Let the average speed of passenger train be x km/hr.
Then the average speed of express train will be (x + 20) km/hr
Time taken by the passenger train to cover distance of 240 km = \(\frac {240}{x}\)hr
Time taken by express train to cover distance of 240 km = \(\frac {240}{x + 20}\) hr
Given, \(\frac {240}{x}\) = \(\frac {240}{x + 20}\) + 1
\(240\left| \frac { 1 }{ x } -\frac { 1 }{ x+20 } \right| \) = 1 gives, \(240\left| \frac { x+20-x }{ x\left( x+20 \right) } \right| \) = 1 we get, 4800 = (x² + 20x)
x² + 20x - 4800 = 0 gives,(x + 80)(x - 60) = 0 we get, x = –80 or 60.
Therefore x = 60 (Rejecting -80 as speed cannot be negative)
Average speed of the passenger train is 60 km/hr
Average speed of the express train is 80 km/hr.

Let the present age of Kumaran be x years.
Two years ago, his age = (x - 2) years
Four years from now, his age =(x + 4) years.
Given, (x − 2)(x + 4) = 1 + 2x
x² + 2x − 8 = 1 + 2x gives (x − 3)(x + 3)= 0 then, x = ±3
Therefore, x = 3 (Rejecting −3 as age cannot be negative)
Kumaran’s present age is 3 years.

\(\sqrt { 16{ x }^{ 2 }+9{ y }^{ 2 }-24xy+24x-18y+9 } \)
= \(\sqrt { { \left( 4x \right) }^{ 2 }+{ \left( -3y \right) }^{ 2 }+{ \left( 3 \right) }^{ 2 }+2{ \left( 4x \right) }\left( -3y \right) +2\left( -3y \right) \left( 3 \right) +2\left( 4x \right) \left( 3 \right) } \)
= \(\sqrt { { \left( 4x+-3y+3 \right) }^{ 2 } } \) = |4x - 3y + 3|

Let f(x) = 2x³ - 5x² + 5x - 3 and g(x) = x³ + x² - x + 2

-7(x² - x + 1) ≠ 0, note that -7 is not a divisor of g(x)
Now dividing, g(x) = x³ + x² - x + 2 by the new remainder x² - x + 1 (leaving the constant factor), we get

Here, we get zero remainder
Therefore, GCD (2x³ - 5x² + 5x - 3, x³ + x² - x + 2) = x² - x + 1.

Let, x + 2y - z = 5... (1)  
x - y + z = -2....(2)  
-5x - 4y + z = -11... (3)


Here we arrive at an identity 0 = 0
Hence the system has an infinite number of solutions.

Let the present age of father be x years and the present age of son be y years
Given, x = 6y … (1)
x + 6 = 4(y + 6) … (2)
Substituting (1) in (2), 6y + 6 = 4(y + 6)
6y + 6 = 4y + 24 gives, y = 9
Therefore, son’s age = 9 years and father’s age = 54 years.

\(f(x)=\frac { 1{ x }^{ 2 } }{ a } -\frac { 2x }{ b } +\frac { 3 }{ c } \)
Sum of the roots \((\alpha +\beta )=\frac { -b }{ a } =-\frac { (-2) }{ 1 } \)
Product of the roots \((\alpha \beta )=\frac { c }{ a } =\frac { 3 }{ 1 } =3\)
∝+2, β+2 are the roots (given)
Sum of the roots = ∝ + 2 + β + 2
= ∝ + β + 4
= 2 + 4 = 6
Product of the roots = (∝ + 2) (β + 2)
= ∝β + 2∝ + 2β + 4
= ∝β + 2(∝ + β) + 4
= 3 + 2 x 2 + 4
= 3 + 4 + 4 = 11
∴ The required equation x² - 6x + 11 = 0.

Let the 
100's digit be 'x'
10's digit be 'y'
Unit's digit be 'z'
Given 100y+ 10x + z - 54 = 3(100x + 10y + z)
Substituting 290x - 70y + 22 = -54 ( /2)
145x - 35y + z = -27          .........(1)
l00x + 10y + z + 198 = 100z + 10y + x
Substituting 99x - 992 = - 198 (99)
x - z = - 2              ...........(2)
y = x + 2 (y - z)
x + y - 2z = 0          ..........(3)
Consider (1) and (3)
145x - 35y + z = -27        .......(1)
35x + 35y - 70z = 0          .......(4)
180x - 69z = -27            .......(5)
Consider (5) and (2)

\(x=\frac{111}{111}=1\)
Substituting x = 1 in         .........(2)
1 - z = -z
z = 1 + 2 = 3
Substituting x - I, z = 3 in (3)
1 + y - 6 = 0
y = 5
solution: x = 1, y = 5, z = 3
The number is 153.

Since A, B are of the same order 3 x 3, subtraction of 4A and 3B is defined.
4A - 3B = \(4\left[ \begin{matrix} 5 & 4 & -2 \\ \frac { 1 }{ 2 } & \frac { 3 }{ 4 } & \sqrt { 2 } \\ 1 & 9 & 4 \end{matrix} \right] -3\left[ \begin{matrix} -7 & 4 & -3 \\ \frac { 1 }{ 4 } & \frac { 7 }{ 2 } & 3 \\ 5 & -6 & 9 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 20 & 16 & -8 \\ 2 & 3 & 4\sqrt { 2 } \\ 4 & 36 & 16 \end{matrix} \right] +\left[ \begin{matrix} 21 & -12 & 9 \\ -\frac { 3 }{ 4 } & -\frac { 21 }{ 2 } & -9 \\ -15 & 18 & -27 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 41 & 4 & 1 \\ \frac { 5 }{ 4 } & -\frac { 15 }{ 2 } & 4\sqrt { 2 } -9 \\ -11 & 54 & -11 \end{matrix} \right] \)

We know that a matrix of order m x n has mn elements. Thus to find all possible orders of a matrix with 16 elements, we will find all ordered pairs of natural numbers whose product is 16.
Such ordered pairs are (1, 16), (16, 1), (4,4), (8,2), (2,8)
Hence possible orders are 1 x 16, 16 x 1, 4 x 4, 2 x 8, 8 x 2

2m² + 19m + 30 = 2m² + 4m + 15m + 30 = 2m(m + 2) + 15(m + 2)
= (m + 2)(2m + 15)
Now, equating the factors to zero we get,
(m + 2)(2m + 15) = 0
m + 2 gives, m = -2 or 2m + 15 = 0 we get, m = \(\frac {-15}{2}\)
Therefore the roots are -2, \(\frac {-15}{2}\)
Some equations which are not quadratic can be solved by reducing them to quadratic equations by suitable substitutions. Such examples are illustrated below.

\(\frac { x-3 }{ { x }^{ 2 }-9 } =\frac { x-3 }{ \left( x+3 \right) \left( x-3 \right) } =\frac { 1 }{ x+3 } \)