CBSE Class 10 Maths HOTs Quadratic Equations Set 19

Question. Find the values of \( h \) and \( k \) for which \( x = -2 \) and \( x = 3/4 \) are solution of the equation \( hx^2 + kx - 6 = 0 \).
Answer: Product of roots \( \frac{-6}{h} = (-2)(\frac{3}{4}) = -\frac{3}{2} \implies h = 4 \).
Sum of roots \( -\frac{k}{h} = -2 + \frac{3}{4} = -\frac{5}{4} \implies \frac{k}{4} = \frac{5}{4} \implies k = 5 \).

Question. Determine the set of values of \( p \) for which the given quadratic equation has real roots:
(i) \( 4x^2 + 8x - p = 0 \)
(ii) \( 4x^2 - 3px + 9 = 0 \).
Answer: (i) \( 8^2 - 4(4)(-p) \geq 0 \implies 64 + 16p \geq 0 \implies p \geq -4 \).
(ii) \( (-3p)^2 - 4(4)(9) \geq 0 \implies 9p^2 - 144 \geq 0 \implies p^2 \geq 16 \implies p \geq 4 \) or \( p \leq -4 \).

Question. A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hour more, it would have taken 30 minutes less for the journey. Find the original speed of the train.
Answer: Let speed = \( v \). \( \frac{90}{v} - \frac{90}{v + 15} = 0.5 \implies \frac{90(15)}{v(v + 15)} = 0.5 \)
\( \implies 2700 = v^2 + 15v \implies v^2 + 15v - 2700 = 0 \implies (v + 60)(v - 45) = 0 \).
Original speed = 45 km/h.

Question. The difference of squares of two natural numbers is 45. The square of the smaller number is four times the larger number. Find the numbers.
Answer: Let larger number be \( x \), smaller be \( y \). \( x^2 - y^2 = 45 \) and \( y^2 = 4x \).
\( \implies x^2 - 4x - 45 = 0 \implies (x - 9)(x + 5) = 0 \implies x = 9 \).
\( y^2 = 36 \implies y = 6 \). Numbers are 9 and 6.

Question. A two digit number is 4 times the sum of its digits and twice the product of its digits. Find the number.
Answer: Let number be \( 10x + y \).
\( 10x + y = 4(x + y) \implies 6x = 3y \implies y = 2x \).
Also \( 10x + y = 2xy \implies 12x = 2x(2x) \implies 12x = 4x^2 \implies x = 3 \).
\( y = 6 \). Number is 36.

Question. An aeroplane takes one hour less for a journey of 1200 km if its speed is increased by 100 km/hour from its usual speed. Find its usual speed.
Answer: \( \frac{1200}{v} - \frac{1200}{v + 100} = 1 \implies \frac{120000}{v(v + 100)} = 1 \)
\( \implies v^2 + 100v - 120000 = 0 \implies (v + 400)(v - 300) = 0 \). Usual speed = 300 km/h.

Question. A two-digit number is 5 times the sum of its digits and is also equal to 5 more than twice the product of its digits. Find the number.
Answer: \( 10x + y = 5(x + y) \implies 5x = 4y \).
\( 10x + y = 2xy + 5 \). Substitute \( y = \frac{5x}{4} \):
\( 10x + \frac{5x}{4} = 2x(\frac{5x}{4}) + 5 \implies \frac{45x}{4} = \frac{5x^2}{2} + 5 \implies 45x = 10x^2 + 20 \implies 2x^2 - 9x + 4 = 0 \)
\( \implies (x - 4)(2x - 1) = 0 \implies x = 4, y = 5 \). Number is 45.

Question. Aeroplane left 30 minutes later than its scheduled time and in order to reach destination 1500 km away in time, it has to increase its speed by 250 km/h from its usual speed. Determine its usual speed.
Answer: \( \frac{1500}{v} - \frac{1500}{v + 250} = 0.5 \implies \frac{1500 \times 250}{v(v + 250)} = 0.5 \)
\( \implies 750000 = v^2 + 250v \implies v^2 + 250v - 750000 = 0 \implies (v + 1000)(v - 750) = 0 \). Usual speed = 750 km/h.

Question. A person on tour has Rs. 360 for his daily expenses. If he exceeds his tour programme by 4 days, he must cut down his daily expenses by Rs. 3 per day. Find the number of days of his tour programme.
Answer: Let days = \( d \). Expense/day = \( \frac{360}{d} \).
\( \frac{360}{d} - \frac{360}{d + 4} = 3 \implies 120(\frac{4}{d(d + 4)}) = 1 \)
\( \implies d^2 + 4d - 480 = 0 \implies (d + 24)(d - 20) = 0 \). Number of days = 20.

Question. An express train makes a run of 240 km at a certain speed. Another train whose speed is 12 km/hr less takes an hour longer to cover the same distance. Find the speed of the express train in km/hr.
Answer: Let the speed of the express train be \( x \) km/hr.
Time taken by express train = \( \frac{240}{x} \) hours.
Speed of the other train = \( (x - 12) \) km/hr.
Time taken by the other train = \( \frac{240}{x - 12} \) hours.
According to the question:
\( \frac{240}{x - 12} - \frac{240}{x} = 1 \)
\( \implies 240 \left[ \frac{x - (x - 12)}{x(x - 12)} \right] = 1 \)
\( \implies 240 \left[ \frac{12}{x^2 - 12x} \right] = 1 \)
\( \implies x^2 - 12x = 2880 \)
\( \implies x^2 - 12x - 2880 = 0 \)
\( \implies x^2 - 60x + 48x - 2880 = 0 \)
\( \implies x(x - 60) + 48(x - 60) = 0 \)
\( \implies (x - 60)(x + 48) = 0 \)
\( \implies x = 60 \) or \( x = -48 \).
Since speed cannot be negative, the speed of the express train is 60 km/hr.

Question. The difference of mother’s age and her daughter’s age is 21 years and the twelfth part of the product of their ages is less than the mother’s age by 18 years. Find their ages.
Answer: Let the mother's age be \( x \) years.
Then, daughter's age = \( (x - 21) \) years.
According to the question:
\( \frac{1}{12}[x(x - 21)] = x - 18 \)
\( \implies x^2 - 21x = 12(x - 18) \)
\( \implies x^2 - 21x = 12x - 216 \)
\( \implies x^2 - 33x + 216 = 0 \)
\( \implies x^2 - 24x - 9x + 216 = 0 \)
\( \implies x(x - 24) - 9(x - 24) = 0 \)
\( \implies (x - 24)(x - 9) = 0 \)
\( \implies x = 24 \) or \( x = 9 \).
Since a mother's age cannot be 9 if the difference is 21, \( x = 24 \).
Mother's age = 24 years, Daughter's age = \( 24 - 21 = 3 \) years.

Question. One side of a rectangle is 21 cm longer than the other. The diagonal is longer than bigger side by 6 cm. Find the area of the rectangle.
Answer: Let the smaller side be \( x \) cm.
Bigger side = \( (x + 21) \) cm.
Diagonal = \( (x + 21) + 6 = (x + 27) \) cm.
Using Pythagoras Theorem:
\( x^2 + (x + 21)^2 = (x + 27)^2 \)
\( \implies x^2 + x^2 + 42x + 441 = x^2 + 54x + 729 \)
\( \implies x^2 - 12x - 288 = 0 \)
\( \implies x^2 - 24x + 12x - 288 = 0 \)
\( \implies x(x - 24) + 12(x - 24) = 0 \)
\( \implies (x - 24)(x + 12) = 0 \)
\( \implies x = 24 \) (neglecting negative value).
Sides are 24 cm and \( 24 + 21 = 45 \) cm.
Area = \( 24 \times 45 = 1080 \text{ cm}^2 \).

Question. A person on tour has Rs. 4200 for his expenses. If he extends his tour for 3 days he has to cut down his daily expenses by Rs. 70. Find the duration of the tour.
Answer: Let the original duration be \( x \) days.
Daily expense = \( \frac{4200}{x} \).
New duration = \( (x + 3) \) days.
New daily expense = \( \frac{4200}{x + 3} \).
According to the question:
\( \frac{4200}{x} - \frac{4200}{x + 3} = 70 \)
\( \implies 4200 \left[ \frac{x + 3 - x}{x(x + 3)} \right] = 70 \)
\( \implies \frac{4200 \times 3}{x^2 + 3x} = 70 \)
\( \implies \frac{180}{x^2 + 3x} = 1 \)
\( \implies x^2 + 3x - 180 = 0 \)
\( \implies x^2 + 15x - 12x - 180 = 0 \)
\( \implies x(x + 15) - 12(x + 15) = 0 \)
\( \implies (x - 12)(x + 15) = 0 \)
\( \implies x = 12 \).
Duration of the tour was 12 days.

Question. A motor boat whose speed is 20 km/h in still water, takes 1 hour more to go 48 km upstream than to return downstream to the same spot. Find the speed of the stream.
Answer: Let the speed of the stream be \( y \) km/h.
Upstream speed = \( (20 - y) \) km/h.
Downstream speed = \( (20 + y) \) km/h.
\( \frac{48}{20 - y} - \frac{48}{20 + y} = 1 \)
\( \implies 48 \left[ \frac{20 + y - (20 - y)}{400 - y^2} \right] = 1 \)
\( \implies 48 \left[ \frac{2y}{400 - y^2} \right] = 1 \)
\( \implies 96y = 400 - y^2 \)
\( \implies y^2 + 96y - 400 = 0 \)
\( \implies y^2 + 100y - 4y - 400 = 0 \)
\( \implies (y + 100)(y - 4) = 0 \)
\( \implies y = 4 \) km/h.

Question. Two pipes running together can fill a cistern in 6 minutes. If one pipe takes 5 minutes more than the other to fill the cistern, find the time in which each pipe would fill the cistern.
Answer: Let the first pipe take \( x \) minutes. Second pipe takes \( (x + 5) \) minutes.
In 1 minute, part filled = \( \frac{1}{x} + \frac{1}{x + 5} = \frac{1}{6} \)
\( \implies \frac{2x + 5}{x^2 + 5x} = \frac{1}{6} \)
\( \implies x^2 + 5x = 12x + 30 \)
\( \implies x^2 - 7x - 30 = 0 \)
\( \implies (x - 10)(x + 3) = 0 \)
\( \implies x = 10 \).
Pipes take 10 minutes and 15 minutes respectively.

Question. If the roots of the equation \( (c^2 - ab)x^2 - 2(a^2 - bc)x + (b^2 - ac) = 0 \) are equal, prove that either \( a = 0 \) or \( a^3 + b^3 + c^3 = 3abc \). [HOTS]
Answer: For equal roots, \( D = 0 \).
\( [2(a^2 - bc)]^2 - 4(c^2 - ab)(b^2 - ac) = 0 \)
\( \implies (a^4 + b^2c^2 - 2a^2bc) - (b^2c^2 - ac^3 - ab^3 + a^2bc) = 0 \)
\( \implies a^4 - 3a^2bc + ac^3 + ab^3 = 0 \)
\( \implies a(a^3 + b^3 + c^3 - 3abc) = 0 \)
\( \implies a = 0 \) or \( a^3 + b^3 + c^3 = 3abc \). Hence proved.

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 \( \frac{a}{b} = \frac{c}{d} \). [HOTS]
Answer: For equal roots, \( D = 0 \).
\( [2(ac + bd)]^2 - 4(a^2 + b^2)(c^2 + d^2) = 0 \)
\( \implies (a^2c^2 + b^2d^2 + 2abcd) - (a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2) = 0 \)
\( \implies 2abcd - a^2d^2 - b^2c^2 = 0 \)
\( \implies -(ad - bc)^2 = 0 \)
\( \implies ad = bc \implies \frac{a}{b} = \frac{c}{d} \). Hence proved.

Question. By a reduction of Rs. 1 per kg in the price of sugar Mohan can buy one kg sugar more for Rs. 56. Find the original price of sugar per kilogram.
Answer: Let original price be \( x \) Rs./kg.
\( \frac{56}{x - 1} - \frac{56}{x} = 1 \)
\( \implies 56 \left[ \frac{x - (x - 1)}{x(x - 1)} \right] = 1 \)
\( \implies \frac{56}{x^2 - x} = 1 \implies x^2 - x - 56 = 0 \)
\( \implies (x - 8)(x + 7) = 0 \implies x = 8 \).
Original price = Rs. 8/kg.

Question. If \( \alpha, \beta \) are roots of \( x^2 + 5x + a = 0 \) and \( 2\alpha + 5\beta = -1 \), then find the value of \( a \).
Answer: Sum of roots: \( \alpha + \beta = -5 \implies \alpha = -5 - \beta \).
Substitute in \( 2\alpha + 5\beta = -1 \):
\( 2(-5 - \beta) + 5\beta = -1 \implies -10 - 2\beta + 5\beta = -1 \implies 3\beta = 9 \implies \beta = 3 \).
Then \( \alpha = -5 - 3 = -8 \).
Product of roots: \( a = \alpha\beta = (-8)(3) = -24 \).

Question. Find the non-integral value of \( x \) so that \( 2^{2x^2 - 7x + 5} = 1 \).
Answer: \( 2^{2x^2 - 7x + 5} = 2^0 \)
\( \implies 2x^2 - 7x + 5 = 0 \)
\( \implies 2x^2 - 5x - 2x + 5 = 0 \)
\( \implies x(2x - 5) - 1(2x - 5) = 0 \)
\( \implies (x - 1)(2x - 5) = 0 \)
\( \implies x = 1, \frac{5}{2} \).
Non-integral value is \( \frac{5}{2} \).

Question. Solve for \( x \): \( x = \frac{1}{2 - \frac{1}{2 - \frac{1}{2 - x}}} \), \( x \neq 2 \).
Answer: Start from the bottom:
\( 2 - \frac{1}{2-x} = \frac{4-2x-1}{2-x} = \frac{3-2x}{2-x} \)
Then \( \frac{1}{\frac{3-2x}{2-x}} = \frac{2-x}{3-2x} \)
Then \( 2 - \frac{2-x}{3-2x} = \frac{6-4x-2+x}{3-2x} = \frac{4-3x}{3-2x} \)
Equation becomes: \( x = \frac{1}{\frac{4-3x}{3-2x}} = \frac{3-2x}{4-3x} \)
\( \implies 4x - 3x^2 = 3 - 2x \)
\( \implies 3x^2 - 6x + 3 = 0 \)
\( \implies x^2 - 2x + 1 = 0 \)
\( \implies (x - 1)^2 = 0 \implies x = 1 \).

Question. Solve for \( x \): \( 6\sqrt{\frac{x}{x + 4}} - 2\sqrt{\frac{x + 4}{x}} = 11 \), \( x \neq 0, -4 \).
Answer: Let \( \sqrt{\frac{x}{x + 4}} = y \). Then \( 6y - \frac{2}{y} = 11 \).
\( \implies 6y^2 - 11y - 2 = 0 \)
\( \implies 6y^2 - 12y + y - 2 = 0 \implies (6y + 1)(y - 2) = 0 \).
Since \( y > 0 \), \( y = 2 \).
\( \implies \frac{x}{x + 4} = 4 \implies x = 4x + 16 \implies 3x = -16 \implies x = -\frac{16}{3} \).

Question. One fourth of a group of people claim they are creative, twice the square root of the group claims to be caring and the remaining 15 claims they are optimistic
(i) Find the total number of people in the group.
(ii) How many persons in the group are creative?
Answer: Let total number be \( x^2 \).
Creative = \( \frac{x^2}{4} \), Caring = \( 2x \), Optimistic = 15.
\( \frac{x^2}{4} + 2x + 15 = x^2 \)
\( \implies x^2 + 8x + 60 = 4x^2 \implies 3x^2 - 8x - 60 = 0 \)
\( \implies 3x^2 - 18x + 10x - 60 = 0 \implies (3x + 10)(x - 6) = 0 \implies x = 6 \).
(i) Total number = \( x^2 = 36 \).
(ii) Creative = \( 36/4 = 9 \) persons.

Question. Solve for \( x \): \( \left(\frac{x}{x + 1}\right)^2 + 6 - 5\left(\frac{x}{x + 1}\right) = 0 \), \( x \neq -1 \).
Answer: Let \( \frac{x}{x + 1} = y \).
\( y^2 - 5y + 6 = 0 \implies (y - 2)(y - 3) = 0 \implies y = 2, 3 \).
Case 1: \( \frac{x}{x+1} = 2 \implies x = 2x + 2 \implies x = -2 \).
Case 2: \( \frac{x}{x+1} = 3 \implies x = 3x + 3 \implies 2x = -3 \implies x = -\frac{3}{2} \).

Question. If one root of \( x^2 - x - k = 0 \) is square that of other, then find the value of \( k \).
Answer: Let roots be \( \alpha \) and \( \alpha^2 \).
\( \alpha + \alpha^2 = 1 \) and \( \alpha \cdot \alpha^2 = -k \implies \alpha^3 = -k \).
From \( \alpha^2 + \alpha - 1 = 0 \), \( \alpha = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \).
Then \( k = -(\alpha)^3 \). Values can be calculated as \( 2 \pm \sqrt{5} \).

Question. Solve for \( x \): \( x + 5 - \frac{8}{x + 5} = 7 \)
Answer: Let \( x + 5 = y \).
\( y - \frac{8}{y} = 7 \implies y^2 - 7y - 8 = 0 \)
\( \implies (y - 8)(y + 1) = 0 \implies y = 8, -1 \).
Case 1: \( x + 5 = 8 \implies x = 3 \).
Case 2: \( x + 5 = -1 \implies x = -6 \).

Question. Determine the value of \( k, k > 0 \) such that equation, \( x^2 + kx + 64 = 0 \) and \( x^2 - 8x + k = 0 \) will have real roots.
Answer: For \( x^2 + kx + 64 = 0 \), \( k^2 - 256 \geq 0 \implies k \geq 16 \) (since \( k > 0 \)).
For \( x^2 - 8x + k = 0 \), \( 64 - 4k \geq 0 \implies k \leq 16 \).
To satisfy both, \( k = 16 \).

Question. Solve the equation for \( x \): \( 3\sqrt{\frac{x}{5}} + 3\sqrt{\frac{5}{x}} = 10 \)
Answer: Let \( \sqrt{\frac{x}{5}} = y \).
\( 3y + \frac{3}{y} = 10 \implies 3y^2 - 10y + 3 = 0 \implies (3y - 1)(y - 3) = 0 \).
\( y = 3 \implies \frac{x}{5} = 9 \implies x = 45 \).
\( y = \frac{1}{3} \implies \frac{x}{5} = \frac{1}{9} \implies x = \frac{5}{9} \).

Question. Find the number which exceeds its positive square root by 20.
Answer: Let the number be \( x \).
\( x - \sqrt{x} = 20 \). Let \( \sqrt{x} = y \).
\( y^2 - y - 20 = 0 \implies (y - 5)(y + 4) = 0 \).
Since square root is positive, \( y = 5 \).
\( \implies x = 25 \).

Question. Evaluate: \( 20 + \frac{1}{20 + \frac{1}{20 + \frac{1}{...}}} \)
Answer: Let the expression be \( y \).
\( y = 20 + \frac{1}{y} \implies y^2 - 20y - 1 = 0 \)
Using quadratic formula: \( y = \frac{20 \pm \sqrt{400 + 4}}{2} = \frac{20 \pm \sqrt{404}}{2} = 10 \pm \sqrt{101} \).
Since the value is positive, \( y = 10 + \sqrt{101} \).

Question. If \( p \) and \( q \) are the roots of the equation \( ax^2 + bx + c = 0 \), then find the value of \( \frac{1}{p^2} + \frac{1}{q^2} \).
Answer: \( \frac{1}{p^2} + \frac{1}{q^2} = \frac{p^2 + q^2}{p^2q^2} = \frac{(p+q)^2 - 2pq}{(pq)^2} \)
Sum \( p+q = -b/a \), Product \( pq = c/a \).
Value = \( \frac{(-b/a)^2 - 2(c/a)}{(c/a)^2} = \frac{b^2 - 2ac}{c^2} \).

Question. If sum of the roots of the equation \( 4x^2 + bx + c = 0 \) is equal to the product of their reciprocal, then find the value of \( bc \).
Answer: Sum of roots = \( -b/4 \).
Product of reciprocals = \( \frac{1}{\alpha \beta} = \frac{1}{c/4} = \frac{4}{c} \).
\( -b/4 = 4/c \implies bc = -16 \).

Question. Find the difference of the roots of the quadratic equation \( 25x^2 - 10x + 1 = 0 \).
Answer: \( (5x - 1)^2 = 0 \).
The roots are \( 1/5 \) and \( 1/5 \).
Difference = \( 1/5 - 1/5 = 0 \).

Question. Solve for \( x \): \( \sqrt{5 - x} = x\sqrt{5 - x} \)
Answer: \( \sqrt{5 - x}(1 - x) = 0 \).
\( \implies \sqrt{5 - x} = 0 \) or \( 1 - x = 0 \).
\( \implies x = 5 \) or \( x = 1 \).

Question. If \( (m^2 + n^2)x^2 - 2(mp + nq)x + p^2 + q^2 = 0 \) has equal roots, then show that \( \frac{m}{p} = \frac{n}{q} \).
Answer: \( D = 0 \implies [2(mp + nq)]^2 - 4(m^2 + n^2)(p^2 + q^2) = 0 \)
\( \implies (m^2p^2 + n^2q^2 + 2mpnq) - (m^2p^2 + m^2q^2 + n^2p^2 + n^2q^2) = 0 \)
\( \implies 2mpnq - m^2q^2 - n^2p^2 = 0 \)
\( \implies -(mq - np)^2 = 0 \implies mq = np \implies \frac{m}{p} = \frac{n}{q} \). Hence proved.

Question. Two trains leave a railway station at the same time. The first train travels due west and the second train travels due north. The first train travels 10 km/hr faster than the second train. If after two hours, they are 100 km apart, find the average speed of each train.
Answer: Let speed of second train be \( v \) km/hr. First train speed = \( v + 10 \) km/hr.
In 2 hours, distances are \( 2v \) and \( 2(v + 10) \).
By Pythagoras: \( (2v)^2 + [2(v + 10)]^2 = 100^2 \)
\( \implies 4v^2 + 4(v^2 + 20v + 100) = 10000 \)
\( \implies 8v^2 + 80v + 400 = 10000 \implies 8v^2 + 80v - 9600 = 0 \)
\( \implies v^2 + 10v - 1200 = 0 \implies (v + 40)(v - 30) = 0 \).
Speeds are 30 km/hr (North) and 40 km/hr (West).

In the following questions, a statement of assertion (A) is followed by a statement of Reason (R). Choose the correct answer out of the following choices.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question. Assertion (A): The value of \( k \) for which the equation \( kx^2 - 12x + 4 = 0 \) has equal roots, is 9.
Reason (R): The equation \( ax^2 + bx + c = 0, (a \neq 0) \) has equal roots, if \( (b^2 - 4ac) > 0 \).
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (c) A is true but R is false.

Question. Assertion (A): \( 2x^2 - 4x + 3 = 0 \) is a quadratic equation.
Reason (R): All polynomials of degree \( n \), when \( n \) is a whole number can be treated quadratic equation.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (c) A is true but R is false.