CBSE Class 10 Maths HOTs Quadratic Equations Set C

In this Chapter...

• Quadratic Equation and its Solutions
• Solution of a Quadratic Equation by Factorisation
• Solution of a Quadratic Equation by Quadratic Formula
• Relationship between Discriminant and Nature of Roots

An equation of the form \( ax^2 + bx + c = 0 \) is called quadratic equation in variable \( x \), where \( a, b \), and \( c \) are real numbers and \( a \neq 0 \).
e.g. \( 2x^2 + x - 100 = 0 \), \( -x^2 + 1 + 300x = 0 \), \( 4x - 3x^2 + 7 = 0 \), \( 4x^2 - 25 = 0 \) are quadratic equations.
The form \( ax^2 + bx + c = 0 \), \( a \neq 0 \) is called the standard form of a quadratic equation.
To express a quadratic equation in its standard form, write the terms of given equation in the descending order of their degrees.
e.g. \( 3x^2 + x + 2 = 0 \) and \( x^2 - 2x + 6 = 0 \), are in standard form whereas, \( x^2 - 3 + 4x = 0 \) and \( x + x^2 + 8 = 0 \) are not in their standard form.

Method to Check Whether a Given Equation is Quadratic or Not

To check whether a given equation is quadratic or not, first write the given equation in its simplest form and then compare the equation with the standard form of a quadratic equation, i.e. \( ax^2 + bx + c = 0, a \neq 0 \).
If the given equation follows the form of quadratic equation \( (ax^2 + bx + c = 0, a \neq 0) \), then it is a quadratic equation otherwise not.

Solutions or Roots of a Quadratic Equation

All the values of variable which satisfy the given quadratic equation, are called roots or zeroes or solutions of given quadratic equation.
In other words, a real number \( \alpha \) is said to be a root or zero or solution of a quadratic equation \( ax^2 + bx + c = 0, a \neq 0 \), if \( a(\alpha)^2 + b(\alpha) + c = 0 \).
Any quadratic equation can have at most two roots.

Method to Check Whether the Given Value is a Solution of the Given Quadratic Equation

Let \( p(x) = 0 \) be the given quadratic equation and \( x = \alpha \) be the given value of \( x \). To check whether \( x = \alpha \) is a solution of the given equation or not, use the following steps:
Case I: Write the given equation in the form, \( p(x) = 0 \).
Case II: Now, put \( x = \alpha \) in \( p(x) \). If \( p(\alpha) = 0 \), then \( x = \alpha \) is the solution of given equation, otherwise not.

Method to Determine An Unknown Constant in a Quadratic Equation when its Solution or Root is Given

I. Sometimes, given quadratic equation involves one unknown constant and its solution or root is given. Then, to find the value of unknown constant, we put the value of root or solution in given quadratic equation and simplify it to get the required unknown constant.
II. Sometimes, quadratic equation involves two unknown constants and its both roots are given. Then, to find unknowns we put both roots one-by-one in the quadratic equation and get two linear equations in two unknowns. On solving these equations, we get the required values of unknown constants.

Solution of a Quadratic Equation by Factorisation

To find the solution of a quadratic equation by factorisation method, we use the following steps:
Step I: Write the given equation in standard form i.e. \( ax^2 + bx + c = 0 \) (if not given in standard form) and find the value of \( a, b \), and \( c \).
Step II: Find the product of \( a \) and \( c \) and write it as a sum of its two factor such that sum is equal to \( b \). i.e. write \( ac = p \times q \) and \( p + q = b \) where, \( p \) and \( q \) are factors of \( ac \).
Step III: Put the value of \( b \) obtained from step II in given equation and write it LHS as product of two linear factors.
Step IV: Now, equate each factor equal to zero and get desired roots of given quadratic equation.

Solution of a Quadratic Equation by Quadratic Formula

In a quadratic equation \( ax^2 + bx + c = 0, a \neq 0 \), if \( b^2 - 4ac \geq 0 \), then the roots of the quadratic equation are given by
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) or \( x = \frac{-b \pm \sqrt{D}}{2a} \)
where, \( D = b^2 - 4ac \) is known as discriminant. This result is known as quadratic formula or Sridharacharya formula.

Relationship between Discriminant and Nature of Roots

The nature of roots depends upon the value of the discriminant \( D \), whereas, \( D \) can be zero, positive or negative, so three cases may arise.
Case I: When \( D = 0 \) i.e. \( b^2 - 4ac = 0 \).
If \( D = b^2 - 4ac = 0 \), then \( x = \frac{-b \pm 0}{2a} \Rightarrow x = -\frac{b}{2a}, -\frac{b}{2a} \)
So, the quadratic equation has two equal real roots or repeated roots or coincident roots.
Case II: When \( D > 0 \) i.e. \( b^2 - 4ac > 0 \).
If \( D = b^2 - 4ac > 0 \), then \( x = \frac{-b + \sqrt{D}}{2a} \) and \( \frac{-b - \sqrt{D}}{2a} \)
So, the quadratic equation has two distinct real roots.
Case III: When \( D < 0 \) i.e. \( b^2 - 4ac < 0 \).
If \( D = b^2 - 4ac < 0 \), then \( \sqrt{D} \) cannot be evaluated as square root of negative value is not defined. So, the quadratic equation has no real roots or imaginary roots or we can say that roots of quadratic equation does not exist.

Method to Determine The Value of Unknown when Nature of Roots is Given

If nature of roots of a quadratic equation is given and quadratic equation involves an unknown. Then to find the value of unknown, first we find the value of discriminant in terms of unknown. After that use the given condition i.e. \( D > 0 \) or \( D = 0 \) or \( D < 0 \) and simplify it.

Some Important Points

• Three consecutive numbers are \( x, (x + 1) \) and \( (x + 2) \), respectively.
• Three consecutive even and odd numbers are \( 2x, (2x + 2), (2x + 4) \) and \( (2x + 1), (2x + 3), (2x + 5) \), respectively.
• Pythagoras theorem, \( (\text{Hypotenuse})^2 = (\text{Perpendicular})^2 + (\text{Height})^2 \)
• Area of triangle = \( \frac{1}{2} \times \text{Base} \times \text{Height} \)
• Area of right angled triangle = \( \frac{1}{2} \times \text{Base} \times \text{Perpendicular} \)
• Area of rectangle = \( \text{Length} \times \text{Breadth} \)
• Perimeter of rectangle = \( 2 \times (\text{Length} + \text{Breadth}) \)
• Speed = \( \frac{\text{Distance}}{\text{Time}} \)
• Two-digit number = \( 10x + y \), where \( x \) and \( y \) are the digits of ten’s place and unit place, respectively. On reversing the digits, new number = \( 10y + x \)
• If speed of stream be \( x \) km/h and speed of boat in still water be \( y \) km/h. Then speed of boat in upstream = \( (y - x) \) km/h and speed of boat in downstream = \( (y + x) \) km/h.

Question. For what value of \( p \), are the roots of the equation \( x^2 - 2x(1 + 3p) + 7(3 + 2p) = 0 \) equal?
Answer: Given, \( x^2 - 2x(1 + 3p) + 7(3 + 2p) = 0 \)
If roots are real and equal, then \( b^2 - 4ac = 0 \)
\( \Rightarrow [-2(1 + 3p)]^2 - 4(1)[7(3 + 2p)] = 0 \)
\( \Rightarrow 4[1 + 3p]^2 - 4(21 + 14p) = 0 \)
\( \Rightarrow 4[1 + 9p^2 + 6p - 21 - 14p] = 0 \)
\( \Rightarrow 9p^2 - 8p - 20 = 0 \Rightarrow 9p^2 - 18p + 10p - 20 = 0 \)
\( \Rightarrow 9p(p - 2) + 10(p - 2) = 0 \Rightarrow (p - 2)(9p + 10) = 0 \)
\( \Rightarrow p = 2, -\frac{10}{9} \).

Question. Solve the equation : \( \frac{4x}{x-2} - \frac{3x}{x-1} = 7\frac{1}{2} \)
Answer: Given, \( \frac{4x}{x-2} - \frac{3x}{x-1} = \frac{15}{2} \)
\( \Rightarrow \frac{4x(x-1) - 3x(x-2)}{(x-2)(x-1)} = \frac{15}{2} \Rightarrow \frac{4x^2 - 3x^2 - 4x + 6x}{x^2 - 3x + 2} = \frac{15}{2} \)
\( \Rightarrow 2[x^2 + 2x] = 15[x^2 - 3x + 2] \)
\( \Rightarrow 13x^2 - 49x + 30 = 0 \Rightarrow 13x^2 - 39x - 10x + 30 = 0 \)
\( \Rightarrow 13x(x - 3) - 10(x - 3) = 0 \Rightarrow (x - 3)(13x - 10) = 0 \)
\( \Rightarrow x = 3 \) or \( x = \frac{10}{13} \)

Question. The product of two successive integral multiples of 5 is 300. Find the multiples.
Answer: Let the two successive integral multiples of 5 be \( 5x \) and \( 5x + 5 \).
ATQ, \( 5x(5x + 5) = 300 \)
\( \Rightarrow 25x^2 + 25x - 300 = 0 \Rightarrow x^2 + x - 12 = 0 \)
\( \Rightarrow x^2 + 4x - 3x - 12 = 0 \Rightarrow x(x + 4) - 3(x + 4) = 0 \)
\( \Rightarrow (x - 3)(x + 4) = 0 \Rightarrow x = 3 \) or \( x = -4 \).
\( \therefore \) Successive multiples of 5 are 15 and 20.

Question. The cost price of an article is \( Rs x \) and is sold at a profit of \( (x + 10)\% \), find the cost price of the article if its selling price is \( Rs (2x - 20) \).
Answer: We know that \( SP = CP \times \frac{P}{100} + CP \)
\( \Rightarrow 2x - 20 = \frac{x(x+10)}{100} + x \Rightarrow 2x - 20 = \frac{x^2+110x}{100} \Rightarrow 200x - 2000 = x^2 + 110x \)
\( \Rightarrow x^2 - 90x + 2000 = 0 \Rightarrow x^2 - 50x - 40x + 2000 = 0 \)
\( \Rightarrow (x - 50)(x - 40) = 0 \Rightarrow x = 50 \) or \( x = 40 \)

Question. Solve for \( x : \frac{a}{x-b} + \frac{b}{x-a} = 2; x \neq a, b
Answer: Given, \( \frac{a}{x-b} + \frac{b}{x-a} = 2 \)
\( \Rightarrow a(x - a) + b(x - b) = 2(x - a)(x - b) \)
\( \Rightarrow (a + b)x - (a^2 + b^2) = 2x^2 - 2(a + b)x + 2ab \)
\( \Rightarrow 2x^2 - 3(a + b)x + (a + b)^2 = 0 \)
\( \Rightarrow 2x^2 - 2(a + b)x - (a + b)x + (a + b)^2 = 0 \)
\( \Rightarrow 2x\{x - (a + b)\} - (a + b)\{x - (a + b)\} = 0 \)
\( \Rightarrow \{x - (a + b)\}\{2x - (a + b)\} = 0 \Rightarrow x = a + b \) or \( x = \frac{a+b}{2} \).

Long answer type Questions (4 Marks)

Question. Find the value of \( p \) for which the quadratic equation \( (2p + 1)x^2 - (7p + 2)x + (7p - 3) = 0 \) has equal roots. Also, find these roots.
Answer: For equal roots \( D = b^2 - 4ac = 0 \); Here \( a = 2p + 1, b = -(7p + 2), c = 7p - 3 \)
\( \therefore (7p + 2)^2 - 4(2p + 1)(7p - 3) = 0 \)
\( \Rightarrow 49p^2 + 4 + 28p - 56p^2 + 24p - 28p + 12 = 0 \)
\( \Rightarrow -7p^2 + 24p + 16 = 0 \Rightarrow 7p^2 - 24p - 16 = 0 \Rightarrow 7p^2 - 28p + 4p - 16 = 0 \)
\( \Rightarrow 7p(p - 4) + 4(p - 4) = 0 \Rightarrow (7p + 4)(p - 4) = 0 \)
\( \Rightarrow p = -\frac{4}{7} \) or \( p = 4 \).
For \( p = -\frac{4}{7} \), the equation becomes \( [2(-\frac{4}{7}) + 1]x^2 - [7(-\frac{4}{7}) + 2]x + [7(-\frac{4}{7}) - 3] = 0 \)
\( \Rightarrow -\frac{x^2}{7} + 2x - 7 = 0 \Rightarrow x^2 - 14x + 49 = 0 \)
\( \Rightarrow (x - 7)^2 = 0 \Rightarrow x = 7, 7 \).
For \( p = 4 \), the equation becomes \( (2 \times 4 + 1)x^2 - (7 \times 4 + 2)x + (7 \times 4 - 3) = 0 \)
\( \Rightarrow 9x^2 - 30x + 25 = 0 \Rightarrow 9x^2 - 15x - 15x + 25 = 0 \)
\( \Rightarrow (3x - 5)(3x - 5) = 0 \Rightarrow x = \frac{5}{3}, \frac{5}{3} \).

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: Hence, \( A = (a - b), B = (b - c), C = (c - a) \)
\( \therefore D = B^2 - 4AC = (b - c)^2 - 4(a - b)(c - a) \)
For equal roots, \( D = 0 \)
\( \Rightarrow (b - c)^2 - 4(a - b)(c - a) = 0 \Rightarrow b^2 + c^2 - 2bc + 4a^2 - 4ab - 4ac + 4bc = 0 \)
\( \Rightarrow 4a^2 + b^2 + c^2 + 2bc - 4ab - 4ac = 0 \)
\( \Rightarrow (-2a)^2 + b^2 + c^2 + 2(-2a)b + 2(-2a)(c) + 2(b)(c) = 0 \)
\( \Rightarrow (-2a + b + c)^2 = 0 \Rightarrow b + c = 2a \). Proved.

Question. Vishal wishes to fit three rods together in the shape of a right triangle. The hypotenuse is to be 2 cm longer than the base and 4 cm longer than the altitude. What should be the lengths of the rods?
Answer: Let the length of hypotenuse = \( x \) cm
Then, base = \( x - 2 \) and altitude = \( x - 4 \)
By Pythagoras theorem \( (x - 4)^2 + (x - 2)^2 = x^2 \)
\( \Rightarrow 2x^2 + 20 - 12x = x^2 \Rightarrow x^2 - 12x + 20 = 0 \)
\( \Rightarrow x^2 - 10x - 2x + 20 = 0 \Rightarrow x(x - 10) - 2(x - 10) = 0 \)
\( \Rightarrow (x - 10)(x - 2) = 0 \Rightarrow x = 2 \) or \( x = 10 \)
\( x = 2 \) will be rejected otherwise base will become 0 (not possible)
\( \therefore x = 10 \text{ cm} \)
Hence, base = 8 cm and altitude = 6 cm.

Question. Solve for \( x : abx^2 = (a + b)^2(x - 1)
Answer: Given, \( abx^2 = (a + b)^2(x - 1) \)
\( \Rightarrow abx^2 - (a + b)^2x + (a + b)^2 = 0 \)
Here \( A = ab, B = -(a + b)^2, C = (a + b)^2 \)
\( \Rightarrow x = \frac{-[-(a + b)^2] \pm \sqrt{[-(a + b)^2]^2 - 4ab(a + b)^2}}{2ab} \)
\( = \frac{(a + b)^2 \pm \sqrt{(a + b)^2[(a + b)^2 - 4ab]}}{2ab} = \frac{(a + b)^2 \pm \sqrt{(a + b)^2(a - b)^2}}{2ab} \)
\( = \frac{(a + b)^2 \pm (a + b)(a - b)}{2ab} = \frac{(a + b)^2 \pm (a^2 - b^2)}{2ab} \)
\( \Rightarrow x = \frac{(a + b)^2 + (a^2 - b^2)}{2ab} \) or \( x = \frac{(a + b)^2 - (a^2 - b^2)}{2ab} \)
\( \Rightarrow x = \frac{2a^2 + 2ab}{2ab} \) or \( x = \frac{2b^2 + 2ab}{2ab} \)
\( \Rightarrow x = \frac{a + b}{b} \) or \( x = \frac{a + b}{a} \)

Question. Solve the following quadratic equation by factorization method : \( \frac{x+3}{x-2} - \frac{1-x}{x} = \frac{17}{4} \)
Answer: Given, \( \frac{x+3}{x-2} - \frac{1-x}{x} = \frac{17}{4} \)
\( \Rightarrow \frac{x(x + 3) - (x - 2)(1 - x)}{x(x - 2)} = \frac{17}{4} \Rightarrow \frac{x^2 + 3x - (x - 2 - x^2 + 2x)}{x^2 - 2x} = \frac{17}{4} \)
\( \Rightarrow \frac{2x^2 + 2}{x^2 - 2x} = \frac{17}{4} \Rightarrow 8x^2 + 8 = 17x^2 - 34x \)
\( \Rightarrow 9x^2 - 34x - 8 = 0 \Rightarrow 9x^2 - 36x + 2x - 8 = 0 \Rightarrow 9x(x - 4) + 2(x - 4) = 0 \)
\( \Rightarrow (x - 4)(9x + 2) = 0 \Rightarrow x = 4 \) or \( x = -\frac{2}{9} \).

Question. Solve for \( x : \frac{1}{2a+b+2x} = \frac{1}{2a} + \frac{1}{b} + \frac{1}{2x}
Answer: We have, \( \frac{1}{2a+b+2x} = \frac{1}{2a} + \frac{1}{b} + \frac{1}{2x} \)
\( \Rightarrow \frac{1}{2a+b+2x} - \frac{1}{2x} = \frac{1}{2a} + \frac{1}{b} \Rightarrow \frac{2x - (2a + b + 2x)}{2x(2a + b + 2x)} = \frac{b + 2a}{2ab} \)
\( \Rightarrow \frac{2x - 2a - b - 2x}{2x(2a + b + 2x)} = \frac{2a + b}{2ab} \Rightarrow \frac{-(2a + b)}{2x(2a + b + 2x)} = \frac{2a + b}{2ab} \)
\( \Rightarrow 2x(2a + b + 2x) = -2ab \Rightarrow 4ax + 2bx + 4x^2 = -2ab \)
\( \Rightarrow 4x^2 + 4ax + 2bx + 2ab = 0 \Rightarrow 4x(x + a) + 2b(x + a) = 0 \)
\( \Rightarrow (4x + 2b)(x + a) = 0 \Rightarrow 4x + 2b = 0 \) or \( x + a = 0 \)
\( \Rightarrow 4x = -2b \) or \( x = -a \Rightarrow x = -\frac{b}{2} \) or \( x = -a \)

Question. Solve for \( x : \frac{a}{x-a} + \frac{b}{x-b} = \frac{2c}{x-c}, x \neq a, b, c.
Answer: We have, \( \frac{a}{x-a} + \frac{b}{x-b} = \frac{2c}{x-c} \)
\( \Rightarrow \frac{a(x - b) + b(x - a)}{(x - a)(x - b)} = \frac{2c}{x - c} \Rightarrow \frac{ax - ab + bx - ab}{x^2 - bx - ax + ab} = \frac{2c}{x - c} \)
\( \Rightarrow \frac{ax + bx - 2ab}{x^2 - ax - bx + ab} = \frac{2c}{x - c} \)
\( \Rightarrow (x - c)(ax + bx - 2ab) = 2c(x^2 - ax - bx + ab) \)
\( \Rightarrow ax^2 - acx + bx^2 - bcx - 2abx + 2abc = 2cx^2 - 2acx - 2bcx + 2abc \)
\( \Rightarrow ax^2 + bx^2 - 2abx - acx - bcx = 2cx^2 - 2acx - 2bcx \)
\( \Rightarrow ax^2 + bx^2 - 2cx^2 - 2abx - acx - bcx + 2acx + 2bcx = 0 \)
\( \Rightarrow ax^2 + bx^2 - 2cx^2 - 2abx + acx + bcx = 0 \Rightarrow (a + b - 2c)x^2 - x(2ab - ac - bc) = 0 \)
\( \Rightarrow x[x(a + b - 2c) - (2ab - ac - bc)] = 0 \Rightarrow x = 0 \) or \( x(a + b - 2c) - (2ab - ac - bc) = 0 \)
\( \Rightarrow x = 0 \) or \( x(a + b - 2c) = 2ab - ac - bc \)
\( \Rightarrow x = 0 \) or \( x = \frac{2ab - bc - ac}{a + b - 2c} \)

Question. Solve for \( x : \frac{x+1}{x-1} + \frac{x-2}{x+2} = 4 - \frac{2x+3}{x-2}; x \neq 1, -2, 2.
Answer: We have, \( \frac{x+1}{x-1} + \frac{x-2}{x+2} = 4 - \frac{2x+3}{x-2} \)
\( \Rightarrow \frac{(x+1)(x+2) + (x-2)(x-1)}{(x-1)(x+2)} = \frac{4(x-2) - (2x+3)}{x-2} \)
\( \Rightarrow \frac{x^2+2x+x+2+x^2-x-2x+2}{x^2+2x-x-2} = \frac{4x-8-2x-3}{x-2} \)
\( \Rightarrow \frac{2x^2+4}{x^2+x-2} = \frac{2x-11}{x-2} \Rightarrow (2x^2 + 4)(x - 2) = (x^2 + x - 2)(2x - 11) \)
\( \Rightarrow 2x^3 - 4x^2 + 4x - 8 = 2x^3 - 11x^2 + 2x^2 - 11x - 4x + 22 \)
\( \Rightarrow 2x^3 - 4x^2 + 4x - 8 = 2x^3 - 9x^2 - 15x + 22 \)
\( \Rightarrow 2x^3 - 2x^3 - 4x^2 + 9x^2 + 4x + 15x - 8 - 22 = 0 \Rightarrow 5x^2 + 19x - 30 = 0 \)
\( \Rightarrow 5x^2 + 25x - 6x - 30 = 0 \Rightarrow 5x(x + 5) - 6(x + 5) = 0 \Rightarrow (5x - 6)(x + 5) = 0 \)
\( \Rightarrow 5x - 6 = 0 \) or \( x + 5 = 0 \Rightarrow 5x = 6 \) or \( x = -5 \Rightarrow x = \frac{6}{5} \) or \( x = -5 \)

Solved Examples

Question. Check whether the following equations are quadratic or not.
(i) \(x + \frac{3}{x} = x^2\)
(ii) \(2x^2 - 5x = x^2 - 2x + 3\)
(iii) \(x^2 - \frac{1}{x^2} = 5\)
(iv) \(x^2 - 3x - \sqrt{x} + 4 = 0\)
Answer: (i) Given that, \(x + \frac{3}{x} = x^2\)
\(\Rightarrow \frac{x^2 + 3}{x} = x^3\)
\(\Rightarrow x^3 - x^2 - 3 = 0\)
Which is not of the form \(ax^2 + bx + c, a \neq 0\).
Thus, the equation is not a quadratic equation.
(ii) Given that, \(2x^2 - 5x = x^2 - 2x + 3\)
\(\Rightarrow 2x^2 - x^2 - 5x + 2x - 3 = 0\)
\(\Rightarrow x^2 - 3x - 3 = 0\)
Which is of the form \(ax^2 + bx + c, a \neq 0\).
Thus, the equation is a quadratic equation.
(iii) Given that, \(x^2 - \frac{1}{x^2} = 5\)
\(\Rightarrow x^4 - 1 = 5x^2\)
\(\Rightarrow x^4 - 5x^2 - 1 = 0\)
Which is not of the form \(ax^2 + bx + c, a \neq 0\).
Thus, the equation is not a quadratic equation.
(iv) Given that, \(x^2 - 3x - \sqrt{x} + 4 = 0\)
Which is not of the form \(ax^2 + bx + c, a \neq 0\).
Thus, the equation is not a quadratic equation.

Question. Which of the following are the roots of \(3x^2 + 2x - 1 = 0\)?
(i) \(x = -1\)
(ii) \(x = \frac{1}{3}\)
(iii) \(x = -\frac{1}{2}\)
(iv) \(x = 2\)
Answer: Given equation is of the form \(p(x) = 0\), where \(p(x) = 3x^2 + 2x - 1\) ...(i)
(i) On putting \(x = -1\) in Eq. (i), we get
\(p(-1) = 3(-1)^2 + 2(-1) - 1 = 3 - 2 - 1 = 0\)
So, \(x = -1\) is a root of the given quadratic equation.
(ii) On putting \(x = \frac{1}{3}\) in Eq. (i), we get
\(p\left(\frac{1}{3}\right) = 3\left(\frac{1}{3}\right)^2 + 2\left(\frac{1}{3}\right) - 1 = \frac{1}{3} + \frac{2}{3} - 1 = \frac{1 + 2 - 3}{3} = 0\)
So, \(x = \frac{1}{3}\) is a root of the given equation.
(iii) On putting \(x = -\frac{1}{2}\) in Eq. (i), we get
\(p\left(-\frac{1}{2}\right) = 3\left(-\frac{1}{2}\right)^2 + 2\left(-\frac{1}{2}\right) - 1 = \frac{3}{4} - 1 - 1 = \frac{3}{4} - 2 = \frac{3 - 8}{4} = \frac{-5}{4} \neq 0\)
So, \(x = -\frac{1}{2}\) is not a root of the given equation.
(iv) On putting \(x = 2\) in Eq. (i), we get
\(p(2) = 3(2)^2 + 2(2) - 1 = 12 + 4 - 1 = 15 \neq 0\)
So, \(x = 2\) is not a root of the given equation.

Question. In each of the following equations, find the value of unknown constant(s) for which the given value(s) is (are) solution of the equations.
(i) \(x^2 - k^2 = 0; x = 0.3\)
(ii) \(3x^2 + 2ax - 3 = 0; x = -\frac{1}{2}\)
Answer: (i) We have, \(x^2 - k^2 = 0\), here \(k\) is unknown.
Since, \(x = 0.3\) is a solution of given equation, so it will satisfy the given equation,
On putting \(x = 0.3\) in the given equation, we get
\((0.3)^2 - k^2 = 0\)
\(\Rightarrow k^2 = (0.3)^2\)
\(\Rightarrow k = \pm 0.3\)
(ii) We have, \(3x^2 + 2ax - 3 = 0\), here \(a\) is unknown.
Since, \(x = -\frac{1}{2}\) is a solution of given equation, so it will satisfy the given equation.
On putting \(x = -\frac{1}{2}\) in the given equation, we get
\(3\left(-\frac{1}{2}\right)^2 + 2a\left(-\frac{1}{2}\right) - 3 = 0\)
\(\Rightarrow \frac{3}{4} - a - 3 = 0\)
\(\Rightarrow a = \frac{3}{4} - 3\)
\(\Rightarrow a = \frac{3 - 12}{4} = -\frac{9}{4}\)

Question. Find the roots of the quadratic equation \(2x^2 + \frac{5}{3}x - 2 = 0\) by factorisation method.
Answer: Given equation is \(2x^2 + \frac{5}{3}x - 2 = 0\)
On multiplying by 3 both sides, we get
\(6x^2 + 5x - 6 = 0\)
\(\Rightarrow 6x^2 + (9x - 4x) - 6 = 0\) [by splitting the middle term]
\(\Rightarrow 6x^2 + 9x - 4x - 6 = 0\)
\(\Rightarrow 3x(2x + 3) - 2(2x + 3) = 0\)
\(\Rightarrow (2x + 3)(3x - 2) = 0\)
Now, \(2x + 3 = 0 \Rightarrow x = -\frac{3}{2}\)
and \(3x - 2 = 0 \Rightarrow x = \frac{2}{3}\)
Hence, the roots of the equation \(2x^2 + \frac{5}{3}x - 2 = 0\) are \(-\frac{3}{2}\) and \(\frac{2}{3}\).

Question. Solve the quadratic equation by factorisation method.
\(3\sqrt{2}x^2 - 5x - \sqrt{2} = 0\)
Answer: Given equation is \(3\sqrt{2}x^2 - 5x - \sqrt{2} = 0\)
\(3\sqrt{2}x^2 - (6x - x) - \sqrt{2} = 0\) [by splitting the middle term]
\(3\sqrt{2}x^2 - 6x + x - \sqrt{2} = 0\)
\(3\sqrt{2}x^2 - 3\sqrt{2} \cdot \sqrt{2}x + x - \sqrt{2} = 0\)
\(\Rightarrow 3\sqrt{2}x(x - \sqrt{2}) + 1(x - \sqrt{2}) = 0\)
\(\Rightarrow (x - \sqrt{2})(3\sqrt{2}x + 1) = 0\)
Now, \(x - \sqrt{2} = 0 \Rightarrow x = \sqrt{2}\)
and \(3\sqrt{2}x + 1 = 0 \Rightarrow x = -\frac{1}{3\sqrt{2}} = -\frac{\sqrt{2}}{6}\)
Hence, the roots of the equation \(3\sqrt{2}x^2 - 5x - \sqrt{2} = 0\) are \(-\frac{\sqrt{2}}{6}\) and \(\sqrt{2}\).

Question. Find the sum of the roots of the equation, \(\left[\frac{1}{x - 3} - \frac{1}{x + 5} = \frac{1}{6}\right]\).
Answer: Given \(\left[\frac{1}{x - 3} - \frac{1}{x + 5}\right] = \frac{1}{6}\)
\(\Rightarrow \frac{(x + 5) - (x - 3)}{(x - 3)(x + 5)} = \frac{1}{6}\)
\(\Rightarrow \frac{x + 5 - x + 3}{(x - 3)(x + 5)} = \frac{1}{6} \Rightarrow 8 \times 6 = (x - 3)(x + 5)\)
\(\Rightarrow 48 = x^2 + 2x - 15\)
\(\Rightarrow x^2 + 2x - 63 = 0\)
\(\Rightarrow x^2 + 9x - 7x - 63 = 0\) [by splitting the middle term]
\(\Rightarrow x(x + 9) - 7(x + 9) = 0\)
\(\Rightarrow (x - 7)(x + 9) = 0\)
\(\Rightarrow x = 7\) and \(x = -9\)
\(\therefore\) Sum of roots = \(7 + (-9) = -2\)

Question. Using the quadratic formula, solve the quadratic equation.
\(x^2 + 2\sqrt{2}x - 6 = 0\).
Answer: Given equation is \(x^2 + 2\sqrt{2}x - 6 = 0\).
On comparing with \(ax^2 + bx + c = 0\), we get
\(a = 1, b = 2\sqrt{2}\) and \(c = -6\)
By quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
\(= \frac{-(2\sqrt{2}) \pm \sqrt{(2\sqrt{2})^2 - 4(1)(-6)}}{2(1)}\)
\(= \frac{-2\sqrt{2} \pm \sqrt{8 + 24}}{2}\)
\(= \frac{-2\sqrt{2} \pm \sqrt{32}}{2} = \frac{-2\sqrt{2} \pm 4\sqrt{2}}{2}\)
\(= \frac{-2\sqrt{2} + 4\sqrt{2}}{2}, \frac{-2\sqrt{2} - 4\sqrt{2}}{2}\)
\(= \sqrt{2}, -3\sqrt{2}\)
So, \(\sqrt{2}\) and \(-3\sqrt{2}\) are the roots of the given equation.

Question. Find discriminant of the quadratic equation \(3x^2 + 4x - 5 = 0\).
Answer: Comparing the given quadratic equation \(3x^2 + 4x - 5 = 0\) with standard quadratic equation \(ax^2 + bx + c = 0\), we get
\(a = 3, b = 4\) and \(c = -5\)
\(\therefore\) Discriminant (D) = \(b^2 - 4ac\)
\(= (4)^2 - 4 \times 3 \times (-5) = 16 + 60 = 76\)

Question. Check whether the quadratic equation has real roots. If real roots exist, find them
\(8x^2 + 2x - 3 = 0\)
Answer: Given equation is \(8x^2 + 2x - 3 = 0\).
On comparing with \(ax^2 + bx + c = 0\), we get
\(a = 8, b = 2\) and \(c = -3\)
\(\therefore\) Discriminant, \(D = b^2 - 4ac\)
\(= (2)^2 - 4(8)(-3) = 4 + 96 = 100 > 0\)
Therefore, the equation \(8x^2 + 2x - 3 = 0\) has two distinct real roots as the discriminant greater than zero.
Thus roots, \(x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-2 \pm \sqrt{100}}{16} = \frac{-2 \pm 10}{16}\)
\(= \frac{-2 + 10}{16}, \frac{-2 - 10}{16}\)
\(= \frac{8}{16}, -\frac{12}{16} = \frac{1}{2}, -\frac{3}{4}\)

Question. Find the nature of roots of the quadratic equation \(3x^2 - 4\sqrt{3}x + 4 = 0\).
If the roots are real, find them. [CBSE 2020 (Standard)]
Answer: Given quadratic equation is \(3x^2 - 4\sqrt{3}x + 4 = 0\)
Compare with standard quadratic equation \(ax^2 + bx + c = 0\), we get
\(a = 3, b = -4\sqrt{3}\) and \(c = 4\)
Now, discriminant = \(b^2 - 4ac\)
\(= (-4\sqrt{3})^2 - 4 \times 3 \times 4 = 48 - 48 = 0\)
Hence, roots are real and equal.
By using Sridharacharya formula, \(x = \frac{-b \pm \sqrt{D}}{2a}\)
\(= \frac{-(-4\sqrt{3}) \pm \sqrt{0}}{2 \times 3} = \frac{4\sqrt{3}}{2 \times 3} = \frac{2\sqrt{3}}{3}\)
Hence, roots of given quadratic equation are \(\frac{2\sqrt{3}}{3}\) and \(\frac{2\sqrt{3}}{3}\).

Question. State whether the following quadratic equations have two distinct real roots. Justify your answer.
(i) \(x^2 - 3x + 4 = 0\)
(ii) \(2x^2 + x - 1 = 0\)
(iii) \(2x^2 - 6x + \frac{9}{2} = 0\)
Answer: (i) Given equation is \(x^2 - 3x + 4 = 0\).
On comparing with \(ax^2 + bx + c = 0\), we get \(a = 1, b = -3\) and \(c = 4\)
\(\therefore\) Discriminant, \(D = b^2 - 4ac = (-3)^2 - 4(1)(4) = 9 - 16 = -7 < 0\)
Hence, the equation \(x^2 - 3x + 4 = 0\) has no real root.
(ii) Given equation is \(2x^2 + x - 1 = 0\)
On comparing with \(ax^2 + bx + c = 0\), we get \(a = 2, b = 1\) and \(c = -1\)
\(\therefore\) Discriminant, \(D = b^2 - 4ac = (1)^2 - 4(2)(-1) = 1 + 8 = 9 > 0\) i.e. \(D > 0\)
Hence, the equation \(2x^2 + x - 1 = 0\) has two distinct real roots.
(iii) Given equation is \(2x^2 - 6x + \frac{9}{2} = 0\).
On comparing with \(ax^2 + bx + c = 0\), we get \(a = 2, b = -6\) and \(c = \frac{9}{2}\)
\(\therefore\) Discriminant, \(D = b^2 - 4ac = (-6)^2 - 4(2)\left(\frac{9}{2}\right) = 36 - 36 = 0\)
i.e. \(D = 0\)
Hence, the equation \(2x^2 - 6x + \frac{9}{2} = 0\) has equal and real roots.

Question. The quadratic equation \(x^2 - 4x + k = 0\) has distinct real roots, if \(k = 4\). Why or why not?
Answer: Given quadratic equation is \(x^2 - 4x + k = 0\)
Compare with standard equation \(ax^2 + bx + c = 0\), we get \(a = 1, b = -4\) and \(c = k\)
The condition for distinct real root is \(b^2 - 4ac > 0\)
\(\Rightarrow (-4)^2 - 4 \times 1 \times k > 0 \Rightarrow 16 - 4k > 0 \Rightarrow 16 > 4k \Rightarrow k < \frac{16}{4} \Rightarrow k < 4\)

Question. Find the value of \(k\), for which the quadratic equation \((k + 4)x^2 + (k + 1)x + 1 = 0\) has equal roots. 
Answer: Given, quadratic equation is \((k + 4)x^2 + (k + 1)x + 1 = 0\)
Compare with \(ax^2 + bx + c = 0\), we get \(a = k + 4, b = k + 1\) and \(c = 1\)
Condition for equal roots, \(b^2 - 4ac = 0\)
\(\therefore (k + 1)^2 - 4(k + 4)(1) = 0\)
\(\Rightarrow k^2 + 1 + 2k - 4k - 16 = 0\) [\(\because (a+b)^2 = a^2 + b^2 + 2ab\)]
\(\Rightarrow k^2 - 2k - 15 = 0\)
\(\Rightarrow k^2 - (5 - 3)k - 15 = 0\) [by splitting middle term]
\(\Rightarrow k^2 - 5k + 3k - 15 = 0\)
\(\Rightarrow k(k - 5) + 3(k - 5) = 0\)
\(\Rightarrow (k + 3)(k - 5) = 0 \Rightarrow k = -3, 5\)

Question. The denominator of a fraction is 3 more than its numerator. The sum of the fraction and its reciprocal is \(\frac{29}{10}\). Find the fraction.
Answer: Let numerator = \(x\)
Then denominator = \(x + 3\)
\(\therefore\) The fraction is the form of \(\frac{x}{x + 3}\)
According to the question, \(\frac{x}{x + 3} + \frac{x + 3}{x} = \frac{29}{10}\)
\(\Rightarrow \frac{x^2 + (x + 3)^2}{x(x + 3)} = \frac{29}{10}\)
\(\Rightarrow 10(x^2 + x^2 + 9 + 6x) = 29(x^2 + 3x)\)
\(\Rightarrow 20x^2 + 60x + 90 = 29x^2 + 87x\)
\(\Rightarrow 9x^2 + 27x - 90 = 0\)
\(\Rightarrow x^2 + 3x - 10 = 0\) [divide by 9]
\(\Rightarrow x^2 + 5x - 2x - 10 = 0\) [by splitting middle term]
\(\Rightarrow x(x + 5) - 2(x + 5) = 0\)
\(\Rightarrow (x + 5)(x - 2) = 0\)
\(\Rightarrow x + 5 = 0\) and \(x - 2 = 0 \Rightarrow x = -5\) and \(x = 2\)

Question. Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.
Answer: Let \(n\) be a required natural number.
Square of a natural number diminished by 84 = \(n^2 - 84\)
And thrice of 8 more than the natural number = \(3(n + 8)\)
Now, by given condition, \(n^2 - 84 = 3(n + 8)\)
\(\Rightarrow n^2 - 84 = 3n + 24\)
\(\Rightarrow n^2 - 3n - 108 = 0\)
\(\Rightarrow n^2 - 12n + 9n - 108 = 0\) [by splitting the middle term]
\(\Rightarrow n(n - 12) + 9(n - 12) = 0\)
\(\Rightarrow (n - 12)(n + 9) = 0 \Rightarrow n = 12\)
[\(\because n \neq -9\) because \(n\) is a natural number]
Hence, the required natural number is 12.

Question. If Zeba were younger by 5 yr than what she really is, then the square of her age (in years) would have been 11 more than five times her actual age, what is her age now?
Answer: Let the actual age of Zeba = \(x\) yr
Her age when she was 5 yr younger = \((x - 5)\) yr
Now, by given condition, Square of her age = 11 more than five times her actual age
\((x - 5)^2 = 5 \times \text{actual age} + 11\)
\(\Rightarrow (x - 5)^2 = 5x + 11\)
\(\Rightarrow x^2 + 25 - 10x = 5x + 11\)
\(\Rightarrow x^2 - 15x + 14 = 0\)
\(\Rightarrow x^2 - 14x - x + 14 = 0\) [by splitting the middle term]
\(\Rightarrow x(x - 14) - 1(x - 14) = 0\)
\(\Rightarrow (x - 1)(x - 14) = 0 \Rightarrow x = 14\)
[here, \(x \neq 1\) because her age is \(x - 5\). So, \(x - 5 = 1 - 5 = -4\) i.e. age cannot be negative]
Hence, required Zeba’s age now is 14 yr.

Question. A two-digit number is such that the product of its digit is 35. When 18 is added to the number the digits interchange their places. Find the number.
Answer: Let the ten’s digit number be \(x\).
According to the question, Product of the digits = 35
i.e. Ten’s digits \(\times\) Unit digit = 35 \(\Rightarrow\) Units digit = \(\frac{35}{x}\)
\(\therefore\) Two digit number = \(10x + \frac{35}{x}\)
Also it is given that if 18 is added to the number, the digits gets interchange.
\(\therefore 10x + \frac{35}{x} + 18 = 10 \times \frac{35}{x} + x\)
\(\Rightarrow \frac{10x^2 + 35 + 18x}{x} = \frac{350 + x^2}{x}\)
\(\Rightarrow 9x^2 + 18x - 315 = 0\)
\(\Rightarrow x^2 + 2x - 35 = 0\) [divide by 9]
\(\Rightarrow x^2 + 7x - 5x - 35 = 0\)
\(\Rightarrow x(x + 7) - 5(x + 7) = 0 \Rightarrow (x - 5)(x + 7) = 0 \Rightarrow x = 5, -7\)
But a digit can never be negative. So, \(x = -7\) is rejected.
\(\therefore\) The required number is \(10 \times 5 + \frac{35}{5} = 50 + 7 = 57\)