CBSE Class 10 Mathematics Quadratic Equation Worksheet Set R

Quadratic Equations

Points to Remember

• The roots of a quadratic equation \( ax^2 + bx + c = 0, a \neq 0 \) can be found by using the formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), provided \( b^2 – 4ac \geq 0 \).
• The quadratic equation \( ax^2 + bx + c = 0, a \neq 0 \) has :
  (i) two distinct real roots, if \( D = b^2 – 4ac > 0 \)
  (ii) two equal or co-identical real roots, if \( D = b^2 – 4ac = 0 \)
  (iii) imaginary or no real roots, if \( D = b^2 – 4ac < 0 \).

Multiple Choice Questions

Question. The roots of the quadratic equation \( x^2 – 0.04 = 0 \) are
(a) \( \pm 0.2 \)
(b) \( \pm 0.02 \)
(c) 0.4
(d) 2
Answer: (a) Given : \( x^2 – 0.04 = 0 \Rightarrow x^2 – (0.2)^2 = 0 \Rightarrow (x + 0.2) (x – 0.2) = 0 \Rightarrow x = – 0.2, 0.2 \)

Question. The roots of the equation \( x^2 – 3x – 9 = 0 \) are :
(a) real and unequal
(b) real and equal
(c) roots are not equal
(d) imaginary roots
Answer: (a) \( D = b^2 – 4ac = (– 3)^2 – 4(1) (– 9) = 9 + 36 = 45 > 0 \). Thus, the roots are real and distinct.

Question. The value of \( x \) in \( x – \frac{18}{x} = 6 \) is :
(a) real and unequal
(b) real and equal
(c) imaginary
(d) none of these
Answer: (a) \( x – \frac{18}{x} = 6 \text{ or, } x^2 – 6x – 18 = 0 \). Thus, \( D = b^2 – 4ac = (– 6)^2 – 4(– 18) = 36 + 72 = 108 > 0 \). Thus, the roots are real and distinct.

Question. The solution of a quadratic equation is as follows : \( x = \frac{8 \pm \sqrt{(-8)^2 - 4(3)(2)}}{2(3)} \). Then the quadratic equation is :
(a) \( 3x^2 – 8x + 2 = 0 \)
(b) \( 2x^2 – 8x + 3 = 0 \)
(c) \( 3x^2 + 8x – 2 = 0 \)
(d) \( 3x^2 + 8x + 2 = 0 \)
Answer: (a) Given, \( x = \frac{8 \pm \sqrt{(-8)^2 - 4(3)(2)}}{2(3)} \). Thus, \( a = 3, b = – 8 \) and \( c = 2 \). The equation is given by \( ax^2 + bx + c = 0 \text{ or } 3x^2 – 8x + 2 = 0 \).

Question. If the roots of \( x^2 – 5x + a = 0 \) are equal, then the value of \( a \) is :
(a) \( \frac{25}{4} \)
(b) \( \pm \frac{25}{4} \)
(c) \( -\frac{25}{4} \)
(d) None of these
Answer: (a) Roots of \( x^2 – 5x + a = 0 \) are equal \( \therefore \) Discriminant = 0 \( \Rightarrow (– 5)^2 – 4 \times 1 \times a = 0 \Rightarrow 25 – 4a = 0 \Rightarrow a = \frac{25}{4} \).

Question. The length and breadth of a room if its area is \( 120\text{ m}^2 \) and perimeter is \( 44\text{ m} \) are :
(a) 11 m and 2 m
(b) 10 m and 2 m
(c) 12 m and 10 m
(d) 12 m and 1 m
Answer: (c) Given, Area = \( 120\text{ m}^2 \) and perimeter = \( 44\text{ m} \). Let the length be \( l \) and the breadth be \( b \). Thus, \( lb = 120 \) and \( 2(l + b) = 44 \Rightarrow l + b = 22 \Rightarrow l = 22 – b \). \( \therefore (22 – b)b = 120 \Rightarrow b^2 – 22b + 120 = 0 \Rightarrow b^2 – 12b – 10b + 120 = 0 \Rightarrow b(b – 12) – 10(b – 12) = 0 \Rightarrow (b – 12) (b – 10) = 0 \Rightarrow b = 10\text{ m} \) or \( 12\text{ m} \). \( \therefore l = 12\text{ m} \) or \( 10\text{ m} \).

Question. Find the marks obtained by Neha in two subjects, if their average is 75 and their product is 5600.
(a) 75 and 75
(b) 70 and 80
(c) 56 and 100
(d) 73 and 77
Answer: (b) Let the marks obtained by Neha in the two subjects be \( x \) and \( y \). So, \( xy = 5600 \) and \( \frac{x + y}{2} = 75 \). Thus, \( x + y = 150 \text{ or } x = 150 – y \). Now, \( xy = 5600 \Rightarrow y(150 – y) = 5600 \Rightarrow 150y – y^2 = 5600 \Rightarrow y^2 – 150y + 5600 = 0 \Rightarrow y^2 – 80y – 70y + 5600 = 0 \Rightarrow y(y – 80) – 70(y – 80) = 0 \Rightarrow (y – 80) (y – 70) = 0 \Rightarrow y = 70 \) or \( 80 \) and \( x = 80 \) or \( 70 \).

Fill in the Blanks

Question. For quadratic equation \( x^2 + 4x + b = 0 \), the discriminant \( D = \dots\dots\dots\dots \)
Answer: \( 16 – 4b \)

Question. The quadratic equation \( x^2 – 10x + 2 = 0 \) has \(\dots\dots\dots\dots \) roots.
Answer: Distinct, real

Question. The linear factors of the quadratic equation \( x^2 – 5x + 6 \) are \(\dots\dots\dots\dots \) and \(\dots\dots\dots\dots \)
Answer: \( (x – 2), (x – 3) \)

Question. The quadratic equation \( 2x^2 + px + 3 = 0 \) has equal roots if \( p = \dots\dots \)
Answer: \( \pm 2 \sqrt{6} \)

Question. Equation \( ax^2 + bx + c = 0 \) represents a quadratic equation if and only if \(\dots\dots\dots\dots \)
Answer: \( a \neq 0 \)

True/False

Question. The sum of the roots of quadratic equation \( 2x^2 – 56x + 23 = 0 \) is 56.
Answer: False Sum of roots = \( -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} = -\frac{(-56)}{2} = 28 \)

Question. The quadratic equation \( x^2 + 2x + 4 = 0 \) has real roots.
Answer: False \( \because \) Discriminant = \( b^2 – 4ac = (2)^2 – 4 \times 1 \times 4 = 4 – 16 = – 12 < 0 \). \( \therefore \) The roots are unreal.

Question. The discriminant of \( 4x^2 – 20x + 25 \) is zero.
Answer: True

Very Short Answer Type Questions

Question. Check whether the following is a quadratic equation: \( (x - 2)^2 + 1 = 2x - 3 \)
Answer: We have, \( (x - 2)^2 + 1 = 2x - 3 \)
\( x^2 - 4x + 4 + 1 = 2x - 3 \)
\( x^2 - 6x + 8 = 0 \)
i.e., It is of the form \( ax^2 + bx + c = 0 \)
\( \therefore \) The given equation is a quadratic equation.

Question. If \( y = 1 \) is a common root of the equations \( ay^2 + ay + 3 = 0 \) and \( y^2 + y + b = 0 \), then find \( ab \).
Answer: The given equations are \( ay^2 + ay + 3 = 0 \) and \( y^2 + y + b = 0 \)
Substituting \( y = 1 \) in them, we have
\( a(1)^2 + a(1) + 3 = 0 \) and \( (1)^2 + (1) + b = 0 \)
\( \Rightarrow 2a + 3 = 0 \) and \( 2 + b = 0 \)
\( \Rightarrow a = - \frac{3}{2} \) and \( b = - 2 \)
Thus, \( ab = - \left(\frac{3}{2}\right) (- 2) = 3 \)

Question. Find the nature of roots of the quadratic equation \( 2x^2 - 4x + 3 = 0 \).
Answer: Given, \( 2x^2 - 4x + 3 = 0 \)
Comparing it with quadratic equation \( ax^2 + bx + c = 0 \)
Here, \( a = 2, b = - 4 \) and \( c = 3 \)
\( \therefore D = b^2 - 4ac \)
\( = (- 4)^2 - 4 \times (2) (3) \)
\( = 16 - 24 \)
\( = - 8 < 0 \)
Hence, \( D < 0 \) this shows that roots will be imaginary.

Question. If the quadratic equation \( ax^2 + bx + c = 0 \) has equal roots, find the value of \( c \).
Answer: For equal roots, discriminant \( = 0 \)
\( \Rightarrow b^2 - 4ac = 0 \)
\( \Rightarrow b^2 = 4ac \)
\( \Rightarrow c = \frac{b^2}{4a} \)

Question. State whether the equation \( (x + 1) (x - 2) + x = 0 \) has two distinct real roots or not.
Answer: Given equation is,
\( (x + 1) (x - 2) + x = 0 \)
\( \Rightarrow x^2 - x - 2 + x = 0 \)
\( \Rightarrow x^2 - 2 = 0 \)
Now, Discriminant, \( D = b^2 - 4ac \)
\( = (0)^2 - 4 \times 1 \times (- 2) \)
\( = + 8 > 0 \)
Hence, the given equation has two distinct real roots.

Question. If \( x = 3 \) in one root of the quadratic equation \( x^2 - 2kx - 6 = 0 \), then find the value of \( k \).
Answer: Let \( \alpha \) be the other root.
\( x^2 - 2kx - 6 = 0 \)
Product \( = \frac{c}{a} = \frac{- 6}{1} = - 6 \)
\( 3 \alpha = - 6 \Rightarrow \alpha = - 2 \)
Sum \( = \frac{-b}{a} = \frac{- (- 2k)}{1} = 2k \)
\( \Rightarrow 3 + (- 2) = 2k \Rightarrow 1 = 2k \Rightarrow k = \frac{1}{2} \).
Value of \( k \) is \( \frac{1}{2} \)

Short Answer Type Questions-I

Question. Find the value of \( ab \) if the roots of equation \( ax^2 + x + b = 0 \) are equal.
Answer: An equation has equal roots, if
Discriminant, \( D = 0 \)
\( \Rightarrow b^2 - 4ac = 0 \)
Here, \( a = a, b = 1, c = b \).
\( \therefore (1)^2 - 4 \times a \times b = 0 \)
\( \Rightarrow 1 - 4ab = 0 \)
\( \Rightarrow 1 = 4ab \)
\( \Rightarrow ab = \frac{1}{4} \).

Question. Find the value of \( k \) for which the roots of the equation \( 3x^2 - 10x + k = 0 \) are reciprocal of each other.
Answer: The given equation is \( 3x^2 - 10x + k = 0 \)
On comparing it with \( ax^2 + bx + c = 0 \), we get
\( a = 3, b = - 10, c = k \)
Let the roots of the equation are \( \alpha \) and \( \frac{1}{\alpha} \).
\( \because \text{Product of the roots} = \frac{c}{a} \)
\( \therefore \alpha \cdot \frac{1}{\alpha} = \frac{k}{3} \)
or \( k = 3 \).

Question. If 2 is a root of the quadratic equation \( x^2 + kx + 12 = 0 \) and the equation \( x^2 + kx + q = 0 \) has equal roots, find the value of \( q \).
Answer: Given : 2 is a root of equation \( x^2 + kx + 12 = 0 \)
\( \therefore (2)^2 + k(2) + 12 = 0 \)
\( \Rightarrow 4 + 2k + 12 = 0 \)
\( \Rightarrow 2k = - 16 \)
\( \Rightarrow k = - 8 \)
Putting the value of \( k \) in equation \( x^2 + kx + q = 0 \), we get
\( x^2 - 8x + q = 0 \)
Now, for equal roots
Discriminant \( = b^2 - 4ac = 0 \)
\( \Rightarrow (- 8)^2 - 4 \times 1 \times q = 0 \)
\( \Rightarrow 64 - 4q = 0 \)
\( \Rightarrow q = \frac{64}{4} = 16 \).

Question. Find the value of \( k \), for which the one root of quadratic equation \( kx^2 - 14x + 8 = 0 \) is six times the other.
Answer: Let \( \alpha \) and \( \beta \) be the roots of given quadratic equation.
Then, \( \alpha = 6 \beta \) [Given]
Now, Sum of roots, \( \alpha + \beta = - \frac{(- 14)}{k} \)
\( \Rightarrow \beta + 6 \beta = \frac{14}{k} \)
\( \Rightarrow 7 \beta = \frac{14}{k} \)
\( \Rightarrow \beta = \frac{2}{k} \dots(i) \)
Also, Product of roots, \( \alpha \beta = \frac{8}{k} \)
\( \Rightarrow 6 \beta \times \beta = \frac{8}{k} \)
\( \Rightarrow 6 \beta^2 = \frac{8}{k} \)
\( \Rightarrow 6 \left(\frac{2}{k}\right)^2 = \frac{8}{k} \) [Using (i)]
\( \Rightarrow \frac{24}{k^2} = \frac{8}{k} \)
\( \Rightarrow k = 3 \).

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

Question. Find the roots of the quadratic equation \( x^2 - 3x - 10 = 0 \) by factorisation.
Answer: We have, \( x^2 - 3x - 10 = 0 \)
\( \Rightarrow x^2 - (5 - 2)x - 10 = 0 \)
\( \Rightarrow x^2 - 5x + 2x - 10 = 0 \)
\( \Rightarrow x(x - 5) + 2(x - 5) = 0 \)
\( \Rightarrow (x + 2) (x - 5) = 0 \)
\( \Rightarrow x + 2 = 0, x - 5 = 0 \)
\( \Rightarrow x = - 2, 5 \).

Question. Solve for \( x \) : \( \sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 \)
Answer: We have, \( \sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 \)
\( \Rightarrow \sqrt{2}x^2 + (5 + 2)x + 5\sqrt{2} = 0 \)
\( \Rightarrow \sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} = 0 \)
\( \Rightarrow x(\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5) = 0 \)
\( \Rightarrow (x + \sqrt{2}) (\sqrt{2}x + 5) = 0 \)

Question. Solve : \( \sqrt{3}x^2 - 2\sqrt{2}x - 2\sqrt{3} = 0 \)
Answer: Given, \( \sqrt{3}x^2 - 2\sqrt{2}x - 2\sqrt{3} = 0 \)
\( \Rightarrow x = \frac{2\sqrt{2} \pm \sqrt{(- 2\sqrt{2})^2 - 4(\sqrt{3})(- 2\sqrt{3})}}{2(\sqrt{3})} \)
Applying the formula : \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( \Rightarrow x = \frac{2\sqrt{2} \pm \sqrt{8 + 24}}{2\sqrt{3}} = \frac{2\sqrt{2} \pm \sqrt{32}}{2\sqrt{3}} \)
\( = \frac{2\sqrt{2} \pm 4\sqrt{2}}{2\sqrt{3}} = \frac{\sqrt{2} \pm 2\sqrt{2}}{\sqrt{3}} \)
\( = \sqrt{6} \) and \( - \sqrt{\frac{2}{3}} \).

Question. Solve : \( 4\sqrt{3}x^2 - 5x - 2\sqrt{3} = 0 \)
Answer: Given, \( 4\sqrt{3}x^2 - 5x - 2\sqrt{3} = 0 \)
\( \Rightarrow 4\sqrt{3}x^2 - 8x + 3x - 2\sqrt{3} = 0 \)
\( \Rightarrow 4\sqrt{3}x(x - 2) + \sqrt{3}(\sqrt{3}x - 2) = 0 \)
\( \Rightarrow (\sqrt{3}x - 2) (4x + \sqrt{3}) = 0 \)
\( \Rightarrow x = \frac{2}{\sqrt{3}} \) and \( - \frac{\sqrt{3}}{4} \).

Question. Solve : \( abx^2 + (b^2 - ac)x - bc = 0 \).
Answer: Given, \( abx^2 + (b^2 - ac)x - bc = 0 \)
\( \Rightarrow abx^2 + b^2x - acx - bc = 0 \)
\( \Rightarrow bx(ax + b) - c(ax + b) = 0 \)
\( \Rightarrow (ax + b) (bx - c) = 0 \)
\( \Rightarrow x = - \frac{b}{a} \) and \( \frac{c}{b} \).

Question. Solve : \( 12abx^2 - (9a^2 - 8b^2)x - 6ab = 0 \).
Answer: Given, \( 12abx^2 - (9a^2 - 8b^2)x - 6ab = 0 \)
\( \Rightarrow 12abx^2 - 9a^2x + 8b^2x - 6ab = 0 \)
\( \Rightarrow 3ax(4bx - 3a) + 2b(4bx - 3a) = 0 \)
\( \Rightarrow (3ax + 2b) (4bx - 3a) = 0 \)
\( \Rightarrow x = - \frac{2b}{3a} \) and \( \frac{3a}{4b} \).

Question. Solve for \( x \) : \( 2x^2 + \sqrt{3}x - 3 = 0 \)
Answer: We have, \( 2x^2 + \sqrt{3}x - 3 = 0 \)
\( \Rightarrow 2x^2 + (2\sqrt{3} - \sqrt{3})x - 3 = 0 \)
\( \Rightarrow 2x^2 + 2\sqrt{3}x - \sqrt{3}x - 3 = 0 \)
\( \Rightarrow 2x(x + \sqrt{3}) - \sqrt{3}(x + \sqrt{3}) = 0 \)
\( \Rightarrow (2x - \sqrt{3}) (x + \sqrt{3}) = 0 \)
\( \Rightarrow 2x - \sqrt{3} = 0 \text{ or } x + \sqrt{3} = 0 \)
\( \Rightarrow x = \frac{\sqrt{3}}{2} \) or \( x = - \sqrt{3} \).

Question. A teacher on attempting to arrange the students for mass drill in the form of a solid square found that 24 students were left. When he increased the square by one row and one column, he was short of 25 students. Find the number of students.
Answer: Let the number of rows be \( x \) and the number of students in each row also be \( x \).
Thus \( x^2 + 24 = (x + 1)^2 - 25 \)
\( \Rightarrow (x + 1)^2 - x^2 = 24 + 25 \)
\( \Rightarrow (x + 1 - x)(x + 1 + x) = 49 \)
\( \Rightarrow 2x + 1 = 49 \)
\( \Rightarrow 2x = 48 \)
\( \Rightarrow x = 24 \)
Thus, the total number of students \( = x^2 + 24 = (24)^2 + 24 = 24(24 + 1) = 24 \times 25 = 600 \).

Question. An equation has been given as \( \frac{c}{x^2} + \frac{k}{x} = 1 \). Find the relation between \( c \) and \( k \), if (i) \( x \) has real values (ii) \( x \) has no real values.
Answer: We have, \( \frac{c}{x^2} + \frac{k}{x} = 1 \)
Multiplying both the sides by \( x^2 \), we get \( c + kx = x^2 \)
\( \Rightarrow x^2 - kx - c = 0 \)
(i) For real values :
\( (- k)^2 - 4.1(- c) \geq 0 \Rightarrow k^2 + 4c \geq 0 \)
\( \Rightarrow k^2 \geq - 4c \Rightarrow - k^2 \leq 4c \Rightarrow c \geq - \frac{k^2}{4} \)
(ii) For non-real values :
\( (- k)^2 - 4.1(- c) \leq 0 \)
\( \Rightarrow k^2 + 4c \leq 0 \Rightarrow k^2 \leq - 4c \)
\( \Rightarrow - k^2 \geq 4c \Rightarrow 4c \leq - k^2 \Rightarrow c \leq - \frac{k^2}{4} \)

Short Answer Type Questions-II

Question. Solve for \( x \) : \( \frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30}, x \neq - 4, 7 \).
Answer: Given : \( \frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30} \)
\( \Rightarrow \frac{(x - 7) - (x + 4)}{(x - 7) (x + 4)} = \frac{11}{30} \)
\( \Rightarrow \frac{- 11}{x^2 - 3x - 28} = \frac{11}{30} \)
\( \Rightarrow x^2 - 3x - 28 = - 30 \)
\( \Rightarrow x^2 - 3x + 2 = 0 \)
\( \Rightarrow x^2 - 2x - x + 2 = 0 \)
\( \Rightarrow x(x - 2) - 1 (x - 2) = 0 \)
\( \Rightarrow (x - 2) (x - 1) = 0 \)
\( \Rightarrow x = 2 \text{ and } x = 1 \)

Question. If the roots of the equation \( (a - b)x^2 + (b - c)x + (c - a) = 0 \) are equal, prove that \( b + c = 2a \).
Answer: Given, \( (a - b)x^2 + (b - c)x + (c - a) = 0 \)
Comparing with \( Ax^2 + Bx + C = 0 \), we get
\( A = a - b, B = b - c \) and \( C = c - a \)
Since, the roots are equal
\( \therefore D = 0 \)
\( \Rightarrow B^2 - 4AC = 0 \)
\( \Rightarrow (b - c)^2 - 4(a - b)(c - a) = 0 \)
\( \Rightarrow b^2 + c^2 - 2bc - 4(ac - bc - a^2 + ab) = 0 \)
\( \Rightarrow b^2 + c^2 - 2bc - 4ac + 4bc + 4a^2 - 4ab = 0 \)
\( \Rightarrow b^2 + c^2 + 4a^2 + 2bc - 4ac - 4ab = 0 \)
\( \Rightarrow (b + c - 2a)^2 = 0 \)
\( \Rightarrow b + c - 2a = 0 \)
\( b + c = 2a \).
Hence Proved.

Question. Solve the equation \( 2x^2 - 5x + 3 = 0 \) by the method of completing the square.
Answer: We have, \( 2x^2 - 5x + 3 = 0 \)
\( \Rightarrow x^2 - \frac{5}{2}x + \frac{3}{2} = 0 \)
\( \Rightarrow x^2 - \frac{5}{2}x + \frac{25}{16} + \frac{3}{2} - \frac{25}{16} = 0 \)
\( \Rightarrow \left\{ (x)^2 - 2 \times \frac{5}{4} \times x + \left(\frac{5}{4}\right)^2 \right\} + \frac{24 - 25}{16} = 0 \)
\( \Rightarrow \left( x - \frac{5}{4} \right)^2 - \left( \frac{1}{4} \right)^2 = 0 \)
\( \Rightarrow \left( x - \frac{5}{4} - \frac{1}{4} \right) \left( x - \frac{5}{4} + \frac{1}{4} \right) = 0 \)
\( \Rightarrow \left( x - \frac{3}{2} \right) (x - 1) = 0 \)
\( \Rightarrow x - \frac{3}{2} = 0 \) or \( x - 1 = 0 \)
\( \Rightarrow x = \frac{3}{2} \) or \( x = 1 \). Ans.

Question. Solve the given equation by the method of completing the squares : \( x^2 + 12x - 45 = 0 \)
Answer: We have, \( x^2 + 12x - 45 = 0 \)
\( \Rightarrow x^2 + 12x + 36 - 45 - 36 = 0 \)
\( \Rightarrow \{ (x)^2 + 2 \times x \times 6 + (6)^2 \} - 81 = 0 \)
\( \Rightarrow (x + 6)^2 - (9)^2 = 0 \)
\( \Rightarrow (x + 6 - 9) (x + 6 + 9) = 0 \)
\( \Rightarrow (x - 3) (x + 15) = 0 \)
\( \Rightarrow x - 3 = 0 \) or \( x + 15 = 0 \)
\( \Rightarrow x = 3 \) or \( x = - 15 \). Ans.

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: Given,
\( (a^2 + b^2) x^2 + 2(ac + bd) x + (c^2 + d^2) = 0 \), \( ad \neq bc \)
\( D = b^2 - 4ac \)
\( = [2(ac + bd)]^2 - 4(a^2 + b^2) (c^2 + d^2) \)
\( = 4 [a^2c^2 + b^2d^2 + 2abcd] - 4[a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2] \)
\( = 4[a^2c^2 + b^2d^2 + 2abcd - a^2c^2 - a^2d^2 - b^2c^2 - b^2d^2] \)
\( = 4[-a^2d^2 - b^2c^2 + 2abcd] \)
\( = - 4(a^2d^2 + b^2c^2 - 2abcd) \)
\( = - 4(ad - bc)^2 \)
\( \because D \) is negative
Hence, given equation has no real roots.
Hence Proved.

Question. Show that if the roots of the following quadratic equation are equal, then \( ad = bc \) : \( x^2 (a^2 + b^2) + 2 (ac + bd)x + (c^2 + d^2) = 0 \)
Answer: Given equation is,
\( x^2 (a^2 + b^2) + 2(ac + bd)x + (c^2 + d^2) = 0 \)
We know, if the roots of a quadratic equation are equal, then, discriminant, \( D = 0 \)
\( \Rightarrow B^2 - 4AC = 0 \)
Here, \( A = (a^2 + b^2), B = 2(ac + bd) \) and \( C = c^2 + d^2 \)
\( \therefore [2(ac + bd)]^2 - 4 \times (a^2 + b^2) \times (c^2 + d^2) = 0 \)
\( \Rightarrow 4a^2c^2 + 4b^2d^2 + 8acbd - 4 (a^2c^2 + b^2c^2 + a^2d^2 + b^2d^2) = 0 \)
\( \Rightarrow 8acbd - 4b^2c^2 - 4a^2d^2 = 0 \)
\( \Rightarrow - 4 (a^2d^2 + b^2c^2 - 2acbd) = 0 \)
\( \Rightarrow (ad - bc)^2 = 0 \)
\( \Rightarrow ad - bc = 0 \)
\( \Rightarrow ad = bc \) Hence Proved.

Question. Find the value of \( k \) such that the equation \( (k - 12)x^2 + 2(k - 12) x + 2 = 0 \) has equal roots.
Answer: Given quadratic equation is,
\( (k - 12)x^2 + 2(k - 12)x + 2 = 0 \)
The above equation has equal roots, if
Discriminant \( = 0 \)
\( \Rightarrow b^2 - 4ac = 0 \)
Here, \( a = (k - 12), b = 2(k - 12) \) and \( c = 2 \)
\( \therefore \{2(k - 12)\}^2 - 4 \times (k - 12) \times 2 = 0 \)
\( \Rightarrow 2(k - 12) \{ (k - 12) - 4 \} = 0 \)
\( \Rightarrow 2(k - 12) (k - 16) = 0 \)
\( \Rightarrow (k - 12) = 0 \) or \( (k - 16) = 0 \)
\( \Rightarrow k = 12 \) or \( 16 \). Ans.

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: Given, equation is \( x^2 + px + 16 = 0 \)
This is of the form \( ax^2 + bx + c = 0 \)
where, \( a = 1, b = p \) and \( c = 16 \)
\( \therefore D = b^2 - 4ac \)
\( = p^2 - 4 \times 1 \times 16 \)
\( = p^2 - 64 \)
for equal roots, we have
\( D = 0 \)
\( p^2 - 64 = 0 \)
\( p^2 = 64 \)
\( p = \pm 8 \)
Putting \( p = 8 \) in given equation we have,
\( x^2 + 8x + 16 = 0 \)
\( (x + 4)^2 = 0 \)
\( x + 4 = 0 \)
\( x = - 4 \)
Now, putting \( p = - 8 \) in the given equation, we get
\( x^2 - 8x + 16 = 0 \)
\( (x - 4)^2 = 0 \)
\( x = 4 \)
\( \therefore \) Required roots are \( - 4 \) and \( - 4 \) or \( 4 \) and \( 4 \). Ans.

Question. Solve : \( \frac{3}{x+1} - \frac{1}{2} = \frac{2}{3x-1} \), where \( x \neq - 1, \frac{1}{3} \).
Answer: Given,
\( \frac{3}{x + 1} - \frac{1}{2} = \frac{2}{3x - 1} \)
\( \Rightarrow \frac{6 - (x + 1)}{2(x + 1)} = \frac{2}{3x - 1} \)
\( \Rightarrow [6 - (x + 1)] (3x - 1) = 4(x + 1) \)
\( \Rightarrow 6(3x - 1) - (3x - 1) (x + 1) = 4(x + 1) \)
\( \Rightarrow 18x - 6 - (3x^2 - x + 3x - 1) = 4x + 4 \)
\( \Rightarrow 14x - 10 - 3x^2 - 2x + 1 = 0 \)
\( \Rightarrow - 3x^2 + 12x - 9 = 0 \)
\( \Rightarrow x^2 - 4x + 3 = 0 \)
\( \Rightarrow x^2 - 3x - x + 3 = 0 \)
\( \Rightarrow x(x - 3) - 1(x - 3) = 0 \)
\( \Rightarrow (x - 3) (x - 1) = 0 \)
\( \Rightarrow x = 1, 3 \). Ans.

Question. Solve for \( x \) : \( \frac{1}{(x-1)(x-2)} + \frac{1}{(x-2)(x-3)} = \frac{2}{3}, x \neq 1, 2, 3 \).
Answer: We have,
\( \frac{1}{(x-1)(x-2)} + \frac{1}{(x-2)(x-3)} = \frac{2}{3}, x \neq 1, 2, 3 \)
\( \frac{3(x - 3) + 3(x - 1)}{(x - 1) (x - 2) (x - 3)} \times \frac{1}{2} \dots \)
[Alternative step simplification]
\( \frac{(x-3) + (x-1)}{(x-1)(x-2)(x-3)} = \frac{2}{3} \)
\( \frac{2x - 4}{(x-1)(x-2)(x-3)} = \frac{2}{3} \)
\( \frac{2(x-2)}{(x-1)(x-2)(x-3)} = \frac{2}{3} \)
\( \frac{2}{(x-1)(x-3)} = \frac{2}{3} \)
\( 3 = (x - 1) (x - 3) \)
\( 3 = x^2 - 3x - x + 3 \)
\( x^2 - 4x = 0 \)
\( x(x - 4) = 0 \)
\( x = 0, 4 \). Ans.

Question. Solve \( \frac{1}{(a+b+x)} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x}, a + b \neq 0 \)
Answer: We have, \( \frac{1}{(a+b+x)} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x} \)
\( \Rightarrow \frac{1}{(a+b+x)} - \frac{1}{x} = \frac{1}{a} + \frac{1}{b} \)
\( \Rightarrow \frac{x - a - b - x}{x(a + b + x)} = \frac{b + a}{ab} \)
\( \Rightarrow \frac{-(a + b)}{x(a + b + x)} = \frac{a + b}{ab} \)
\( \Rightarrow \frac{-1}{x(a + b + x)} = \frac{1}{ab} \)
\( \Rightarrow - ab = ax + bx + x^2 \)
\( \Rightarrow x^2 + ax + bx + ab = 0 \)
\( \Rightarrow x(x + a) + b(x + a) = 0 \)
\( \Rightarrow (x + a) (x + b) = 0 \)
\( \Rightarrow x + a = 0 \) or \( x + b = 0 \)
\( \Rightarrow x = - a \) or \( x = - b \). Ans.

Question. Divide 27 into two parts such that the sum of their reciprocals is \( \frac{3}{20} \).
Answer: Let the two parts be \( x \) and \( (27 - x) \).
Then, according to the question,
\( \frac{1}{x} + \frac{1}{27 - x} = \frac{3}{20} \)
\( \Rightarrow \frac{27 - x + x}{x(27 - x)} = \frac{3}{20} \)
\( \Rightarrow \frac{27}{x(27 - x)} = \frac{3}{20} \)
\( \Rightarrow \frac{9}{x(27 - x)} = \frac{1}{20} \)
\( \Rightarrow 180 = 27x - x^2 \)
\( \Rightarrow x^2 - 27x + 180 = 0 \)
\( \Rightarrow x^2 - (15 + 12)x + 180 = 0 \)
\( \Rightarrow x^2 - 15x - 12x + 180 = 0 \)
\( \Rightarrow x(x - 15) - 12(x - 15) = 0 \)
\( \Rightarrow (x - 12) (x - 15) = 0 \)
\( \Rightarrow x - 12 = 0 \) or \( x - 15 = 0 \)
\( \Rightarrow x = 12 \) or \( x = 15 \)
If \( x = 12 \), then \( 27 - x = 27 - 12 = 15 \)
If \( x = 15 \), then \( 27 - x = 27 - 15 = 12 \)
Thus, the two parts are 12 and 15. Ans.

Question. The sum of a number and its reciprocal is \( \frac{10}{3} \). Find the number.
Answer: Let the number be \( x \). So, its reciprocal is \( \frac{1}{x} \).
Now, \( x + \frac{1}{x} = \frac{10}{3} \)
\( \Rightarrow \frac{x^2 + 1}{x} = \frac{10}{3} \)
\( \Rightarrow 3x^2 + 3 = 10x \)
\( \Rightarrow 3x^2 - 10x + 3 = 0 \)
\( \Rightarrow 3x^2 - 9x - x + 3 = 0 \)
\( \Rightarrow 3x(x - 3) - 1(x - 3) = 0 \)
\( \Rightarrow (x - 3) (3x - 1) = 0 \)
\( \Rightarrow x = 3 \) or \( \frac{1}{3} \). Ans.

Question. Amita was given a test in Mathematics by her tutor. Square of the actual marks scored by her was 9 times the marks reported by her to her parents. If the actual marks scored were 10 less than the score reported by her to the parents, then find the actual score of Amita.
Answer: Suppose the actual score of Amita be \( x \).
So, according to the question,
\( x^2 = 9(x + 10) \)
\( \Rightarrow x^2 = 9x + 90 \)
\( \Rightarrow x^2 - 9x - 90 = 0 \)
\( \Rightarrow x^2 - 15x + 6x - 90 = 0 \)
\( \Rightarrow x(x - 15) + 6(x - 15) = 0 \)
\( \Rightarrow (x - 15)(x + 6) = 0 \)
\( \Rightarrow x = 15 \) (\( \because x > 0 \))
Hence, the actual score of Amita is 15. Ans.

Question. If \( \alpha \) and \( \beta \) are roots of a quadratic equation such that \( \alpha + \beta = 2 \) and \( \alpha^4 + \beta^4 = 272 \), then find the equation.
Answer: \( \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2\alpha^2\beta^2 \)
\( = \{(\alpha + \beta)^2 - 2\alpha\beta\}^2 - 2\alpha^2\beta^2 \)
\( = (2^2 - 2\alpha\beta)^2 - 2\alpha^2\beta^2 \)
Let \( \alpha\beta = t \)
Then \( (4 - 2t)^2 - 2t^2 = 272 \)
\( \Rightarrow 16 - 16t + 4t^2 - 2t^2 = 272 \)
\( \Rightarrow 2t^2 - 16t - 256 = 0 \)
\( \Rightarrow t^2 - 8t - 128 = 0 \)
\( \Rightarrow t^2 - 16t + 8t - 128 = 0 \)
\( \Rightarrow t(t - 16) + 8(t - 16) = 0 \)
\( \Rightarrow (t - 16)(t + 8) = 0 \)
\( \Rightarrow t = 16 \) or \( - 8 \)
For \( t = 16 \),
\( \alpha + \beta = 2 \) and \( \alpha\beta = 16 \)
\( \Rightarrow x^2 - 2x + 16 = 0 \)
For \( t = - 8 \),
\( \alpha + \beta = 2 \) and \( \alpha\beta = - 8 \)
\( \Rightarrow x^2 - 2x - 8 = 0 \) Ans.

Passage Based Questions

Read the following passage and answer the questions that follows :
A teacher told 6 students to write equations in one variable on their notebook. Students wrote
(a) \( x^2 + 2x + 1 = 0 \)
(b) \( 3x^2 - 2x - 1 = 0 \)
(c) \( x^3 + 3x^2 - x + 1 = 0 \)
(d) \( x^2 - 5x - 6 = 0 \)
(e) \( x + 1 = 0 \)
(f) \( 2x - 1 = 0 \)

Question. (i) Which of the following are the quadratic equations? (ii) For the quadratic equation, \( ax^2 + bx + c = 0 \), write the formula to find the roots. (iii) What is the condition for roots of a quadratic equation to be unreal ?
Answer: (i) \( x^2 + 2x + 1 = 0, 3x^2 - 2x - 1 = 0, x^2 - 5x - 6 = 0 \)
(ii) \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
(iii) For roots to be unreal, Discriminant < 0

An industry produces a certain number of toys in a day. On a particular day, the cost of production of each toy was 9 less than twice the number of toys produced on that day. The total cost of production on that day was ₹ 143.
Based on the given information, answer the following questions :

Question. (i) Find the number of toys produced in the industry on that day. (ii) What is the cost of each toy?
Answer: (i) Let the number of toys be \( n \).
Then, according to the question,
\( n(2n - 9) = 143 \)
\( \Rightarrow 2n^2 - 9n - 143 = 0 \)
\( \Rightarrow (n + \frac{13}{2})(n - 11) = 0 \)
\( \Rightarrow n = -\frac{13}{2} \) or \( n = 11 \)
Since the number of toys cannot be a negative number or a fraction.
\( \therefore n = 11 \)
Hence, 11 toys were produced on that day.
(ii) The cost of each toy = ₹ \( ((2 \times 11) - 9) \) = ₹ 13.

Long Answer Type Questions

Question. Solve for \( x \) : \( \frac{1}{x-2} + \frac{2}{x-1} = \frac{6}{x}, x \neq 0, 1, 2. \)
Answer: We have,
\( \frac{1}{x - 2} + \frac{2}{x - 1} = \frac{6}{x} \)
\( \Rightarrow \frac{x - 1 + 2(x - 2)}{(x - 2)(x - 1)} = \frac{6}{x} \)
\( \Rightarrow \frac{3x - 5}{x^2 - 3x + 2} = \frac{6}{x} \)
\( \Rightarrow 3x^2 - 5x = 6x^2 - 18x + 12 \)
\( \Rightarrow 3x^2 - 13x + 12 = 0 \)
\( \Rightarrow 3x^2 - (9 + 4)x + 12 = 0 \)
\( \Rightarrow 3x^2 - 9x - 4x + 12 = 0 \)
\( \Rightarrow 3x (x - 3) - 4 (x - 3) = 0 \)
\( \Rightarrow (3x - 4) (x - 3) = 0 \)
\( \Rightarrow 3x - 4 = 0 \) or \( x - 3 = 0 \)
\( \Rightarrow x = \frac{4}{3} \) or \( x = 3 \). Ans.

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

Question. Check whether the equation \( 5x^2 - 6x - 2 = 0 \) has real roots and if it has, find them by the method of completing the square. Also verify that roots obtained satisfy the given equation.
Answer: Given quadratic equation is,
\( 5x^2 - 6x - 2 = 0 \)
A quadratic equation has real roots, if
Discriminant > 0
\( \Rightarrow b^2 - 4ac > 0 \)
Here, \( a = 5; b = - 6; c = - 2 \)
So, Discriminant = \( (- 6)^2 - 4 \times 5 \times (- 2) \)
\( = 36 + 40 \)
\( = 76 > 0 \)
Hence, the given equation has real roots.
Now,
\( 5x^2 - 6x - 2 = 0 \)
\( \Rightarrow x^2 - \frac{6}{5}x - \frac{2}{5} = 0 \)
\( \Rightarrow x^2 - \frac{6}{5}x + \frac{9}{25} - \frac{2}{5} - \frac{9}{25} = 0 \)
\( \Rightarrow \left\{ (x)^2 - 2 \times \frac{3}{5} \times x + \left(\frac{3}{5}\right)^2 \right\} - \frac{10 + 9}{25} = 0 \)
\( \Rightarrow \left( x - \frac{3}{5} \right)^2 - \left( \frac{\sqrt{19}}{5} \right)^2 = 0 \)
\( \Rightarrow \left( x - \frac{3}{5} - \frac{\sqrt{19}}{5} \right) \left( x - \frac{3}{5} + \frac{\sqrt{19}}{5} \right) = 0 \)
\( \Rightarrow x = \frac{3 + \sqrt{19}}{5} \) or \( x = \frac{3 - \sqrt{19}}{5} \)
Verification :
For \( x = \frac{3 + \sqrt{19}}{5} \)
The value of quadratic equation is,
\( 5 \left( \frac{3 + \sqrt{19}}{5} \right)^2 - 6 \left( \frac{3 + \sqrt{19}}{5} \right) - 2 = 0 \)
\( \Rightarrow \frac{1}{5} (9 + 19 + 6\sqrt{19}) - \frac{6}{5} (3 + \sqrt{19}) - 2 = 0 \)
\( \Rightarrow 28 + 6\sqrt{19} - 18 - 6\sqrt{19} - 10 = 0 \)
\( \Rightarrow 0 = 0 \)
i.e., L.H.S. = R.H.S.