Let the two zeroes of the polynomial f(t) = kt² + 2t + 3k be α and β.we know that 
Sum of the roots = {tex}-\frac ba{/tex}
Product of the roots = {tex}\frac ca{/tex}
And we have a = k, b = 2 and c = 3k.
Sum of the zeroes = α + β = {tex}-\frac {{ 2}}{k}{/tex}
Product of the zeroes = α {tex}\times{/tex} β = 3
Now,
{tex}\frac{{ - 2}}{k} ={/tex} 3 {tex}\Rightarrow{/tex} 3k = - 2 {tex}\Rightarrow k = \frac{{ - 2}}{3}{/tex}
Hence, required value of k is {tex} \frac{{ - 2}}{3}{/tex}.

Given integers are 468 and 222 where 468 > 222.
By applying Euclid’s division lemma, we get 468 = 222 × 2 + 24 …(i)
Since remainder ≠ 0, apply division lemma on division 222 and remainder 24 222 = 24 × 9 + 6 …(ii)
Since remainder ≠ 0, apply division lemma on division 24 and remainder 6 24 = 6 × 4 + 0 …(iii)
We observe that the remainder = 0, so the last divisor 6 is the HCF of the 468 and 222
From (ii) we have
6 = 222 – 24 × 9
⇒ 6 = 222 – [468 – 222 × 2] × 9 [Substituting 24 = 468 – 222 × 2 from (i)]
⇒ 6 = 222 – 468 × 9 – 222 × 18
⇒ 6 = 222y + 468x, where x = −9 and y = 19
⇒ 6 = 222 × 19 – 468 × 9

The given equations are
3x - y - 5 = 0 ......... (i)
6x - 2y + k = 0 .......... (ii)
The system of linear equations 
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
Compare (i) and (ii), we get
a₁ = 3, b₁ = -1, c₁ = -5
and a₂ = 6, b₂ = -2 , c₂ = k
The equations has no solution if {tex}\frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } \neq \frac { c _ { 1 } } { c _ { 2 } }{/tex}
{tex}\Rightarrow \frac { 3 } { 6 } = \frac { - 1 } { - 2 } \neq \frac { - 5 } { k }{/tex}
So,{tex}- \mathrm { k } \neq 10{/tex}
{tex} \Rightarrow \mathrm { k } \neq - 10{/tex}

So, 5005 = 5 {tex}\times{/tex} 7{tex}\times{/tex} 11 {tex}\times{/tex} 13.

72 = 8 x = 2³ x 3³

So, 5005 = 5 {tex}\times{/tex} 7{tex}\times{/tex} 11 {tex}\times{/tex} 13.

n³ - n = n (n² - 1) = n (n - 1) (n + 1) 

Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.
∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.
If n = 3p, then n is divisible by 3.
If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.
If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.
So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.
⇒ n (n – 1) (n + 1) is divisible by 3.
 
Similarly, whenever a number is divided by 2, the remainder obtained is 0 or 1.
∴ n = 2q or 2q + 1, where q is some integer.
If n = 2q, then n is divisible by 2.
If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.
So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.
⇒ n (n – 1) (n + 1) is divisible by 2.
Since, n (n – 1) (n + 1) is divisible by 2 and 3.

∴ n (n-1) (n+1) = n³ - n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6) 
 

Let the number be (3q + r)
{tex}n = 3 q + r \quad 0 \leq r < 3{/tex}
{tex}\text { or } 3 q , 3 q + 1,3 q + 2{/tex}
{tex}\text { If } n = 3 q \text { then, numbers are } 3 q , ( 3 q + 1 ) , ( 3 q + 2 ){/tex}
{tex}3 q \text { is divisible by } 3{/tex}.
{tex}\text { If } n = 3 q + 1 \text { then, numbers are } ( 3 q + 1 ) , ( 3 q + 3 ) , ( 3 q + 4 ){/tex}
{tex}( 3 q + 3 ) \text { is divisible by } 3{/tex}.
{tex}\text { If } n = 3 q + 2 \text { then, numbers are } ( 3 q + 2 ) , ( 3 q + 4 ) , ( 3 q + 6 ){/tex}
{tex}( 3 q + 6 ) \text { is divisible by } 3{/tex}.
{tex}\therefore \text { out of } n , ( n + 2 ) \text { and } ( n + 4 ) \text { only one is divisible by } 3{/tex}.

• All rational and all irrational number makes the collection of real numbers. It is denoted by the letter R
• We can represent real numbers on the number line. The square root of any positive real number exists and that also can be represented on number line
• The sum or difference of a rational number and an irrational number is an irrational number.
• The product or division of a rational number with an irrational number is an irrational number.

Let n be an arbitrary positive integer.
On dividing n by 3, let q be the quotient and r be the remainder.
Then, by Euclid's division lemma, we have
n = 3q + r, where {tex}0 \leq r < 3{/tex}.
The possibilities of remainder = 0,1 or 2
n² = (3q + r)² [∵ (a + b) ² = a² + 2ab + b²]
{tex}\therefore{/tex} n² = 9q² + r² + 6qr ... ....(i), where {tex}0 \leq r < 3{/tex}.
Case I When r = 0.
Putting r = 0 in (i), we get
n² = 9q²
= 3(3q²)
n² =  3m, where m = 3q² is an integer.
Case II When r = 1.
Putting r = 1 in (i), we get
n² = (9q² + 1 + 6 q)
= 3(3q² + 2q) + 1
n²=  3 m + 1, where m = (3q² + 2q) is an integer.
Case lll When r = 2.
Putting r = 2 in (i), we get
n² = (9q² + 4 + 12q)
= 3(3q² + 4q + 1) + 1
n²= 3m + 1, where m = (3q² + 4q + 1) is an integer.
From all the above cases it is clear that the square of any positive integer is of the form 3m or (3m + 1) for some integer m.

First of all, rationalise the denominator of the reciprocal of 3 + 2√2.

{tex}\sf 3 + 2 \sqrt{2} \\ \\ \sf \frac{1}{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } \\ \\ \sf \frac{3 - 2 \sqrt{2} }{(3 ){}^{2} - (2 \sqrt{2} ) {}^{2} } \\ \\ \sf \frac{3 - 2 \sqrt{2} }{9 - 8} \\ \\ \bf 3 - 2 \sqrt{2}{/tex}

After rationalising its denominator, we get ( 3 - 2√2 ) as a result.

Now, let us assume that ( 3 - 2√2 ) is an irrational number. So, taking a rational number i.e., 3 and subtracting from it.

We have ;

[ 3 - 2√2 - 3 ]

⇒ - 2√2

As a result, we get ( - 2√2 ) which is an irrational number.

Hence, the reciprocal of ( 3 + 2√2) is an irrational number.

Every composite number can be expressed (factorised ) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

This theorem also says that the prime factorisation of a natural number is unique, except for the order of its factors.

For example 20 can be expressed as 2×2×5

Using this theorem the LCM and HCF of the given pair of positive integers can be calculated.

LCM = Product of the greatest power of each prime factor, involved in the numbers.

HCF = Product of the smallest power of each common prime factor in the numbers.

3x - 5y - 4 = 0  .....(1)
9x = 2y + 7  .......(2)
So multiply equation 1 by 3 to get
9x - 15y - 12 = 0 ......... (3)

Put (2) in (3)
2y + 7 - 15y - 12 = 0
- 13y - 5 = 0
- 13y = 5
y = 5/-13
Put value of y in (2)
9x = 2(5/-13) + 7
9x = - 10/13 + 7
9x = (-10+ 91)/13
9x = 81/13
x = 9/13

x + y = 5 ... (i)
we get
x = 5 - y Putting the value of x in equation (ii) we get
5 – y – 3y = 4
-4y = - 1
y = 1/4 
Putting the value of y in equation (i) we get
x = 5 – 1/4
x = 19/4
Hence, x = 19/4 and y = 1/4

Let the speed of train be x kmph and car be y kmph
Condition(i):
Given that Ram travelled a distance of 760 km
Distance travelled by train = 160 km
Recall, time taken = distance / speed
Hence time taken to travel by train = 160/x
Distance travelled by car = 600 km
Time taken to travel by car = 600/y
Total time taken = 8 hours
⇒ 160/x + 600/y = 8 .......... (i)
Condition (ii):
Distance travelled by train = 240 m
Time taken to travel by train = 240 / x
Distance travelled by car = 760 - 240 = 520 km
Time taken to travell by car = 520 / y
Total time taken = 8 hrs + 12 minutes
                         = 8 + (12/60)
                         = 8.2 hours
⇒ 240/x + 520/y = 8.2 ........ (2)
Put 1/x = a and 1/y = b
Hence equations (1) and (2) becomes
160a + 600 b = 8
⇒ 20a + 75b = 1 ........ (3)
240a + 520 b = 8.2
⇒ 120a + 260b = 4.1
⇒ 1200 a + 2600b = 41 ......... (4)
Solving (3) and (4), we get
1900 b = 19
⇒ b = 1/100
⇒ 1/y = 1/100
⇒ y = 100
Put y = 100 in (1), we get
160/x + 600/100 = 8
⇒ 600/x = 8 - 6 = 2
⇒ x = 300
Thus speed of train is 300 kmph and speed of car is 100 kmph.