CBSE Class 10 Maths HOTs Quadratic Equations Set L

Very Short Answer Type Questions

Question. If 1 is a root of the equations \( ay^2 + ay + 3 = 0 \) and \( y^2 + y + b = 0 \), then find the value of \( ab \).
Answer: Given, 1 is the root of \( ay^2 + ay + 3 = 0 \) and \( y^2 + y + b = 0 \)
\( \therefore y = 1 \) will satisfy these equations.
i.e., \( a(1)^2 + a \times 1 + 3 = 0 \Rightarrow 2a + 3 = 0 \Rightarrow a = \frac{-3}{2} \)
\( (1)^2 + 1 + b = 0 \Rightarrow 2 + b = 0 \Rightarrow b = -2 \)
\( \therefore ab = \frac{-3}{2} \times (-2) = 3 \)

Question. If the quadratic equation \( mx^2 + 2x + m = 0 \) has two equal roots, then find the values of \( m \).
Answer: For roots of \( mx^2 + 2x + m = 0 \) to be equal, Discriminant, \( D = 0 \)
\( \therefore (2)^2 - 4(m)(m) = 0 \Rightarrow 4 - 4m^2 = 0 \Rightarrow m^2 = 1 \Rightarrow m = \pm 1 \).

Question. Form the quadratic equation from \( x = \sqrt{6 + \sqrt{6 + \sqrt{6 + \dots}}} \), where \( x \) is a natural number.
Answer: We have, \( x = \sqrt{6 + \sqrt{6 + \sqrt{6 + \dots}}} \)
Squaring both sides, we get \( x^2 = 6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + \dots}}} \)
\( \Rightarrow x^2 = 6 + x \Rightarrow x^2 - x - 6 = 0 \), which is the required quadratic equation.

Question. Find the signs of the roots of the quadratic equation \( x^2 + kx + k = 0 \), where \( k > 0 \).
Answer: Let \( \alpha \) and \( \beta \) be the roots of \( x^2 + kx + k = 0 \).
Then, \( \alpha + \beta = -k \) and \( \alpha\beta = k \).
This is possible only when \( \alpha \) and \( \beta \) are both negative.

Question. Check whether the equation \( (x^2 + 1)(x + 2) = (x + 3)^2 \) is quadratic or not?
Answer: We have, \( (x^2 + 1)(x + 2) = (x + 3)^2 \)
\( \Rightarrow x^3 + 2x^2 + x + 2 = x^2 + 6x + 9 \)
\( \Rightarrow x^3 + x^2 - 5x - 7 = 0 \),
which is not of the form \( ax^2 + bx + c = 0, a \neq 0 \).
Hence, given equation is not a quadratic equation.

Question. Find the roots of the quadratic equation \( 2x^2 - 7x - 85 = 0 \) by factorisation.
Answer: We have, \( 2x^2 - 7x - 85 = 0 \)
\( \Rightarrow 2x^2 + 10x - 17x - 85 = 0 \)
\( \Rightarrow 2x(x + 5) - 17(x + 5) = 0 \)
\( \Rightarrow (x + 5)(2x - 17) = 0 \Rightarrow x = -5 \) or \( x = \frac{17}{2} \)

Question. If a number is added to twice its square, then the resultant is 21. Find the number.
Answer: Let the number be \( x \).
According to question, \( x + 2x^2 = 21 \)
\( \Rightarrow 2x^2 + x - 21 = 0 \Rightarrow 2x^2 - 6x + 7x - 21 = 0 \)
\( \Rightarrow 2x(x - 3) + 7(x - 3) = 0 \)
\( \Rightarrow (x - 3)(2x + 7) = 0 \Rightarrow x = 3 \) or \( x = \frac{-7}{2} \)

Question. Find the discriminant of the quadratic equation \( x^2 + px + 2q = 0 \).
Answer: The given equation is, \( x^2 + px + 2q = 0 \)
Here, \( a = 1, b = p \) and \( c = 2q \)
\( \therefore D = b^2 - 4ac = p^2 - 4 \times 1 \times 2q = p^2 - 8q \)

Question. Check whether \( x = 0 \) and \( x = 1 \) are the solutions of the equation \( x^2 + x + 1 = 0 \) or not?
Answer: Given equation is, \( x^2 + x + 1 = 0 \)
When \( x = 0 \), L.H.S. = \( 0^2 + 0 + 1 = 1 \neq 0 = \) R.H.S.
\( \therefore x = 0 \) is not the solution of given equation.
When \( x = 1 \), L.H.S. = \( 1^2 + 1 + 1 = 3 \neq 0 = \) R.H.S.
\( \therefore x = 1 \) is not the solution of given equation.

Question. Find the value(s) of \( k \) so that the quadratic equation \( x^2 - 4kx + k = 0 \) has equal roots.
Answer: Given, \( x^2 - 4kx + k = 0 \)
Since, given equation has equal roots. \( \therefore D = 0 \)
\( \Rightarrow (-4k)^2 - 4(1)(k) = 0 \Rightarrow 16k^2 - 4k = 0 \)
\( \Rightarrow 4k(4k - 1) = 0 \Rightarrow k = 0 \) or \( k = \frac{1}{4} \)

Short Answer Type Questions

Question. Find the value of \( p \), for which one root of the quadratic equation \( px^2 - 14x + 8 = 0 \) is 6 times the other.
Answer: Given equation is, \( px^2 - 14x + 8 = 0 \).
Let roots of equation be \( \alpha \) and \( \beta \) such that
\( \beta = 6\alpha \Rightarrow 6\alpha - \beta = 0 \) ...(i)
Now, sum of roots = \( \alpha + \beta = -\left(\frac{-14}{p}\right) = \frac{14}{p} \) ...(ii)
and product of roots = \( \alpha\beta = \frac{8}{p} \) ...(iii)
Solving (i) and (ii), we get \( \alpha = \frac{2}{p} \) and \( \beta = \frac{12}{p} \)
Putting these values in (iii) we get
\( \left(\frac{2}{p}\right) \times \left(\frac{12}{p}\right) = \frac{8}{p} \Rightarrow 8p = 24 \Rightarrow p = 3 \quad (\because p \neq 0) \)

Question. Solve for \( x \) (in terms of \( a \) and \( b \)): \( \frac{a}{x - b} + \frac{b}{x - a} = 2 \), \( x \neq a, b \).
Answer: We have, \( \frac{a}{x - b} + \frac{b}{x - a} = 2 \)
\( \Rightarrow \frac{a(x - a) + b(x - b)}{(x - b)(x - a)} = 2 \)
\( \Rightarrow ax - a^2 + bx - b^2 = 2(x^2 - bx - ax + ab) \)
\( \Rightarrow 2x^2 - 3bx - 3ax + 2ab + a^2 + b^2 = 0 \)
\( \Rightarrow 2x^2 - 3(a + b)x + (a + b)^2 = 0 \)
\( \Rightarrow 2x^2 - 2(a + b)x - (a + b)x + (a + b)^2 = 0 \)
\( \Rightarrow 2x[x - (a + b)] - (a + b)[x - (a + b)] = 0 \)
\( \Rightarrow [x - (a + b)] [2x - (a + b)] = 0 \)
\( \Rightarrow x = a + b \) or \( x = \frac{a + b}{2} \)

Question. Find the value of \( m \) so that the quadratic equation \( mx(5x - 6) + 9 = 0 \) has two equal roots.
Answer: Given, \( mx(5x - 6) + 9 = 0 \)
\( \therefore 5mx^2 - 6mx + 9 = 0 \)
For equal roots, discriminant, \( D = 0 \)
\( \therefore (-6m)^2 - 4(5m)(9) = 0 \)
\( \Rightarrow 36m^2 - 180m = 0 \Rightarrow 36m(m - 5) = 0 \)
\( \Rightarrow m = 5 \quad (\because m \neq 0) \)

Question. If \( x = 2/3 \) and \( x = -3 \) are roots of the quadratic equation \( ax^2 + 7x + b = 0 \), then find the values of \( a \) and \( b \).
Answer: Given, roots of quadratic equation \( ax^2 + 7x + b = 0 \) are \( \frac{2}{3} \) and \( -3 \).
\( \therefore \) Sum of roots = \( \frac{-7}{a} \)
\( \Rightarrow \frac{2}{3} + (-3) = \frac{-7}{a} \Rightarrow \frac{-7}{3} = \frac{-7}{a} \Rightarrow a = 3 \)
Product of roots = \( \frac{b}{a} \)
\( \Rightarrow \left(\frac{2}{3}\right)(-3) = \frac{b}{3} \Rightarrow -2 = \frac{b}{3} \Rightarrow b = -6 \)
\( \therefore a = 3, b = -6 \)

Question. For what value of \( k \) does the quadratic equation \( (k - 5)x^2 + 2(k - 5)x + 2 = 0 \) have equal roots?
Answer: We have, \( (k - 5)x^2 + 2(k - 5)x + 2 = 0 \)
Given equation has equal roots. \( \therefore \) Discriminant, \( D = 0 \)
\( \Rightarrow [2(k - 5)]^2 - 4(k - 5)(2) = 0 \)
\( \Rightarrow k^2 + 25 - 10k - 2k + 10 = 0 \)
\( \Rightarrow k^2 - 12k + 35 = 0 \Rightarrow k^2 - 7k - 5k + 35 = 0 \)
\( \Rightarrow (k - 7)(k - 5) = 0 \Rightarrow k = 7 \quad [\because k \neq 5] \)

Question. Find the value of \( p \) for which the roots of the equation \( px(x - 2) + 6 = 0 \), are equal.
Answer: Given, \( px(x - 2) + 6 = 0 \)
\( \Rightarrow px^2 - 2px + 6 = 0 \)
Since, the roots are equal \( \therefore D = 0 \)
\( \Rightarrow (-2p)^2 - 4 \times p \times 6 = 0 \Rightarrow 4p^2 - 24p = 0 \)
\( \Rightarrow 4p(p - 6) = 0 \Rightarrow 4p = 0 \) or \( p - 6 = 0 \)
\( \Rightarrow p = 6 \quad [\because p \neq 0] \)

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 three consecutive natural numbers be \( x, x + 1, x + 2 \).
According to question,
\( (x + 1)^2 - [(x + 2)^2 - (x)^2] = 60 \)
\( \Rightarrow (x + 1)^2 - [x^2 + 4 + 4x - x^2] = 60 \)
\( \Rightarrow x^2 + 1 + 2x - 4 - 4x = 60 \)
\( \Rightarrow x^2 - 2x - 63 = 0 \Rightarrow x^2 - 9x + 7x - 63 = 0 \)
\( \Rightarrow x(x - 9) + 7(x - 9) = 0 \Rightarrow (x + 7)(x - 9) = 0 \)
\( \Rightarrow x = 9 \quad (\because x \neq -7 \) as \( x \) is a natural number)
\( \therefore \) The numbers are 9, 10, 11.

Question. If 1 is a root of the quadratic equation \( 3x^2 + ax - 2 = 0 \) and the quadratic equation \( a(x^2 + 6x) - b = 0 \) has equal roots, find the value of \( b \).
Answer: Since, 1 is a root of \( 3x^2 + ax - 2 = 0 \)
\( \therefore 3(1)^2 + a(1) - 2 = 0 \)
\( \Rightarrow a + 1 = 0 \Rightarrow a = -1 \)
Putting \( a = -1 \) in \( a(x^2 + 6x) - b = 0 \), we get
\( x^2 + 6x + b = 0 \) ...(i)
Since, (i) has equal roots. \( \therefore D = 0 \)
\( \Rightarrow 6^2 - 4(b) = 0 \Rightarrow 36 = 4b \Rightarrow b = 9 \)

Question. Solve the quadratic equation \( 16x^2 - 24x - 1 = 0 \) by using the quadratic formula.
Answer: Here, \( a = 16, b = -24, c = -1 \)
\( \therefore D = b^2 - 4ac = (-24)^2 - 4(16)(-1) = 576 + 64 = 640 > 0 \)
\( \therefore \sqrt{D} = \sqrt{640} = \sqrt{64 \times 10} = 8\sqrt{10} \)
By quadratic formula, we have
\( x = \frac{-b \pm \sqrt{D}}{2a} = \frac{24 \pm 8\sqrt{10}}{32} = \frac{3 \pm \sqrt{10}}{4} \)
Therefore, the roots are \( \frac{3 + \sqrt{10}}{4} \) and \( \frac{3 - \sqrt{10}}{4} \).

Question. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore. If the average speed of the express train is 11 km/hr more than that of the passenger train, form the quadratic equation to find the average speed of express train.
Answer: Let the average speed of express train be \( x \) km/hr.
\( \therefore \) The average speed of passenger train \( = (x - 11) \) km/hr
Time taken by express train to travel 132 km \( = \frac{132}{x} \) hr
Time taken by passenger train to travel 132 km \( = \frac{132}{x - 11} \) hr
According to question, \( \frac{132}{x - 11} - \frac{132}{x} = 1 \)
\( \Rightarrow 132x - 132x + 132 \times 11 = x(x - 11) \)
\( \Rightarrow 1452 = x^2 - 11x \)
\( \Rightarrow x^2 - 11x - 1452 = 0 \), which is the required equation.

Question. Solve for \( x \): \( \frac{63}{x} + \frac{72}{x + 6} = 3 \)
Answer: Given, \( \frac{63}{x} + \frac{72}{x + 6} = 3 \)
\( \Rightarrow \frac{7}{x} + \frac{8}{x + 6} = \frac{3}{9} = \frac{1}{3} \Rightarrow \frac{7(x + 6) + 8x}{x(x + 6)} = \frac{1}{3} \)
\( \Rightarrow 3(7x + 42 + 8x) = x^2 + 6x \)
\( \Rightarrow 45x + 126 = x^2 + 6x \)
\( \Rightarrow x^2 - 39x - 126 = 0 \Rightarrow x^2 - 42x + 3x - 126 = 0 \)
\( \Rightarrow x(x - 42) + 3(x - 42) = 0 \)
\( \Rightarrow (x - 42)(x + 3) = 0 \)
\( \Rightarrow \) Either \( x - 42 = 0 \) or \( x + 3 = 0 \)
\( \Rightarrow x = 42 \) or \( x = -3 \)

Question. If the sum of two natural numbers is 8 and their product is 15, then find the numbers.
Answer: Let the two natural numbers be \( a \) and \( b \).
So, \( a + b = 8 \) (Given) \( \Rightarrow a = 8 - b \) ...(1)
and \( ab = 15 \) (given)
\( \Rightarrow (8 - b)b = 15 \quad [\text{Using (1)}] \)
\( \Rightarrow 8b - b^2 = 15 \Rightarrow b^2 - 8b + 15 = 0 \)
\( \Rightarrow b^2 - 5b - 3b + 15 = 0 \Rightarrow b(b - 5) - 3(b - 5) = 0 \)
\( \Rightarrow (b - 3)(b - 5) = 0 \Rightarrow b = 3 \) or \( b = 5 \)
When \( b = 3 \) then \( a = 8 - 3 = 5 \). When \( b = 5 \) then \( a = 8 - 5 = 3 \).

Question. A two digit number is such that the product of the digits is 24. When 18 is added to the number, the digits interchange their places. Formulate the quadratic equation whose roots are the digits of the number.
Answer: Let the digit at ten’s place be \( y \) and digit at unit’s place be \( x \).
\( \therefore \) Number \( = 10y + x \)
Reversed Number \( = 10x + y \)
According to the question, \( 10x + y = 18 + 10y + x \)
\( \Rightarrow 10x + y - 10y - x = 18 \Rightarrow 9x - 9y = 18 \Rightarrow x - y = 2 \)
Also, \( xy = 24 \)
Now, \( (x + y)^2 = (x - y)^2 + 4xy = 2^2 + 4(24) = 100 \Rightarrow x + y = 10 \quad (\because x, y > 0) \)
\( \therefore \) Required equation is \( t^2 - 10t + 24 = 0 \)

Question. If the sum of first \( n \) even natural numbers is 420, then find the value of \( n \).
Answer: We have, \( 2 + 4 + 6 + 8 + \dots \) to \( n \) terms = 420
\( \Rightarrow \frac{n}{2}[2 \times 2 + (n - 1) \times 2] = 420 \)
\( \Rightarrow n(2 + n - 1) = 420 \Rightarrow n(n + 1) = 420 \)
\( \Rightarrow n^2 + n - 420 = 0 \Rightarrow n^2 + 21n - 20n - 420 = 0 \)
\( \Rightarrow n(n + 21) - 20(n + 21) = 0 \)
\( \Rightarrow (n + 21)(n - 20) = 0 \Rightarrow n = 20, -21 \)
\( n \) is a natural number \( \therefore n > 0 \Rightarrow n = 20 \)

Question. If the ratio of the roots of the equation \( lx^2 + nx + n = 0 \) is \( p : q \), prove that \( \sqrt{\frac{p}{q}} + \sqrt{\frac{q}{p}} + \sqrt{\frac{n}{l}} = 0 \).
Answer: Roots of the equation \( lx^2 + nx + n = 0 \) are \( \frac{-n + \sqrt{n^2 - 4nl}}{2l} \) and \( \frac{-n - \sqrt{n^2 - 4nl}}{2l} \)
Also, \( \frac{p}{q} = \frac{-n + \sqrt{n^2 - 4nl}}{-n - \sqrt{n^2 - 4nl}} \quad [\text{Given}] \)
Now, \( \sqrt{\frac{p}{q}} + \sqrt{\frac{q}{p}} + \sqrt{\frac{n}{l}} = \sqrt{\frac{-n + \sqrt{n^2 - 4nl}}{-n - \sqrt{n^2 - 4nl}}} + \sqrt{\frac{-n - \sqrt{n^2 - 4nl}}{-n + \sqrt{n^2 - 4nl}}} + \sqrt{\frac{n}{l}} \)
\( = \frac{-n + \sqrt{n^2 - 4nl} - n - \sqrt{n^2 - 4nl}}{\sqrt{(-n - \sqrt{n^2 - 4nl})(-n + \sqrt{n^2 - 4nl})}} + \sqrt{\frac{n}{l}} \)
\( = \frac{-2n}{\sqrt{n^2 - (n^2 - 4nl)}} + \sqrt{\frac{n}{l}} = \frac{-2n}{2\sqrt{nl}} + \sqrt{\frac{n}{l}} = -\sqrt{\frac{n}{l}} + \sqrt{\frac{n}{l}} = 0 \)

Question. Is the following situation possible? If so, then determine the present ages of the mother and her daughter in the problem given below:
The sum of the ages of a mother and her daughter is 25 years. Five years ago, the product of their ages was 58.
Answer: Let the present age of the daughter be \( x \) years.
Then, the present age of the mother \( = (25 - x) \) years. Five years ago, the ages of the daughter and the mother was \( (x - 5) \) years and \( (25 - x - 5) \) years respectively.
According to the given condition, we have
\( (x - 5)(25 - x - 5) = 58 \Rightarrow (x - 5)(20 - x) = 58 \)
\( \Rightarrow -x^2 + 20x + 5x - 100 = 58 \)
\( \Rightarrow x^2 - 25x + 158 = 0 \)
Here, \( a = 1, b = -25, c = 158 \)
\( \therefore D = b^2 - 4ac = (-25)^2 - 4 \times 1 \times 158 = 625 - 632 = -7 < 0 \)
Thus, no real value of \( x \) is possible in the above problem. Hence, the given situation is not possible.

Question. Check whether the following equations are quadratic or not
(i) \( x(2x + 3) = x + 2 \)
(ii) \( y(2y + 15) = 2(y^2 + y + 8) \).
Answer: (i) Given equation \( x(2x + 3) = x + 2 \) can be written as
\( 2x^2 + 3x = x + 2 \Rightarrow 2x^2 + 3x - x - 2 = 0 \)
\( \Rightarrow 2x^2 + 2x - 2 = 0 \), which of the standard form \( ax^2 + bx + c = 0, a \neq 0 \).
\( \therefore \) It is a quadratic equation.
(ii) Given equation, \( y(2y + 15) = 2(y^2 + y + 8) \) can be written as
\( 2y^2 + 15y = 2y^2 + 2y + 16 \)
\( \Rightarrow 2y^2 - 2y^2 + 15y - 2y - 16 = 0 \)
\( \Rightarrow 13y - 16 = 0 \), which is not of the standard form \( ax^2 + bx + c = 0, a \neq 0 \).
\( \therefore \) It is not a quadratic equation.

Question. A polygon of \( n \) sides has \( \frac{n(n-3)}{2} \) diagonals. How many sides a polygon has with 54 diagonals?
Answer: Given, when number of sides is \( n \), then the number of diagonals is \( \frac{n(n-3)}{2} \).
It is given that the number of diagonals = 54
\( \Rightarrow \frac{n(n-3)}{2} = 54 \Rightarrow n^2 - 3n = 108 \)
\( \Rightarrow n^2 - 3n - 108 = 0 \Rightarrow n^2 - 12n + 9n - 108 = 0 \)
\( \Rightarrow n(n - 12) + 9(n - 12) = 0 \Rightarrow (n - 12)(n + 9) = 0 \)
\( \Rightarrow n = 12 \) or \( n = -9 \Rightarrow n = 12 \quad (\because n \neq -9, \text{ as number of sides cannot be negative}) \)
\( \therefore \) The number of sides of the polygon is 12.

Question. If we buy 2 tickets from station A to station B and 3 from station A to C, we have to pay rs 795. But 3 tickets from A to B and 5 tickets from A to C cost a total of ₹ 1300. What is the fare from A to B and from A to C?
Answer: Let fare from station A to B be \( rs x \) and fare from station A to C be \( rs y \).
Then, according to the question,
\( 2x + 3y = 795 \) ...(i)
\( 3x + 5y = 1300 \) ...(ii)
Multiplying (i) by 3 and (ii) by 2, we get
\( 6x + 9y = 2385 \) ...(iii)
\( 6x + 10y = 2600 \) ...(iv)
Subtracting (iii) from (iv), we get \( y = 215 \)
Putting \( y = 215 \) in (i), we get
\( 2x + 3(215) = 795 \Rightarrow 2x = 150 \Rightarrow x = 75 \)
Hence, fare from station A to B is \( rs 75 \) and fare from station A to C is \( rs 215 \).

Question. If roots of quadratic equation \( x^2 + 2px + mn = 0 \) are real and equal, then show that the roots of the quadratic equation \( x^2 - 2(m + n)x + (m^2 + n^2 + 2p^2) = 0 \) are also equal.
Answer: \( x^2 + 2px + mn = 0 \) ...(i)
and \( x^2 - 2(m + n)x + (m^2 + n^2 + 2p^2) = 0 \) ...(ii)
Equation (i) has equal roots.
\( \therefore D = 0 \Rightarrow (2p)^2 - 4mn = 0 \Rightarrow 4p^2 = 4mn \Rightarrow p^2 = mn \) ...(iii)
Now, for equation (ii), discriminant, \( D = [-2(m + n)]^2 - 4 (m^2 + n^2 + 2p^2) \)
\( = 4(m^2 + n^2 + 2mn) - 4(m^2 + n^2 + 2mn) \quad [\text{Using (iii)}] \)
\( = 0 \)
Thus, equation (ii) also has equal roots.

Question. The sum of the squares of two consecutive multiples of 7 is 637. Find the multiples.
Answer: Let two consecutive multiples of 7 be \( k, k + 7 \).
According to question, \( (k)^2 + (k + 7)^2 = 637 \)
\( \Rightarrow k^2 + k^2 + 49 + 14k = 637 \)
\( \Rightarrow 2k^2 + 14k - 588 = 0 \Rightarrow k^2 + 7k - 294 = 0 \)
\( \Rightarrow k^2 + 21k - 14k - 249 = 0 \)
\( \Rightarrow k(k + 21) - 14(k + 21) = 0 \)
\( \Rightarrow (k + 21)(k - 14) = 0 \Rightarrow k = 14 \) or \( k = -21 \)
Hence, required multiples are 14, 21 or -14, -21.

Question. A farmer wishes to fence a 100 m² rectangular vegetable garden. Since he has with him only 30 m barbed wire, he fences three sides of the rectangular garden, letting compound wall of adjoining house act as the fourth side fence. Find the dimensions of his garden.
Answer: Let the length of one side of the garden be \( x \) m.
Then, \( 2x + \text{length of other side} = 30 \) m \( \Rightarrow \text{Length of other side} = (30 - 2x) \) m
\( \therefore x(30 - 2x) = 100 \quad [\because \text{Area} = 100 \text{ m}^2] \)
\( \Rightarrow 2x^2 - 30x + 100 = 0 \Rightarrow x^2 - 15x + 50 = 0 \)
\( \Rightarrow x^2 - 10x - 5x + 50 = 0 \Rightarrow x(x - 10) - 5(x - 10) = 0 \)
\( \Rightarrow (x - 10)(x - 5) = 0 \Rightarrow x = 10 \) or \( x = 5 \)
Hence, dimensions of the rectangular garden are 20 m and 5 m.

Question. The ratio of the roots of the equation \( ax^2 + bx + c = 0 \) is same as the ratio of the roots of the equation \( px^2 + qx + r = 0 \). If \( D_1 \) and \( D_2 \) are the discriminants of \( ax^2 + bx + c = 0 \) and \( px^2 + qx + r = 0 \) respectively, then \( D_1 : D_2 = \)
Answer: Let \( \alpha_1 \) and \( \beta_1 \) be the roots of \( ax^2 + bx + c = 0 \). \( \alpha_1 + \beta_1 = \frac{-b}{a}, \alpha_1\beta_1 = \frac{c}{a} \) and \( D_1 = b^2 - 4ac \). Let \( \alpha_2 \) and \( \beta_2 \) be the roots of \( px^2 + qx + r = 0 \). \( \alpha_2 + \beta_2 = \frac{-q}{p}, \alpha_2\beta_2 = \frac{r}{p}, D_2 = q^2 - 4pr \).
Given \( \frac{\alpha_1}{\beta_1} = \frac{\alpha_2}{\beta_2} \Rightarrow \frac{\alpha_1 + \beta_1}{\alpha_1 - \beta_1} = \frac{\alpha_2 + \beta_2}{\alpha_2 - \beta_2} \) (Applying componendo and dividendo rule)
\( \Rightarrow \frac{\alpha_1 + \beta_1}{\sqrt{(\alpha_1 + \beta_1)^2 - 4\alpha_1\beta_1}} = \frac{\alpha_2 + \beta_2}{\sqrt{(\alpha_2 + \beta_2)^2 - 4\alpha_2\beta_2}} \)
\( \Rightarrow \frac{-b/a}{\sqrt{D_1/a^2}} = \frac{-q/p}{\sqrt{D_2/p^2}} \Rightarrow \frac{-b}{\sqrt{D_1}} = \frac{-q}{\sqrt{D_2}} \Rightarrow \frac{D_1}{D_2} = \frac{b^2}{q^2} \)

Question. If 8 and 2 are roots of \( x^2 + ax + b = 0 \), whereas 3 and -3 are roots of \( x^2 + ax - b = 0 \), then find the roots of \( x^2 + ax + b = 0 \).
Answer: Since 8 and 2 are roots of \( x^2 + ax + b = 0 \)
\( \therefore 8^2 + a(8) + b = 0 \) and \( 2^2 + a(2) + b = 0 \Rightarrow 64 + 8a + b = 0 \) and \( 4 + 2a + b = 0 \)
Solving both equations, we get \( 60 + 6a = 0 \Rightarrow a = -10 \) ...(i)
Also, 3 and -3 are roots of \( x^2 + ax - b = 0 \).
\( \therefore 3^2 + a(3) - b = 0 \) and \( (-3)^2 + a(-3) - b = 0 \Rightarrow 9 + 3a - b = 0 \) and \( 9 - 3a - b = 0 \)
Solving both equations, we get \( 18 - 2b = 0 \Rightarrow b = 9 \) ...(ii)
Now, consider \( x^2 + ax + b = 0 \Rightarrow x^2 - 10x + 9 = 0 \quad [\text{From (i) and (ii)}] \)
\( \Rightarrow (x - 9)(x - 1) = 0 \Rightarrow x = 9 \) or \( x = 1 \)

Question. In a class test, the sum of Kamal’s marks in Maths and English is 40. Had he got 3 marks more in Maths and 4 marks less in English, the product of their marks would have been 360. Find his marks in two subjects.
Answer: Let the marks obtained in Maths be \( x \), then marks obtained in English \( = 40 - x \)
By the given condition, we have \( (x + 3) (40 - x - 4) = 360 \Rightarrow (x + 3) (36 - x) = 360 \)
\( \Rightarrow -x^2 + 33x + 108 = 360 \Rightarrow x^2 - 33x + 252 = 0 \)
\( \Rightarrow (x - 21) (x - 12) = 0 \Rightarrow x = 21 \) or \( x = 12 \)
If \( x = 21 \), then marks in Maths = 21 and English = 19. If \( x = 12 \), then Maths = 12 and English = 28.

Long Answer Type Questions 

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 \( x \) km/h.
Speed of the boat while going upstream \( = (20 - x) \) km/h.
Speed of the boat while going downstream \( = (20 + x) \) km/h.
According to question, \( \frac{48}{20 - x} - \frac{48}{20 + x} = 1 \Rightarrow \frac{960 + 48x - 960 + 48x}{(20 - x)(20 + x)} = 1 \)
\( \Rightarrow 400 - x^2 = 96x \Rightarrow x^2 + 96x - 400 = 0 \)
\( \Rightarrow (x + 100)(x - 4) = 0 \Rightarrow x = 4 \quad [\because \text{Speed cannot be negative}] \)

Question. Find the roots of the equation \( \frac{1}{2x - 3} + \frac{1}{x - 5} = 1, x \neq \frac{3}{2}, 5 \).
Answer: We have, \( \frac{1}{2x - 3} + \frac{1}{x - 5} = 1 \Rightarrow \frac{(x - 5) + (2x - 3)}{(2x - 3)(x - 5)} = 1 \)
\( \Rightarrow 3x - 8 = 2x^2 - 13x + 15 \Rightarrow 2x^2 - 16x + 23 = 0 \)
Using quadratic formula, \( x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4 \times 2 \times 23}}{2 \times 2} = \frac{16 \pm \sqrt{72}}{4} = \frac{8 \pm 3\sqrt{2}}{2} \)
\( \therefore x = \frac{8 + 3\sqrt{2}}{2} \) or \( \frac{8 - 3\sqrt{2}}{2} \)

Question. Solve for \( x \): \( \frac{1}{(x - 1)(x - 2)} + \frac{1}{(x - 2)(x - 3)} + \frac{1}{(x - 3)(x - 4)} = \frac{1}{6} \)
Answer: We have, \( \left[\frac{1}{(x - 1)(x - 2)} + \frac{1}{(x - 2)(x - 3)}\right] + \frac{1}{(x - 3)(x - 4)} = \frac{1}{6} \)
\( \Rightarrow \frac{x - 3 + x - 1}{(x - 1)(x - 2)(x - 3)} + \frac{1}{(x - 3)(x - 4)} = \frac{1}{6} \Rightarrow \frac{2}{(x - 1)(x - 3)} + \frac{1}{(x - 3)(x - 4)} = \frac{1}{6} \)
\( \Rightarrow \frac{2x - 8 + x - 1}{(x - 1)(x - 3)(x - 4)} = \frac{1}{6} \Rightarrow \frac{3(x - 3)}{(x - 1)(x - 3)(x - 4)} = \frac{1}{6} \Rightarrow (x - 1)(x - 4) = 18 \)
\( \Rightarrow x^2 - 5x - 14 = 0 \Rightarrow (x - 7)(x + 2) = 0 \Rightarrow x = 7, -2 \)

Question. By increasing the list price of a book by rs 10 a person can buy 10 books less for rs 1200. Find the original list price of the book.
Answer: Let the original price of the book be \( rs x \).
According to question, \( \frac{1200}{x} - \frac{1200}{x + 10} = 10 \Rightarrow 1200 \left[\frac{10}{x(x + 10)}\right] = 10 \)
\( \Rightarrow x(x + 10) = 1200 \Rightarrow x^2 + 10x - 1200 = 0 \)
\( \Rightarrow (x + 40)(x - 30) = 0 \Rightarrow x = 30 \quad [\because x \neq -40] \)
Hence, the original list price is \( rs 30 \).

Question. A piece of cloth costs rs 200. If the piece was 5 m longer and each metre of cloth costs rs 2 less, then the cost of the piece would have remained unchanged. How long is the piece and what is the original rate per metre?
Answer: Let the length of piece of cloth be \( x \) m. Rate per metre \( = rs \frac{200}{x} \).
New rate per metre \( = rs \frac{200}{x + 5} \).
According to the question, \( \frac{200}{x} - \frac{200}{x + 5} = 2 \Rightarrow \frac{x + 5 - x}{x(x + 5)} = \frac{2}{200} \)
\( \Rightarrow x^2 + 5x = 500 \Rightarrow x^2 + 5x - 500 = 0 \Rightarrow (x + 25)(x - 20) = 0 \)
\( \Rightarrow x = 20 \quad [\because x \neq -25] \)
So, \( x = 20 \) m. Rate per metre \( = \frac{200}{20} = rs 10 \).