac{x + 5 - x}{x(x + 5)} \right) = \frac{4}{5} \)
\( \Rightarrow \frac{1800}{x^2 + 5x} = \frac{4}{5} \Rightarrow \frac{450}{x^2 + 5x} = \frac{1}{5} \)
\( \Rightarrow x^2 + 5x - 2250 = 0 \Rightarrow (x + 50)(x - 45) = 0 \).
\( x = 45 \) (as speed cannot be negative). Hence, original speed \( = 45 \text{ km/h} \).

Question. Two water taps together can fill a tank in \( 1 \frac{7}{8} \) hours. The tap with longer diameter takes 2 hours less than the tap with smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.
Answer: Let time taken by smaller tap \( = x \) h. Then larger tap \( = (x - 2) \) h.
Total time taken together \( = \frac{15}{8} \) h.
In 1 hour, portion filled \( = \frac{1}{x} + \frac{1}{x - 2} = \frac{8}{15} \)
\( \Rightarrow \frac{x - 2 + x}{x(x - 2)} = \frac{8}{15} \Rightarrow 15(2x - 2) = 8x(x - 2) \)
\( \Rightarrow 30x - 30 = 8x^2 - 16x \Rightarrow 8x^2 - 46x + 30 = 0 \)
\( \Rightarrow 4x^2 - 23x + 15 = 0 \Rightarrow 4x^2 - 20x - 3x + 15 = 0 \)
\( \Rightarrow (4x - 3)(x - 5) = 0 \Rightarrow x = 5 \) (neglecting \( x = 3/4 \) as time cannot be negative/invalid).
So, smaller tap takes 5 h and larger tap takes 3 h.

Question. In a class test, the sum of Arun’s marks in Hindi and English is 30. Had he got 2 marks more in Hindi and 3 marks less in English, the product of the marks would have been 210. Find his marks in the two subjects.
Answer: Let marks in Hindi be \( x \). Then marks in English \( = 30 - x \).
According to question, \( (x + 2)(30 - x - 3) = 210 \)
\( \Rightarrow (x + 2)(27 - x) = 210 \Rightarrow 27x - x^2 + 54 - 2x = 210 \)
\( \Rightarrow x^2 - 25x + 156 = 0 \Rightarrow (x - 12)(x - 13) = 0 \)
Either \( x = 12 \) or \( x = 13 \).
If Hindi \( = 12 \), English \( = 18 \). If Hindi \( = 13 \), English \( = 17 \).
Hence, marks are (12, 18) or (13, 17).

Question. The total cost of a certain length of a piece of cloth is ₹ 200. If the piece was 5 m longer and each metre of cloth costs ₹ 2 less, the cost of the piece would have remained unchanged. How long is the piece and what is its original rate per metre ?
Answer: Let total length of cloth \( = l \) m. Rate per metre \( = \frac{200}{l} \).
According to question, \( (l + 5) \left( \frac{200}{l} - 2 \right) = 200 \)
\( \Rightarrow (l + 5)(200 - 2l) = 200l \)
\( \Rightarrow 200l - 2l^2 + 1000 - 10l = 200l \)
\( \Rightarrow l^2 + 5l - 500 = 0 \Rightarrow (l + 25)(l - 20) = 0 \)
\( \Rightarrow l = 20 \) (neglecting negative). Length \( = 20 \text{ m} \).
Original rate \( = \frac{200}{20} = ₹ 10 \text{ per metre.} \)

Question. A shopkeeper buy certain number of books in ₹ 80. If he buy 4 more books then new cost price of each book is reduced by ₹ 1. Find the number of books initially he buy.
Answer: Let the no. of books bought be '\( n \)'. Total money spent \( = ₹ 80 \).
Cost of each book \( = \frac{80}{n} \).
New no. of books \( = n + 4 \). New cost of each book \( = \frac{80}{n + 4} \).
Given: \( \frac{80}{n} - \frac{80}{n + 4} = 1 \Rightarrow 80 \left( \frac{1}{n} - \frac{1}{n + 4} \right) = 1 \)
\( \Rightarrow \frac{n + 4 - n}{n(n + 4)} = \frac{1}{80} \Rightarrow n^2 + 4n - 320 = 0 \)
\( \Rightarrow n = \frac{-4 \pm \sqrt{16 + 1280}}{2} = \frac{-4 \pm 36}{2} \)
\( n = \frac{32}{2} = 16 \) (neglecting negative). Initial number of books \( = 16 \).