Let p(u) = 4u² + 8u = 4u(u+2)
To find zeroes, put p(u) = 0
\(\Rightarrow\) 4u(u+2) = 0 \(\Rightarrow\) u = 0 or u + 2 = 0 [\(\because\) 4 \(\neq\)0]
\(\Rightarrow\) u = 0 or u = -2
Hence, zeroes of the given polynonial are 0 and -2.
Verification
Here, sum of zeroes = 0 - 2 = -2 = -(8/4)
=-\(\frac{Coefficient \quad of \quad u}{Coefficient \quad of \quad u^{2}}\)
and product of zeroes
=0 \(\times\)-2 = 0 = (0/4) = \(\frac{Constant \quad term}{Coefficient \quad of \quad u^{2}}\)
so, the relationship between the zeroes and its coefficients is verified.

Given, a pair of linear equations is :
s - t = 3          
\(\Rightarrow\) s = t + 3        ....(i)
and \(\frac { s }{ 3 } +\frac { t }{ 2 } =6\) ...(ii)
On substituting s = t + 3, from egn. (i) in eqn. (ii),
we get
\(\\ \frac { t+3 }{ 3 } +\frac { t }{ 2 } =6\)
\(\Rightarrow\) 2(t + 3) + 3t = 36
\(\Rightarrow\) 5t + 6 = 36
\(\Rightarrow\) 5t = 30
\(\Rightarrow\) t = 6
From eqn., (i), s = 6 + 3 = 9
Hence, s = 9, t = 6

Let us assume to the contrary that \(3+2\sqrt { 5 } \)  is a rational number. Then, it can be expressed in the form \(\frac{a}{b}\), where a, b are coprime integers and \(b\neq 0\)
Now,  \(3+2\sqrt { 5 } \) = a/b, where a,b are integers and \(b\neq 0\)
On rearranging, we get
\(2\sqrt { 5 } =\frac { a }{ b } -3\quad or\quad \sqrt { 5 } =\frac { a }{ 2b } -\frac { 3 }{ 2 } \)
Since, a, b are integers and \(b\neq 0\) , therefore \(\frac{a}{2b}\) is rational number and so \(\frac{a}{2b}\) - \(\frac{3}{2}\) is a rational number.
[since, difference of two rational numbers is also a rational number]
\(\Rightarrow \sqrt { 5 } \) is a rational number. But \(\sqrt { 5 } \) is an irrational number.
This shows that our assumption is incorrect.
So, \(3+2\sqrt { 5 } \) is irrational.

Let the two consecutive integers be x and x+1
ATQ x²+(x+1)2=365
\(\Rightarrow\) x²+x2+2x+1=365 \(\Rightarrow\) 2x²+2x-364=0
\(\Rightarrow\) x²+x-182=0 \(\Rightarrow\) x²+14x-13x-182=0
\(\Rightarrow\) x(x+14)-13(x+14)=0 \(\Rightarrow\) (x-13)(x+14)=0
\(\Rightarrow\) x=13, -14 (-14 is rejected because it is a negative integer)
Hence, the two consecutive positive integers are 13 and 13+1=14

By elimination method:
Given, 3x - 5y = 4 .....(i)
and 9x = 2y + 7   ......(ii)
On multiplying eqn. (i) by 3 and eqn. (ii) by 1,
9x -15y = 12       ......(iii)
9x - 2y = 7         .......(iv)
On subtracting eqn. (iv) from eqn. (iii),
(9x - 15y) - (9x - 2y) = 12 - 7
\(\Rightarrow\) -15y + 2y = 5
\(\Rightarrow\) -13y = 5
\(\therefore \quad y=-\frac { 5 }{ 13 } \)
On substituting y = \(\\ \\ \frac { -5 }{ 13 } \) in eqn. (i),
\(3x-5\left( \frac { -5 }{ 13 } \right) =4\)
\(\Rightarrow\) 39x + 25 = 52
\(\Rightarrow\) 39x = 27
\(\therefore \quad x=\frac { 9 }{ 13 } \)
Hence \(x=\frac { 9 }{ 13 }\) and \( \ y=-\frac { 5 }{ 13 } \)
By substitution method:
Given, 3x - 5y = 4   ....(i)
and    9x = 2y + 7  ....(ii)
From eqn. (ii), 2y = 9x - 7
\(\Rightarrow \quad y=\frac { 9x-2 }{ 2 } \)     ....(iii)
On substituting y from eqn. (iii) in eqn. (i),
\(3x-5\times \left( \frac { 9x-7 }{ 2 } \right) =4\)
\(\Rightarrow\) 6x - 45x + 35 = 8
\(\Rightarrow\) -39x = -27
\(\therefore \quad x=\frac { 9 }{ 13 } \)
On substituting x = \(\frac { 9 }{ 13 } \) in eqn. (iii),
\(y=\frac { 9\times \frac { 9 }{ 13 } -7 }{ 2 } \)
\(=\frac { 81-91 }{ 2\times 13 } \)
\(=-\frac { 10 }{ 26 } \)
\(\therefore \quad y=-\frac { 5 }{ 13 } \)
Hence, \(\\ x=\frac { 9 }{ 13 } \) and \( \ y=-\frac { 5 }{ 13 } \) 

Let p(y)=y²+92y+1920=(y+32)(y+60)
Now, for zeroes of p(y), put p(y)=0
Zeroes y=-32, -60

Given, HCF = 113, LCM = 56952 and one number = 904
Let another number = x
\(\therefore \quad HCF=\frac { x\times 904 }{ LCM } \)
\(\Rightarrow \quad 113=\frac { x\times 904 }{ 56952 } \)
\(\Rightarrow \quad x=\frac { 113\times 56952 }{ 904 } =7119\)

Let the speed of the passenger train be x km/hr. Then,
Speed of the express train = (x + 11)km/hr
Time taken by the passenger train to cover 132 km between Mysore to Bangalore \(\frac{132}{x} \mathrm{hr}\)
Time taken by the express train to cover 132 km between Mysore to Bangalore = \(=\frac{132}{x+11} \mathrm{hr}\)
Therefore,
\( \frac{132}{x}-\frac{132}{x+11}=1 \)
\( \frac{132(x+11)-132 x}{x(x+11)}=1 \)
\(\frac{132 x+1452-132 x}{x^{2}+11}=1 \)
\(\frac{1452}{x^{2}+11}=1 \)
1452 = x² + 11
x² + 11 - 1452 = 0
x2 - 33x + 44x - 1452 = 0
x(x - 33) + 44(x - 33) = 0
(x - 33)(x + 44) = 0
So, either 
x - 33 = 0
x = 33
Or
x + 44 = 0
x = -44
But, the speed of the train can never be negative.
Thus, when x = 33 then speed of express train
= x + 11
= 33 + 11
= 44
e speed of the passenger train is x = 33 km/hr
and the speed of the express train is x = 44 km/hr respectively.

\(\frac{4}{3} x+2 y=8 ; 2 x+3 y=12\)
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.

510 and 92
510 = 2 x 255
= 2 x 3 x 85
= 2 x 3 x 5 x 17
= 2 x 3 x 5 x 17 x 1
Therefore 510 = 2 x 3 x 5 x 17 x 1 .....(A)
92 = 2 x 46
= 2 x 2 x 23
Hence 92 = 2 x 2 x 23 .....(B)
From (A) and (B) HCF of 510 and 92 is = 2 and their 
LCM is 2 x 2 x 3 x 5 x 17 x 23 = 23460
Product of the LCM and HCF = 2 x 23460 = 46920
Product of the two numbers = 510 x 92 = 46920
Therefore it is proved that LCM x HCF = Product of the two numbers.

(i) The number of zeroes is 1 as the graph intersects the x-axis at one point only.
(ii) The number of zeroes is 2 as the graph intersects the x-axis at two points.
(iii) The number of zeroes is 3. (Why?)
(iv) The number of zeroes is 1. (Why?)
(v) The number of zeroes is 1. (Why?)
(vi) The number of zeroes is 4. (Why?)

Let breadth of a rectangular mango grove be x m
Then, length of a rectangular mango grove = 2x m
According to the question,
Area of rectangular mango grove = 800 m²
\(\Rightarrow\) 2x(x) = 800  [\(\because\) Area = length \(\times\) breadth]
\(\Rightarrow\) 2x² = 800 \(\Rightarrow\) x² = 400
\(\Rightarrow\) x = \(\pm\)20
But, x = -20 is not possible because breadth can never be negative. so, x = 20.
Thus, length = 2x = 40 m and breadth = 20 m.

(i) (a): 1^st situation can be represented as x + 7y = 650 ...(i) and
2^nd situation can be represented as x + 11y = 970 ...(ii)
(ii) (b): Subtracting equations (i) from (ii), we get 
\(4 y=320 \Rightarrow y=80\)
\(\therefore\) Proportional expense for each person is Rs 80.
(iii) (c): Puttingy = 80 in equation (i), we get
x + 7 x 80 = 650 \(\Rightarrow\) x = 650 - 560 = 90
\(\therefore\) Fixed expense for the party is Rs 90
(iv) (d): If there will be 15 guests, then amount that Mr Jindal has to pay = Rs (90 + 15 x 80) = Rs 1290
(v) (a): We have a₁ = 1, b₁ = 7, c₁ = -650 and 
\(a_{2}=1, b_{2}=11, c_{2}=-970 \)
\(\therefore \frac{a_{1}}{a_{2}}=1, \frac{b_{1}}{b_{2}}=\frac{7}{11}, \frac{c_{1}}{c_{2}}=\frac{-650}{-970}=\frac{65}{97}\)
\(\text { Here, } \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Thus, system of linear equations has unique solution.