\(\frac{4}{3} x+2 y=8 ; 2 x+3 y=12\)
Here,\(\frac{a_{1}}{a_{2}}=\frac{4}{3 \times 2}=\frac{2}{3}, \frac{b_{1}}{b_{2}}=\frac{2}{3}, \frac{c_{1}}{c_{2}}=\frac{-8}{-12}=\frac{2}{3}\) 
\(\because \quad \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\) 
\(\therefore\)  Pair of equations is consistent with infinitely many solutions.

Let the polynomial be ax² + bx + c, and its zeroes be α and ß
Given \(\alpha+\beta=\sqrt{2}=\frac{-b}{a}\) 
\(\alpha \beta=\frac{1}{3}=\frac{c}{a}\) 
If a = 3, then \(b=\sqrt{2}\) and  c = 1/3
Therefore, the quadratic polynomial is \(3 x^{2}-3 \sqrt{2} x+1\).

The prime factorisation of 96 and 404 gives :
96 = 2⁵ x 3, 404 = 2² x 101
Therefore, the HCF of these two integers is 2² = 4.
Also,LCM (96, 404) = \(\frac{96 \times 404}{\mathrm{HCF}(96,404)}=\frac{96 \times 404}{4}=9696\)

By property of rectangle, we know that its opposite sides are of equal lengths.
i.e. DC=AB \(\Rightarrow\) x+3y=13   ...(i)
and AD=BC \(\Rightarrow\) 3x+y=7   .....(ii)
On multiplying Eq. (ii) by 3 and then subtracting Eq. (i), we get

\(\Rightarrow\) x=1
On putting x=1 in Eq. (i), we get
3y=12 \(\Rightarrow\) y=4
Hence, x=1 and y=4.

For parallel lines, \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
k=-6

-2 and -3

We have, 2ⁿ x 5ⁿ
If n = 1, then 2 x 5 = 10
If n = 2, then 2² x 5² = (2 x 5)² = 10² = 100 and soon.
Thus, for any value of n, we get that 2ⁿ x 5ⁿ ends with zero (always). So, for no value of n, 2ⁿ x 5ⁿ ends with 5.

\(3 x^{2}-4 \sqrt{3} x+4=0\)
Comparing it with ax² + bx + c = 0, we get
a = 3, b = \(-4 \sqrt{3}\)  and c = 4
Discriminant = b² - 4ac
\(=(-4 \sqrt{3})^{2}-4(3)(4)\)
= 48 - 48 = 0
As b² - 4ac = 0,
Therefore, real roots exist for the given equation and they are equal to each other.
And the roots will be \(\frac{-b}{2 a}\)
Therefore, the roots are \(\frac{2}{\sqrt{3}} \text { and } \frac{2}{\sqrt{3}}\)

\(\frac { 1 }{ x+4 } -\frac { 1 }{ x-7 } =\frac { 11 }{ 30 } \quad \Rightarrow \quad \frac { x-7-(x+4) }{ (x+4)(x-7) } =\frac { 11 }{ 30 } \)
\(\Rightarrow \quad \frac { x-7-x-4 }{ { x }^{ 2 }-7x+4x-28 } =\frac { 11 }{ 30 } \quad \Rightarrow \quad \frac { -11 }{ { x }^{ 2 }-3x-28 } =\frac { 11 }{ 30 } \quad or\quad \frac { -1 }{ { x }^{ 2 }-3x-28 } =\frac { 1 }{ 30 } \)
\(\Rightarrow\) x²-3x-28+30=0 \(\Rightarrow\) x²-3x+2=0
This is of the form ax²+bx+c=0, where a=1, b=-3 and c=2
Discriminant(D)=b²-4ac=(-3)²-4 x 1 x 2=9-8=1
Let roots be \(\alpha\)and \(\beta\),  \(​​\alpha =\frac { -b+\sqrt { D } }{ 2a } =\frac { -(-3)+\sqrt { 1 } }{ 2\times 1 } =\frac { 3+1 }{ 2 } =\frac { 4 }{ 2 } =2\)
\(\beta =\frac { -b-\sqrt { D } }{ 2a } =\frac { -(-3)-\sqrt { 1 } }{ 2\times 1 } =\frac { 3-1 }{ 2 } =\frac { 2 }{ 2 } =1\)
Hence roots are 2 and 1.

(2x-1)(x-3)=(x+5)(x-1)
2x²-6x-x+3=x²-x+5x-5
2x²-6x-x+3-x²+x-5x+5=0
x²-11x+8=0
Which is of the form ax²+bx+c=0. Hence the given equation is a quadratic equation.

we found that if the breadth of the hall is x m, then x satisfies the equation 2x² + x - 300 = 0. Applying the factorisation method, we write this equation as
2x² - 24x + 25x - 300 = 0
2x (x - 12) + 25 (x - 12) = 0
i.e., (x - 12)(2x + 25) = 0
So, the roots of the given equation are x = 12 or x = - 12.5. Since x is the breadth of the hall, it cannot be negative.
Thus, the breadth of the hall is 12 m. Its length = 2x + 1 = 25 m.

6x² - 3 - 7x = 6x² - 7x - 3
=6x² -9x + 2x -3
=3x(2x - 3) + 1(2x - 3)
=(2x - 3)(3x + 1)
\(=2\left(x-\frac{3}{2}\right) 3\left(x+\frac{1}{2}\right)\) 
\(=6\left(x-\frac{3}{2}\right)\left(x+\frac{1}{3}\right)\) 
The zeroes of the polynomial are {3/2, -1/3}
Relationship between the zeroes and the coeffieicients of the polynomial
Also sum of the zeroes = \(\frac{3}{2}-\frac{1}{3}=\frac{9-2}{6}=\frac{7}{6}\) 
Also product of the zeroes = \(\frac{3}{2} \times-\frac{1}{3}=-\frac{1}{2}\) 
Hence Verified.

Step 1 : Multiply Equation (1) by 2 and Equation (2) by 1 to make the coefficients of x equal. Then we get the equations as
4x + 6y = 16    (3)
4x + 6y = 7       (4)
Step 2 : Subtracting Equation (4) from Equation (3),
(4x - 4x) + (6y - 6y) = 16 - 7
i.e., 0 = 9, which is a false statemnt.
Therefore, the pair of equations has no solution.

Suppose, \(\sqrt { 5 } \) is a rational number. Then, \(\sqrt { 5 } \) can be expressed in the form \(\frac{a}{b}\), where a and b are coprime integers and \(b\neq 0\).
\(\therefore \sqrt { 5 } =\frac { a }{ b } \)
On squaring both sides, we get
\(5=\frac { { a }^{ 2 } }{ { b }^{ 2 } } \quad \Rightarrow { a }^{ 2 }=5{ b }^{ 2 }\)     ....(i)
\(\Rightarrow \)  5 divides a².
\(\Rightarrow \)  5 divides a. [by theorem 1]   ...(ii)
So, we can take a = 5m
\(\Rightarrow \)   a²  = 25m² [squaring both sides]
On putting the value of a² in Eq. (i), we get
\(\Rightarrow \)  5 divides b²
\(\Rightarrow \)  5 divides b. [by theorem 1]  ... (iii)
Thus, from Eq.(ii), 5 divides a and from Eq. (iii), 5 divides b. It means 5 is a common factor of a and b. This contradicts that there is no common factor of a and b.
This contradiction arises by assuming that \(\sqrt { 5 } \) is rational.
Hence, \(\sqrt { 5 } \) is irrational number.

Consider an arbitrary positive integer y, then y is of the form 3q, (3q + 1) or (3q + 2), where q is some integer.
Consider, y = 3q and  taking cube on both sides, we get
y³ = (3q)³ = 27q³ = 9(3q³) = 9m .... (i)
[taking 3q³ = m, where m is some integer.]
Consider, y = 3q + 1 and taking cube on both sides, we get
y³ = (3q + 1)³ = 27q³ + 27q² + 9q + 1
[\(\because \)  (a + b)³ = a³ + 3a²b + 3ab² + b³]
\(\Rightarrow \)  y³ = 9(3q³ + 3q² + q) + 1 = 9m + 1 ..... (ii)
[taking 3q³ + 3q² + q = m, where m is some integer.]
Consider, y = 3q + 2 and taking cube on both sides, we get
y³ = (3q + 2)³ \(\Rightarrow \)  y³ + 27q³ + 54q² + 36q + 8
= 9(3q³ + 6q² + 4q) + 8 = 9m + 8  .... (iii)
[taking 3q³ + 6q² + 4q = m, where m is some integer]
From Eqs. (i), (ii) and (iii), we get
y³ = 9m, (9m + 1) or (9m + 8)
Thus, the cube of any positive integer is either of the form 9m, (9m + 1) or (9m + 8).

Speed of motor boat in still water is 24 km/h
 Let speed of the stream be x km/h
Speed of the boat in downstream direction 
=(24+x)km/h
Distance covered in downstream direction
= 32 km
Time taken in downstream direction 
\(=\frac { 32 }{ 24+x } \) hours
Speed of the boat in upstream direction (24 +x) km/h
Distance covered in upstream direction  = 32 km 
ATQ \(\frac { 32 }{ 24-x } -\frac { 32 }{ 24+x } =1\)
\(\Rightarrow \frac { 32(24+x)-32(24-x) }{ (24-x)(24+x) } =1\)
\(\Rightarrow 64x=576-x^{ 2 }\)
\(\Rightarrow x^{ 2 }+64x-576=0\)
\(\Rightarrow x^{ 2 }+72x-8x-576=0\)
\(\Rightarrow (x+72)(x-8)=0\)
x=-72 (rejected) or x=8
\(\therefore \) x=8 speed of stream = 8 km/hr