CBSE Class 10 Maths HOTs Quadratic Equations Set I

Very Short Answer Type Questions

Question. If \((x^2 + y^2)(a^2 + b^2) = (ax + by)^2\). Prove that \(\frac{x}{a} = \frac{y}{b}\).
Answer: We have \((x^2 + y^2)(a^2 + b^2) = (ax + by)^2\)
\(x^2a^2 + x^2b^2 + y^2a^2 + y^2b^2 = a^2x^2 + b^2y^2 + 2abxy\)
\(x^2b^2 + y^2a^2 - 2abxy = 0\)
\((xb - ya)^2 = 0\)
\(xb = ya\)
Thus \(\frac{x}{a} = \frac{y}{b}\) Hence Proved.

Question. \[ \frac{7x+1}{x^2-1} = \frac{29}{4x-1} \]
Answer: \( (7x+1)(4x-1) = 29(x^2 - 29) \)
\( 28x^2 - 7x + 4x - 1 = 29x^2 - 29 \)
\( -3x = x^2 - 28 \)
\( x^2 + 3x - 28 = 0 \)
\( x^2 + 7x - 4x - 28 = 0 \)
\( x(x + 7) - 4(x + 7) = 0 \)
\( (x + 7)(x - 4) = 0 \)
Hence, \( x = 4, -7 \)

Question. Find for \( x \) : \( \frac{1}{x-2} + \frac{2}{x-1} = \frac{6}{x} \); \( x \neq 0, 1, 2 \)
Answer: [Board Outside Delhi Compt. Set-I, II 2017]
We have \( \frac{1}{x-2} + \frac{2}{x-1} = \frac{6}{x} \)
\( \frac{x-1 + 2x - 4}{(x-2)(x-1)} = \frac{6}{x} \)
\[3x^2 - 5x = 6x^2 - 18x + 12\] \[3x^2 - 13x + 12 = 0\] \[3x^2 - 4x - 9x + 12 = 0\] \[x(3x - 4) - 3(3x - 4) = 0\] \[(3x - 4)(x - 3) = 0\] \[x = \frac{4}{3} \text{ and } 3\] Hence, \(x = 3, \frac{4}{3}\)

Question. The perimeter of a rectangular field is 82 m and its area is 400 square metre. Find the length and breadth of the rectangle.
Answer: We have Perimeter = \(2(l + b) = 82 \text{ m}\)
or, \(l + b = 41 \text{ m}\)
Let length be \(x \text{ m}\), then breadth = \((41 - x) \text{ m}\).

Question. Find the values of \(k\) for which the equation \((3k + 1)x^2 + 2(k + 1)x + 1 = 0\) has equal roots. Also find the roots.
Answer: We have \((3k + 1)x^2 + 2(k + 1)x + 1 = 0\)
Compare with \(Ax^2 + Bx + C = 0\) we get
\(A = (3k + 1), B = 2(k + 1), C = 1\)

Question. Find the positive root of \(\sqrt{3x^2 + 6} = 9\).
Answer: We have \(\sqrt{3x^2 + 6} = 9\)
\(3x^2 + 6 = 81\)
\(3x^2 = 81 - 6 = 75\)
\(x^2 = \frac{75}{3} = 25\)
Thus \(x = \pm 5\)
Hence 5 is positive root.

Question. If \(x = -\frac{1}{2}\), is a solution of the quadratic equation \(3x^2 + 2kx - 3 = 0\), find the value of \(k\).
Answer: We have \(3x^2 + 2kx - 3 = 0\)
Putting \(x = -\frac{1}{2}\) we get
\(3(-\frac{1}{2})^2 + 2k(-\frac{1}{2}) - 3 = 0\)
\(\frac{3}{4} - k - 3 = 0\)
\(k = \frac{3}{4} - 3\)
\(= \frac{3 - 12}{4} = -\frac{9}{4}\)
Hence \(k = -\frac{9}{4}\)

Question. Find the roots of the quadratic equation \(\sqrt{3}x^2 - 2x - \sqrt{3} = 0\).
Answer: We have \(\sqrt{3}x^2 - 2x - \sqrt{3} = 0\)
\(\sqrt{3}x^2 - 3x + x - \sqrt{3} = 0\)
\(\sqrt{3}x(x - \sqrt{3}) + 1(x - \sqrt{3}) = 0\)
\((x - \sqrt{3})(\sqrt{3}x + 1) = 0\)
Thus \(x = \sqrt{3}, -\frac{1}{\sqrt{3}}\)

Question. If one root of the quadratic equation \(6x^2 - x - k = 0\) is \(\frac{2}{3}\), then find the value of \(k\).
Answer: We have \(6x^2 - x - k = 0\)
Substituting \(x = \frac{2}{3}\), we get
\(6(\frac{2}{3})^2 - \frac{2}{3} - k = 0\)
\(6 \times \frac{4}{9} - \frac{2}{3} - k = 0\)
\(k = 6 \times \frac{4}{9} - \frac{2}{3} = \frac{24 - 6}{9} = \frac{18}{9} = 2\)
Thus \(k = 2\).

Question. Find the roots the quadratic equation \(6x^2 - x - 2 = 0\).
Answer: We have \(6x^2 - x - 2 = 0\)
\(6x^2 + 3x - 4x - 2 = 0\)
\(3x(2x + 1) - 2(2x + 1) = 0\)
\((2x + 1)(3x - 2) = 0\)
\(3x - 2 = 0\) or \(2x + 1 = 0\)
\(x = \frac{2}{3}\) or \(x = -\frac{1}{2}\)
Hence roots of equation are \(\frac{2}{3}\) and \(-\frac{1}{2}\).

Question. Find the roots of the following quadratic equation: \(15x^2 - 10\sqrt{6}x + 10 = 0\)
Answer: We have \(15x^2 - 10\sqrt{6}x + 10 = 0\)
\(3x^2 - 2\sqrt{6}x + 2 = 0\)
\(3x^2 - \sqrt{6}x - \sqrt{6}x + 2 = 0\)
\(\sqrt{3}x(\sqrt{3}x - \sqrt{2}) - \sqrt{2}(\sqrt{3}x - \sqrt{2}) = 0\)
\((\sqrt{3}x - \sqrt{2})(\sqrt{3}x - \sqrt{2}) = 0\)
Thus \(x = \frac{\sqrt{2}}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}}\)

Short Answer Type Questions

Question. Solve for \(x : \frac{3}{x+1} + \frac{4}{x-1} = \frac{29}{4x-1}; x = \neq -1, 1, \frac{1}{4}\)
Answer: We have \(\frac{3}{x+1} + \frac{4}{x-1} = \frac{29}{4x-1}\)
\(\frac{3x - 3 + 4x + 4}{x^2 - 1} = \frac{29}{4x - 1}\)
\(\frac{7x + 1}{x^2 - 1} = \frac{29}{4x - 1}\)
\((7x + 1)(4x - 1) = 29(x^2 - 1)\)
\(28x^2 - 7x + 4x - 1 = 29x^2 - 29\)
\(28x^2 - 3x - 1 = 29x^2 - 29\)
\(x^2 + 3x - 28 = 0\)
\(x^2 + 7x - 4x - 28 = 0\)
\(x(x + 7) - 4(x + 7) = 0\)
\((x + 7)(x - 4) = 0\)
\(x = -7\) or \(x = 4\)

Question. Two pipes running together can fill a tank in \( 11\frac{1}{9} \) minutes. If one pipe takes 5 minutes more than the other to fill the tank, find the time in which each pipe would fill the tank separately.
Answer: Let time taken by pipe A be \( x \) minutes and time taken by pipe B be \( x + 5 \) minutes.
In one minute pipe A will fill \( \frac{1}{x} \) tank.
In one minute pipe B will fill \( \frac{1}{x+5} \) tank.
Thus pipes \( A + B \) will fill \( \frac{1}{x} + \frac{1}{x+5} \) tank in one minute.
As per question, two pipes running together can fill a tank in \( 11\frac{1}{9} = \frac{100}{9} \) minutes, in one minute \( \frac{9}{100} \) tank will be filled.
Now according to the question we have
\( \frac{1}{x} + \frac{1}{x+5} = \frac{9}{100} \)
\( \frac{x+5+x}{x(x+5)} = \frac{9}{100} \)
\( 100(2x + 5) = 9x(x + 5) \)
\( 200x + 500 = 9x^2 + 45x \)
\( 9x^2 - 155x - 500 = 0 \)
\( 9x^2 - 180x + 25x - 500 = 0 \)
\( 9x(x - 20) + 25(x - 20) = 0 \)
\( (x - 20)(9x + 25) = 0 \)
\( x = 20, \frac{-25}{9} \)
As time can't be negative we take \( x = 20 \) minutes
and \( x + 5 = 25 \) minutes
Hence pipe A will fill the tank in 20 minutes and pipe B will fill it in 25 minutes.

Question. The time taken by a person to cover 150 km was \( 2\frac{1}{2} \) hours more than the time taken in the return journey. If he returned at a speed of 10 km/hour more than the speed while going, find the speed per hour in each direction.
Answer: Let the speed while going be \( x \) km/h
Speed while returning = \( (x + 10) \) km/h
According to question we have
\( \frac{150}{x} - \frac{150}{x+10} = \frac{5}{2} \)
\( x^2 + 10x - 600 = 0 \)
\( (x + 30)(x - 20) = 0 \)
\( x = 20 \)
Speed while going is 20 km/h and speed while returning will be \( 20 + 10 = 30 \) km/h

Question. The denominator of a fraction is one more than twice its numerator. If the sum of the fraction and its reciprocal is \( 2\frac{16}{21} \), find the fraction.
Answer: Let numerator be \( x \) then fraction will be \( \frac{x}{2x+1} \)
As per the question we have
\( \frac{x}{2x+1} + \frac{2x+1}{x} = 2\frac{16}{21} = \frac{58}{21} \)
\( 21[x^2 + (2x + 1)^2] = 58(2x^2 + x) \)
or, \( 11x^2 - 26x - 21 = 0 \)
\( 11x^2 - 33x + 7x - 21 = 0 \)
\( x = 3, -\frac{7}{11} \) (rejected)
We reject \( x = -\frac{7}{11} \), thus \( x = 3 \) and fraction will be \( \frac{3}{6+1} = \frac{3}{7} \)

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. Find the time in which each tap can separately fill the tank.
Answer: Let the tap with smaller diameter fills the tank in \( x \) hours, then the other tap fills the tank in \( (x - 8) \) hours.
In one hour small tap will fill \( \frac{1}{x} \) tank.
In one hour large tap will fill \( \frac{1}{x-8} \) tank.
Thus both tap will fill \( \frac{1}{x} + \frac{1}{x-8} \) tank in one hour.
9 hours 36 minutes \( = 9 + \frac{36}{60} = 9 + \frac{3}{5} = \frac{48}{3} \)
Since two water taps together can fill a tank in \( \frac{48}{5} \) hour, tank fill by both pipe in one hour is \( 1 / \frac{48}{5} = \frac{5}{48} \).
Thus \( \frac{1}{x} + \frac{1}{x-8} = \frac{5}{48} \)
\( \frac{x-8+x}{x(x-8)} = \frac{5}{48} \)
\( 5x(x - 8) = (2x - 8)48 \)
\( 5x^2 - 136x + 384 = 0 \)
\( x = \frac{136 \pm \sqrt{(136)^2 - 4 \times 5 \times 384}}{2 \times 5} \)
\( = \frac{136 \pm \sqrt{18496 - 7680}}{10} \)
\( x = \frac{136 \pm 104}{10} = 24, \frac{16}{5} \)
There is no possibility of \( x = \frac{16}{5} \) because it is less than 8 Hours.
Thus smaller tap can fill the tank in 24 hours and larger tap can fill in 16 hrs.

Question. The denominator of a fraction is two more than its numerator. If the sum of the fraction and its reciprocal is \( \frac{34}{15} \), find the fraction.
Answer: Let numerator be \( x \), then denominator will be \( x + 2 \).
and fraction \( = \frac{x}{x+2} \)
Now \( \frac{x}{x+2} + \frac{x+2}{x} = \frac{34}{15} \)
\( 15(x^2 + x^2 + 4x + 4) = 34(x^2 + 2x) \)
\( 30x^2 + 60x + 60 = 34x^2 + 68x \)
\( 4x^2 + 8x - 60 = 0 \)
\( x^2 + 2x - 15 = 0 \)
\( x^2 + 5x - 3x - 15 = 0 \)
\( x(x + 5) - 3(x + 5) = 0 \)
\( (x + 5)(x - 3) = 0 \)
We reject the \( x = -5 \). Thus \( x = 3 \) and fraction \( = \frac{3}{5} \)

Question. Solve for \( x \): \( \frac{2}{x+1} + \frac{3}{2(x-2)} = \frac{33}{5x} \); \( x \neq 0, -1, 2 \)
Answer: We have \( \frac{2}{x+1} + \frac{3}{2(x-2)} = \frac{23}{5x} \)
\( \frac{2 \times 2x(x-2) + 3x(x+1)}{2(x+1)(x-2)} = \frac{23}{5} \)
\( \frac{4x^2 - 8x + 3x^2 + 3x}{2(x^2 - x - 2)} = \frac{23}{5} \)
\( \frac{7x^2 - 5x}{2(x^2 - x - 2)} = \frac{23}{5} \)
\( 35x^2 - 25x = 46x^2 - 46x - 92 \)
\( 11x^2 - 21x - 92 = 0 \)
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{21 \pm \sqrt{(-21)^2 - 4(11)(-92)}}{2 \times 11} \)
\( = \frac{21 \pm \sqrt{441 + 4048}}{22} \)
\( = \frac{21 \pm 67}{22} \)
\( x = \frac{21+67}{22} \) or \( x = \frac{21-67}{22} \)
Thus \( x = 4, -\frac{23}{11} \)

Question. The numerator of a fraction is 3 less than its denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and original fraction is \( \frac{29}{20} \). Find the original fraction.
Answer: Let the denominator be \( x \), then numerator will \( x - 3 \)
So the fraction will be \( \frac{x-3}{x} \)
By the given condition, new fraction will \( \frac{x-3+2}{x+2} = \frac{x-1}{x+2} \)
Now \( \frac{x-3}{x} + \frac{x-1}{x+2} = \frac{29}{20} \)
\( 20[(x - 3)(x + 2) + x(x - 1)] = 29(x^2 + 2x) \)
\( 20(x^2 - x - 6 + x^2 - x) = 29x^2 + 58x \)
\( 20(2x^2 - 2x - 6) = 29x^2 + 58x \)
\( 40x^2 - 40x - 240 = 29x^2 + 58x \)
\( 11x^2 - 98x - 120 = 0 \)
\( 11x^2 - 110x + 12x - 120 = 0 \)
\( (11x + 20)(x - 10) = 0 \),
We take \( x = 10 \) and fraction will be \( \frac{10-3}{10} = \frac{7}{10} \).

Question. A train travels at a certain average speed for a distance of 54 km and then travels a distance of 63 km at an average speed of 6 km/h more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed ?
Answer: Let the speed of the train be \( x \) km/hr. for first 54 km. and for next 63 km, speed = \( (x + 6) \) km/hr.
According to the question
\( \frac{54}{x} + \frac{63}{x + 6} = 3 \)
\( \frac{54(x + 6) + 63x}{x(x + 6)} = 3 \)
\( 54x + 324 + 63x = 3x(x + 6) \)
\( 117x + 324 = 3x^2 + 18x \)
\( 3x^2 - 99x - 324 = 0 \)
\( x^2 - 33x - 108 = 0 \)
\( x^2 - 36x + 3x - 108 = 0 \)
\( x(x - 36) + 3(x - 36) = 0 \)
\( (x - 36)(x + 3) = 0 \)
\( x = -3, 36 \)
Negative value is rejected, thus first speed of train is 36 km/h.

Question. A truck covers a distance of 150 km at a certain average speed and then covers another 200 km at average speed which is 20 km per hour more than the first speed. If the truck covers the total distance in 5 hours, find the first speed of the truck.
Answer: Let the average speed of the truck be \( x \) km/hr. for first 150 km and for next 200 km, speed \( (x + 20) \) km/hr.
Now
\( \frac{150}{x} + \frac{200}{x + 20} = 5 \)
\( 150x + 3000 + 200x = 5x(x + 20) \)
\( x^2 - 50x - 600 = 0 \)
\( x^2 - 60x + 10x - 600 = 0 \)
\( x(x - 60) + 10(x - 60) = 0 \)
\( (x - 60)(x + 10) = 0 \)
Negative value is rejected, thus first speed of truck is 60 km/h.

Question. The total cost of a certain length of cloth is Rs 200. If the piece was 5 m longer and each metre of cloth coast Rs 2 less, the cost of the piece would have remained unchanged. How longer is the piece and what is its original rate per metre ?
Answer: Let the length of the cloth be \( x \) m.
cost per metre = \( \frac{200}{x} \)
New length of the cloth = \( (x + 5) \) m
New cost per metre = \( (\frac{200}{x} - 2) \)
Since cost of the piece have remained unchanged,
\( (x + 5)(\frac{200}{x} - 2) = 200 \)
\( \frac{(x + 5)(200 - 2x)}{x} = 200 \)
\( 200x - 2x^2 + 1000 - 10x = 200x \)
\( x^2 + 5x - 500 = 0 \)
\( (x + 25)(x - 20) = 0 \)
\( x = -25, 20 \)
Negative value is rejected, thus length of the piece is 20 m.
Original cost per metre is \( \frac{200}{20} = 10 \) Rs.

Question. A motorboat whose speed in still water is 18 km/h, takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
Answer: Let the speed of stream be \( x \) km/h
Then the speed of boat upstream = \( (18 - x) \) km/h
Speed of boat downstream = \( (18 + x) \) km/h
According to the question,
\( \frac{24}{18 - x} - \frac{24}{18 + x} = 1 \)
\( \frac{24[(18 + x) - (18 - x)]}{18^2 - x^2} = 1 \)
\( 48x = 324 - x^2 \)
\( x^2 + 48x - 324 = 0 \)
\( x^2 + 54x - 6x - 324 = 0 \)
\( x(x + 54) - 6(x + 54) = 0 \)
\( (x + 54)(x - 6) = 0 \)
\( x + 54 = 0, x - 6 = 0 \)
\( x = -54, x = 6 \)
Since speed cannot be negative, we reject \( x = -54 \).
The speed of stream is 6 km/h.

Question. Solve for \( x \) : \( \frac{x-3}{x-4} + \frac{x-5}{x-6} = \frac{10}{3} \); \( x \neq 4, 6 \)
Answer: We have \( \frac{x-3}{x-4} + \frac{x-5}{x-6} = \frac{10}{3} \)
\( \frac{(x-3)(x-6) + (x-4)(x-5)}{(x-4)(x-6)} = \frac{10}{3} \)
\( \frac{x^2 - 9x + 18 + x^2 - 9x + 20}{x^2 - 10x + 24} = \frac{10}{3} \)
\( 3(2x^2 - 18x + 38) = 10(x^2 - 10x + 24) \)
\( 6x^2 - 54x + 114 = 10x^2 - 100x + 240 \)
\( 4x^2 - 46x + 126 = 0 \)
\( 2x^2 - 23x + 63 = 0 \)
\( 2x^2 - 14x - 9x + 63 = 0 \)
\( 2x(x - 7) - 9(x - 7) = 0 \)
\( (2x - 9)(x - 7) = 0 \)
\( 2x - 9 = 0, x - 7 = 0 \)
\( x = \frac{9}{2}, x = 7 \)

Question. A motor boat 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: Let the speed of stream be \( x \) km/h
Then the speed of boat upstream = \( (24 - x) \) km/h
Speed of boat downstream = \( (24 + x) \) km/h
According to the question,
\( \frac{32}{24 - x} - \frac{32}{24 + x} = 1 \)
\( 32 \left[ \frac{1}{24 - x} - \frac{1}{24 + x} \right] = 1 \)
\( 32 \left[ \frac{24 + x - 24 + x}{576 - x^2} \right] = 1 \)
\( 32(2x) = 576 - x^2 \)
\( 64x = 576 - x^2 \)
\( x^2 + 64x - 576 = 0 \)
\( x^2 + 72x - 8x - 576 = 0 \)
\( x(x + 72) - 8(x + 72) = 0 \)
\( (x - 8)(x + 72) = 0 \)
\( x = 8, -72 \)
Since speed cannot be negative, we reject \( x = -72 \).
The speed of steam is 8 km/h.

Question. A student scored a total of 32 marks in class tests in mathematics ans science. Had he scored 2 marks less in science and 4 more in mathematics, the product of his marks would have been 253. Find his marks in two subjects.
Answer: Let marks in Mathematics be \( x \), then marks in science will be \( 32 - x \)
According to question,
\( (32 - x - 2)(x + 4) = 253 \)
\( (30 - x)(x + 4) = 253 \)
\( 26x - x^2 + 120 = 253 \)
\( x^2 - 26x + 133 = 0 \)
\( x^2 - 19x - 7x + 133 = 0 \)
\( x(x - 19) - 7(x - 19) = 0 \)
or, \( x = 7 \) or \( x = 19 \)
If \( x = 7 \) then marks in mathematics = 7, and marks in science = 25
If \( x = 19 \), then marks in mathematics = 19 and marks in science = 13.

Question. The sum of squares of two consecutive multiples of 7 is 637. Find the multiples.
Answer: Let \( 7x \) and \( 7x + 7 \) be two consecutive multiples of 7.
According to question,
\( (7x)^2 + (7x + 7)^2 = 637 \)
\( 49x^2 + 49x^2 + 49 + 98x = 637 \)
\( 98x^2 + 98x - 588 = 0 \)
\( x^2 + x - 6 = 0 \)
\( (x + 3)(x - 2) = 0 \)
\( x = -3, 2 \)
Neglecting negative value, \( x = 2 \)
Therefore multiples are 14 and 21.

Question. Solve for \( x \) : \( \frac{x-1}{2x+1} + \frac{2x+1}{x-1} = 2 \) where \( x \neq -\frac{1}{2}, 1 \)
Answer: We have \( \frac{x-1}{2x+1} + \frac{2x+1}{x-1} = 2 \)
Let \( \frac{x-1}{2x+1} \) be \( y \) so \( \frac{2x+1}{x-1} = \frac{1}{y} \)
Substituting this value we obtain
\( y + \frac{1}{y} = 2 \)
\( y^2 + 1 = 2y \)
\( y^2 - 2y + 1 = 0 \)
\( (y - 1)^2 = 0 \)
\( y = 1 \)
Putting \( y = \frac{x-1}{2x+1} \) we have
\( \frac{x-1}{2x+1} = 1 \) or \( x - 1 = 2x + 1 \)
or \( x = -2 \)

Question. The difference of two numbers is 5 and the difference of their reciprocals is \(\frac{1}{10}\). Find the numbers
Answer: Let the first number be \(x\), then second number will be \(x + 5\)
Now according to the question
\(\frac{1}{x} - \frac{1}{x+5} = \frac{1}{10}\)
\(\frac{x+5-x}{x(x+5)} = \frac{1}{10}\)
\(50 = x^2 + 5x\)
\(x^2 + 5x - 50 = 0\)
\(x^2 + 10x - 5x - 50 = 0\)
\(x(x + 10) - 5(x + 10) = 0\)
\((x + 10)(x - 5) = 0\)
\(x = 5, -10\)
Rejecting the negative value, numbers are 5 and 10.

Question. The sum of squares of two consecutive even numbers is 340. Find the numbers.
Answer: Let the number be \(x\) and \(x + 2\)
Now \((x)^2 + (x + 2)^2 = 340\)
\(x^2 + x^2 + 4 + 4x = 340\)
\(2x^2 + 4x - 336 = 0\)
\(x^2 + 2x - 168 = 0\)
\((x + 14)(x - 12) = 0\)
\(x = 12\)
Thus numbers are 12 and 14.

Question. The sum of the squares of two consecutive odd numbers is 394. Find the numbers.
Answer: Let the odd number be \(2x + 1\), then consecutive odd number will be \(2x + 1 + 2 = 2x + 3\)
Now, according to question
\((2x + 1)^2 + (2x + 3)^2 = 394\)
\(4x^2 + 4x + 1 + 4x^2 + 12x + 9 = 394\)
\(8x^2 + 16x - 384 = 0\)
\(x^2 + 2x - 48 = 0\)
\(x^2 + 8x - 6x - 48 = 0\)
\(x(x + 8) - 6(x + 8) = 0\)
\(x = -8, 6\)
Rejecting the negative value,
Ist number = \(2 \times 6 + 1 = 13\)
and second odd number = 15

Question. Sum of the areas of two squares is \(400\text{ cm}^2\). If the difference of their perimeters is 16 cm, find the sides of the two squares.
Answer: Let the sides of two squares be \(a\) and \(b\),
then \(a^2 + b^2 = 400\) ... (1)
and \(4(a - b) = 16\)
\(a - b = 4\)
\(a = 4 + b\) ... (2)
From equations (1) and (2), we obtain
\((4 + b)^2 + b^2 = 400\)
\(16 + b^2 + 8b + b^2 = 400\)
\(2b^2 + 8b - 384 = 0\)
\(b^2 + 4b - 192 = 0\)
\(b^2 + 16b - 12b - 192 = 0\)
\(b(b + 16) - 12(b + 16) = 0\)
\((b + 16)(b - 12) = 0\)
\(b = -16, 12\)
Rejecting the negative value, \(b = 12 \text{ cm}\)
then \(a = 4 + 12 = 16 \text{ cm}\).

Question. A takes 6 days less than the time taken by B to finish a piece of work. If both A and B together can finish it in 4 days, find the time taken by B to finish the work.
Answer: Suppose B alone finish the work in \(x\) days and A alone takes \((x - 6)\) days.
B’s one day work = \(\frac{1}{x}\)
A’s one day work = \(\frac{1}{x - 6}\)
and (A+B)’s one day work = \(\frac{1}{4}\)
According to the question,
\(\frac{1}{x} + \frac{1}{x - 6} = \frac{1}{4}\)
\(x^2 - 14x + 24 = 0\)
\(x^2 - 12x - 2x + 24 = 0\)
\(x(x - 12) - 2(x - 12) = 0\)
\((x - 12)(x - 2) = 0\)
\(x = 12 \text{ or } x = 2\)
But \(x\) cannot be less than 6. So \(x = 12\)
Hence B can finish the work in 12 days.

Question. The product of Tanay’s age (in years) five years ago and his age ten years later is 16. Determine Tanay’s present age.
Answer: Let the present age of Tanay’s be \( x \) years.
According to question we have
\( (x - 5)(x + 10) = 16 \)
\( x^2 + 5x - 50 = 16 \)
\( x^2 + 5x - 66 = 0 \)
\( x^2 + 11x - 6x - 66 = 0 \)
\( x(x + 11) - 6(x + 11) = 0 \)
\( (x + 11)(x - 6) = 0 \)
\( x = -11, 6. \)
As age cannot be negative, we reject \( x = -11 \). Thus present age of Tanay is 6 years.

Question. Solve for \( x \) : \( \frac{x+3}{x-2} - \frac{1-x}{x} = \frac{17}{4} \); \( x \neq 0, 2 \)
Answer: We have
\( \frac{x+3}{x-2} - \frac{1-x}{x} = \frac{17}{4} \)
\( \frac{x(x+3) - (1-x)(x-2)}{x(x-2)} = \frac{17}{4} \)
\( \frac{(x^2+3x) - (-x^2+3x-2)}{x^2-2x} = \frac{17}{4} \)
\( \frac{2x^2+2}{x^2-2x} = \frac{17}{4} \)
\( 8x^2 + 8 = 17x^2 - 34x \)
\( 9x^2 - 34x - 8 = 0 \)
\( 9x^2 - 36x + 2x - 8 = 0 \)
\( 9x(x - 4) + 2(x - 4) = 0 \)
\( (x - 4)(9x + 2) = 0 \)
\( x = 4 \) or \( x = -\frac{2}{9} \)
Hence, \( x = 4, -\frac{2}{9} \)

Question. Solve for \( x \) : \( 4x^2 + 4bx - (a^2 - b^2) = 0 \)
Answer: We have \( 4x^2 + 4bx - (a^2 - b^2) = 0 \)
Compare with \( Ax^2 + Bx + C = 0 \) we get
\( A = 4, B = 4b \) and \( C = -(a^2 - b^2) = b^2 - a^2 \)
\( x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \)
\( = \frac{-4b \pm \sqrt{(4b)^2 - 4 \cdot 4(b^2 - a^2)}}{2 \cdot 4} \)
\( = \frac{-4b \pm \sqrt{16b^2 - 16b^2 + 16a^2}}{8} \)
\( = \frac{-4b \pm \sqrt{16a^2}}{8} \)
\( = \frac{-4b \pm 4a}{8} \)
\( = -\frac{(a+b)}{2}, \frac{(a-b)}{2} \)
Hence the roots are \( -\frac{(a+b)}{2} \) and \( \frac{(a-b)}{2} \)

Question. Find \( k \) so that the quadratic equation \( (k+1)x^2 - 2(k+1)x + 1 = 0 \) has equal roots.
Answer: We have \( (k+1)x^2 - 2(k+1)x + 1 = 0 \)
Compare with \( Ax^2 + Bx + C = 0 \) we get
\( A = (k+1), B = -2(k+1), C = 1 \)
If roots are equal, \( D = 0 \), i.e.
\( B^2 = 4AC \)
\( 4(k+1)^2 = 4(k+1) \)
\( k^2+2k+1 = k+1 \)
\( k^2+k = 0 \)
\( k(k+1) = 0 \)
\( k = 0, -1 \)
\( k = -1 \) does not satisfy the equation, thus \( k = 0 \)

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: We have \( x^2 + kx + 12 = 0 \)
If 2 is the root of above equation, it must satisfy it.
\( (2)^2 + 2k + 12 = 0 \)
\( 2k + 16 = 0 \)
\( k = -8 \)
Substituting \( k = -8 \) in \( x^2 + kx + q = 0 \) we have
\( x^2 - 8x + q = 0 \)
For equal roots,
\( (-8)^2 - 4(1)q = 0 \)
\( 64 - 4q = 0 \)
\( 4q = 64 \)
\( q = 16 \)

Question. Find the values of \( k \) for which the quadratic equation \( 9x^2 - 3kx + k = 0 \) has equal roots.
Answer: We have \( 9x^2 - 3kx + k = 0 \)
Compare with \( ax^2 + bx + c = 0 \) we get
\( a = 9, b = -3k, c = k \)
Since roots of the equation are equal, \( b^2 - 4ac = 0 \)
\( (-3k)^2 - (4 \times 9 \times k) = 0 \)
\( 9k^2 - 36k = 0 \)
\( k^2 - 4k = 0 \)
\( k(k - 4) = 0 \)
\( k = 0 \) or \( k = 4 \)
Hence, \( k = 4 \).

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: We have \( 3x^2 + px - 8 = 0 \)
Since 2 is a root of above equation, it must satisfy it.
Substituting \( x = 2 \) in \( 3x^2 + px - 8 = 0 \) we have
\( 12 + 2p - 8 = 0 \)
\( p = -2 \)
Since \( 4x^2 - 2px + k = 0 \) has equal roots,
or \( 4x^2 + 4x + k = 0 \) has equal roots,
\( D = b^2 - 4ac = 0 \)
\( 4^2 - 4(4)(k) = 0 \)
\( 16 - 16k = 0 \)
\( 16k = 16 \)
Thus \( k = 1 \)