9x² - 24x + 16 = 0
Here, a = 9, b = −24, c = 16
Now, Δ = b² - 4ac = (-24)² - 4(9)(16) = 0
Here, Δ = 0. So, the equation will have real and equal roots

Given points (2, 3) and (- 7, - 1)
Equation of the line passing through (x₁ , y₁) and (x₂, y₂) is
\( \frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}} \)
\(\frac{y-3}{-1-3}=\frac{x-2}{-7-2} \)
9 (y - 3) - 4 (x - 2)
9y - 27 = 4 x - 8
4 x - 9y +19 = 0

The 17^th term can be obtained by subtracting the sum of first 16 terms from the sum of first 17 terms
\({ S }_{ 17 }=\frac { 5\times \left( 17 \right) ^{ 2 } }{ 2 } +\frac { 3\times 17 }{ 2 } =\frac { 1445 }{ 2 } +\frac { 51 }{ 2 } =748\)
\({ s }_{ 16 }=\frac { 5\times \left( 16 \right) ^{ 2 } }{ 2 } +\frac { 3\times 16 }{ 2 } =\frac { 1280 }{ 2 } +\frac { 48 }{ 2 } =664\)
Now, t₁₇ = S₁₇ - S₁₆ = 748 - 664 = 84

Th e slope of line r is m₁  \(=\frac { 8-2 }{ 5+2 } =\frac { 6 }{ 7 } \)
The slope of line θ is m₂  \(=\frac { 0-7 }{ -2+8 } =\frac { -7 }{ 6 } \)
The product of slopes \(=\frac { 6 }{ 7 } \times \frac { -7 }{ 6 } =-1\)
That is, m₁m₂ = -1

To find a₁₁, Since 11 is odd, we put n = 11 in a_n = n (n + 3)
Thus, the eleventh term a₁₁ = 11(11 + 3) = 154
To find a₁₈, Since 18 is even, we put n = 18 in a_n = n² +1
Thus the eighteenth term a₁₈ = 18² + 1 = 325

A x B = {(3,1), (3,7), (3,10), (4,1), (4,7), (4,10), (7,1), (7,7), (7,10), (8,1), (8,7), (8,10)}
We note that, R₁ ⊆  A x B. Thus, R₁ is a relation from A to B.

Let, f(x) = 6x³ - 30x² + 60x - 48 = 6(x³ - 5x² + 10x - 8) and
g(x) = 3x³ - 12x² + 21x - 18 = 3(x³ - 4x² + 7x - 6)
Now, we shall find the GCD of x³ - 5x² + 10x - 8 and x³ - 4x² + 7x - 6

GCD of leading coefficients 3 and 6 is 3.
Thus, GCD [(6x³ - 30x² + 60x - 48, 3x³ - 12x² + 21x - 18)] = 3(x - 2).

In \(\triangle\) ABC we have DE || BC.
By Thales theorem, we have \(\frac { AD }{ DB } =\frac { AE }{ EC } \)
\(\frac { x }{ x-2 } =\frac { x+2 }{ x-1 } \) gives x(x - 1) = (x - 2)(x + 2)
When x = 4, AD = 4, DB = x - 2, AE + x + 2 = 6, EC = x - 1 = 3 
Hence, AB = AD + DB = 4 + 2 = 6, AC = AE + EC = 6 + 3 = 9
Therefore, AB = 6, AC = 9

A = {1, 2, 3, 4} ; B = {2, 5, 8,11,14}; f(x) = 3x − 1
f(1) = 3(1) –1 = 3 – 1 = 2; f(2) = 3(2) –1 = 6 –1 = 5
f(3) = 3(3) –1 = 9 –1 = 8; f(4) = 4(3) –1 = 12 –1 = 11
(i) Arrow diagram
Let us represent the function f :A ⟶ B by an arrow diagram  

(ii) Table form
The given function f can be represented in a tabular form as given below

(iii) Set of ordered pairs
The function f can be represented as a set of ordered pairs as
f = {(1,2),(2,5),(3,8),(4,11)}
(iv) Graphical form
In the adjacent xy -plane the points
(1,2), (2,5), (3,8), (4,11) are plotted (Fig.1.20).

Step 1: Draw the graph of y = 2x² by preparing the table of values as below

Step 2 : To solve 2x² - x - 6 = 0, subtract 2x² - x - 6 = 0 from y = 2x²

The equation y = x + 6 represents a straight line. Draw the graph of y = x + 6 by forming table of values as below

Step 3 : Mark the points of intersection of the curve y = 2x² and the line y = x + 6. That is, (–1.5, 4.5) and (2,8)
Step 4 : The x coordinates of the respective points forms the solution set {–1.5,2} for 2x² - x - 6 = 0